Heat and thermodynamics

Class 11 Physics, unit 8, book pages 127–153

Practise 82 MCQs Take the unit quiz Review 64 cards

Thermodynamics studies how heat is transformed into other forms of energy, above all into work. This unit starts from thermal equilibrium, internal energy and the ideal gas equation, then uses kinetic theory, the work done by a gas and the gas laws to state the first law of thermodynamics and apply it to isothermal, adiabatic, isochoric and isobaric processes. It goes on to heat engines, reversible and irreversible processes, the second law, the Carnot engine and its efficiency, the refrigerator, and entropy as the measure of disorder and the degradation of energy.

Learning outcomes (23)

State that regions of equal temperatures are in thermal equilibrium.
Relate a rise in temperature of an object to an increase in its internal energy.
Apply the equation of state for an ideal gas.
State that the Boltzmann constant is given by $k = R / N_{A}$ .
Describe the basic assumptions of the kinetic theory of gases.
Use $W = p Δ V$ for the work done when the volume of a gas changes at constant pressure.
Describe the difference between the work done by a gas and the work done on a gas.
Define and use the first law of thermodynamics.
Explain qualitatively, in terms of particles, the relationship between the pressure, temperature and volume of a gas.
Use the equation, including a graphical representation of the relationship between pressure and volume for a gas at constant temperature.
Justify how the first law of thermodynamics expresses the conservation of energy.
Relate a rise in temperature of a body to an increase in its internal energy.
State the working principle of a heat engine.
Describe the concept of reversible and irreversible processes.
State and explain the second law of thermodynamics.
State the working principle of Carnot's engine.
Describe that a refrigerator is a heat engine operating in reverse as that of an ideal heat engine.
Explain that an increase in temperature increases the disorder of the system.
Explain that an increase in entropy means degradation of energy.
Explain that energy is degraded during all natural processes.
Identify that a system tends to become less orderly over time.
Explain that entropy, $S$ , is a thermodynamic quantity that relates to the degree of disorder of the particles in a system.
State that the Carnot cycle sets a limit for the efficiency of a heat engine at the temperatures of its heat reservoirs given by: $Efficiency = 1 - \frac{T _{cold reservoir}}{T _{hot reservoir}}$ .

8.1Thermal equilibrium

The idea of temperature starts from the feelings of "hot" and "cold" that we get from touch. A body that feels hot usually has a higher temperature than a similar body that feels cold. Many properties of matter that we can measure depend on temperature: the length of a metal rod, the steam pressure in a boiler, the ability of a wire to conduct an electric current and the colour of a very hot glowing object. Temperature is also related to the kinetic energies of the molecules of a material.

Heat: The form of energy associated with the kinetic energies of all the molecules in a substance. Its unit is the joule.
Temperature: Describes the degree of hotness and coldness of an object. It is the measure of the average kinetic energy of the molecules, and its unit is the kelvin.

Thermal equilibrium

Thermal equilibrium: Two systems are in thermal equilibrium with each other if they are at the same temperature.

Temperature is the main property that decides whether two systems are in thermal equilibrium. Join two bodies at different temperatures by a diathermic substance and heat starts to flow from the higher temperature body to the lower temperature body. The flow goes on until their temperatures become equal. After that, no net heat flows between them, and they are said to be in thermal equilibrium (Fig. 8.1(a)). A system is in thermal equilibrium with its surroundings if each part of the system and of the surroundings are at the same temperature.

Figure 8.1(a): Thermal equilibrium. Heat (coloured arrow) flows from the hot region to the cold region until both reach the same temperature, after which no more energy is transferred.

System, surroundings and boundary

Heat and work are discussed for a chosen part of the universe. The book gives these terms.

System: The matter or region of space that is being studied, for example a gas in a cylinder.
Surroundings: Everything outside the system.
Boundary of the system: The surface that separates a system from its surroundings. All energy changes, caused by work done or heat exchange between the system and its surroundings, take place through this boundary.

Types of systems

Closed system: A system in which heat energy can enter or leave but mass cannot. Example: a container sealed on all sides.
Open system: A system in which both heat energy and mass can enter or leave it. Example: a glass beaker with an open top, which allows matter (water) to be added or removed, as well as heat.
Isolated system: A system in which there is no transfer of mass and heat energy across its boundary. Example: a thermos flask filled with hot water.

The three types of system.
Type of system	Heat can cross the boundary	Mass can cross the boundary	Example
Closed	Yes	No	A container sealed on all sides
Open	Yes	Yes	An open glass beaker
Isolated	No	No	A thermos flask of hot water

8.1.1Internal energy

Internal energy: The sum of all the forms of molecular energies (such as kinetic and potential energy) of a substance.

An ideal gas is a gas in which all collisions between atoms or molecules are perfectly elastic and there are no intermolecular attractive forces between them. The internal energy of an ideal gas is therefore just the translational kinetic energy of its molecules. The molecules of a gas are always in random motion. According to the kinetic theory of gases, the average kinetic energy of a gas molecule is

⟨ \frac{1}{2} m v^{2} ⟩ = \frac{3}{2} k T

8.1

where $k$ is Boltzmann's constant. If the temperature increases, the average kinetic energy of the molecules also increases, so the internal energy increases with a rise in temperature. Since the temperature is directly proportional to the average kinetic energy ( $T \propto ⟨ K . E . ⟩$ ), the internal energy of an ideal gas is directly proportional to its temperature.

How the internal energy of a system is increased

The internal energy of a system can be increased by heating and by doing mechanical work.

When we heat a substance, the energy associated with its atoms or molecules increases: heat is converted into internal energy.
When two objects are rubbed together, their internal energy increases because of the mechanical work done. The rise in temperature of the object shows the rise in its internal energy.
When an object slides over a surface and comes to rest because of friction, the mechanical work done on or by the system is partly converted into internal energy.

In the study of heat changes and heat flow we are interested in the changes in the internal energy of a substance. A change in temperature shows a change in internal energy.

Internal energy is a function of state

In thermodynamics the internal energy is a function of state. It does not depend on the path but only on the initial and final states of the system. Suppose a system changes from pressure and volume $P_{1}$ and $V_{1}$ to $P_{2}$ and $V_{2}$ , by any process whatever, as in Fig. 8.1(b). The change in internal energy is the same for path $C_{1}$ and path $C_{2}$ , and it depends only on the temperature difference between the two points. Like gravitational potential energy, it is the change in internal energy, and not its absolute value, that matters.

Path independence of the change in internal energy

V_{1}

V_{2}

P_{1}

P_{2}

V

P

A (P_{1}, V_{1}, T_{1})

B (P_{2}, V_{2}, T_{2})

C_{1}

C_{2}

Figure 8.1(b):

P V

graph of a system. The system goes from

A (P_{1}, V_{1}, T_{1})

B (P_{2}, V_{2}, T_{2})

by path

C_{1}

or by path

C_{2}

. The change in internal energy is the same for both.

In a real gas the molecules exert attractive forces on one another. The internal energy of a real gas is therefore partly kinetic energy and partly intermolecular potential energy, so it is a function of temperature and volume.

Sign conventions

Let the initial internal energy be $U_{1}$ at the initial temperature $T_{1}$ and the final internal energy be $U_{2}$ at the final temperature $T_{2}$ . Then the change in internal energy is

Δ U = U_{2} - U_{1}

Sign of the change in internal energy.
Change in temperature	Change in internal energy	Sign of $Δ U$
Temperature rises	$U_{2} > U_{1}$ : internal energy increases	Positive
Temperature falls	$U_{2} < U_{1}$ : internal energy decreases	Negative
Temperature stays constant	$U_{2} = U_{1}$ : internal energy is unchanged	Zero

8.1.2Ideal gas equation and its applications

The ideal gas equation is an equation of state of an ideal gas. It relates the pressure, volume, quantity of gas and temperature:

P V = n R T

8.2

Here $P$ is the pressure of the ideal gas, $V$ its volume, $n$ its number of moles, $R$ the universal gas constant ( $R = 8.314 J mo l^{- 1} K^{- 1}$ ) and $T$ its temperature (in kelvin).

Derivation Ideal gas equation in terms of the number of molecules

Aim. Write $P V = n R T$ using the number of molecules $N$ and Boltzmann's constant $k$ .

The number of moles is the total number of molecules $N$ divided by Avogadro's number $N_{A}$ .
$n = \frac{N}{N _{A}}$
Substitute this $n$ in $P V = n R T$ (8.2).
$P V = \frac{N}{N _{A}} R T or P V = N (\frac{R}{N _{A}}) T$
The ratio $R / N_{A}$ is a constant. It is called Boltzmann's constant $k$ .
$k = \frac{R}{N _{A}}$
Put $k$ in place of $R / N_{A}$ .
$P V = N k T$

Result

P V = N k T (8.3)

The same law can be used with moles ( $n$ and $R$ ) or with molecules ( $N$ and $k$ ). $R$ is the gas constant per mole; $k$ is the gas constant per molecule.

Value and dimensions of Boltzmann's constant

The numerical value of $k$ is estimated from $R$ and Avogadro's number:

k = \frac{R}{N _{A}} = \frac{8.314 J K ^{- 1} mo l ^{- 1}}{6.022140857 \times 1 0 ^{23} mo l ^{- 1}} = 1.3806452 \times 1 0^{- 23} J K^{- 1}

The dimensions of the Boltzmann constant are $[M L^{2} T^{- 2} K^{- 1}]$ .

Example 8.1: average translational kinetic energy

Calculate the average translational kinetic energy of molecules in a gas at room temperature.

Given

$T = 2 5^{\circ} C = 25 + 273 = 298 K$
$k = 1.38 \times 1 0^{- 23} J K^{- 1}$

To find

average translational kinetic energy $⟨ K . E . ⟩$

The average translational kinetic energy of a molecule is given by (8.1).
$⟨ K . E . ⟩ = \frac{3}{2} k T$
Substitute $k$ and $T$ .
$⟨ K . E . ⟩ = \frac{3}{2} \times 1.38 \times 1 0^{- 23} \times 298$
Multiply out.
$⟨ K . E . ⟩ = 6.17 \times 1 0^{- 21} J$

⟨ K . E . ⟩ = 6.17 \times 1 0^{- 21} J

Assignment 8.1

Estimate the total number of air molecules (including molecules of oxygen, nitrogen, water vapour and other constituents) in a room of capacity

20.0 m^{3}

at a temperature of

2 7^{\circ} C

and

1

atm pressure.

Given

$V = 20.0 m^{3}$
$T = 27 + 273 = 300 K$
$P = 1 atm = 1.013 \times 1 0^{5} Pa$
$k = 1.38 \times 1 0^{- 23} J K^{- 1}$

To find

total number of molecules $N$

All the different gases in the air together obey $P V = N k T$ , where $N$ is the total number of molecules. Make $N$ the subject.
$N = \frac{P V}{k T}$
Substitute the values.
$N = \frac{1.013 \times 1 0 ^{5} \times 20.0}{1.38 \times 1 0 ^{- 23} \times 300} = \frac{2.026 \times 1 0 ^{6}}{4.14 \times 1 0 ^{- 21}}$
Divide.
$N \approx 4.9 \times 1 0^{26} molecules$
The same answer comes from $n = P V / R T \approx 812 mol$ and $N = n N_{A}$ .

N \approx 4.9 \times 1 0^{26}

8.2Kinetic theory of gases

In the 19th century James Clerk Maxwell, Rudolf Clausius and others developed the kinetic theory of gases to explain the behaviour of gases. According to the theory, a gas is made of a number of tiny, hard spheres (molecules) that collide with one another and with the walls of the container. The molecules move in accordance with Newton's laws of motion. The theory explains how the motion of molecules influences the macroscopic properties of gases, such as temperature and pressure. It relates the macroscopic properties ( $T$ , $P$ , $V$ etc.) of gases to microscopic properties (kinetic energy etc.).

Figure 8.2: Gas molecules in a container. Each molecule moves in a straight line until it collides with another molecule or with a wall.

Assumptions of the kinetic theory of gases

A finite volume of gas consists of a very large number of molecules.
The size of the molecules is much smaller than the separation between them.
The gas molecules are in random motion and may change their direction of motion after every collision. Collisions between gas molecules and with the walls of the container are assumed to be perfectly elastic.
Molecules exert no force on each other except during a collision.

Condition for a real gas to behave as an ideal gas

Derivation Density of a gas from the gas equation

Aim. Find the relation between the density of a gas, its pressure and its temperature, and use it to see when a real gas is ideal.

Start from the general gas equation.
$P V = n R T$
The number of moles is the mass of gas divided by its molar mass $M$ .
$n = \frac{mass of gas}{molar mass} = \frac{m}{M}$
Put this $n$ in the gas equation.
$P V = (\frac{m}{M}) R T or P M = (\frac{m}{V}) R T$
The mass per unit volume $m / V$ is the density $ρ$ of the gas.
$P M = ρR T$
Make $ρ$ the subject.
$ρ = \frac{P M}{R T}$

Result

ρ = \frac{P M}{R T} (8.4)

For a given mass of gas (fixed $M$ ), $ρ \propto \frac{P}{T}$ : the density rises with pressure and falls with temperature.

When a real gas behaves ideally.
Condition	Density of the gas	Distance between molecules	Intermolecular forces	Behaviour
Low pressure and high temperature	Low	Large	Negligible	A real gas behaves as an ideal gas
High pressure and low temperature	High	Small	Stronger	A real gas cannot behave as an ideal gas

8.2.1Work in thermodynamicsOften examined

Heat and work both correspond to the transfer of energy. One of the main ways in which energy enters or leaves a system is through work. The work done by a constant force is

Work = Force \times Displacement

In thermodynamics we are mainly concerned with the work that gases do when they expand or contract.

Pressure-volume work

Derivation Work done by a gas, $W = P Δ V$

Aim. Find the work done by a gas that expands against a constant pressure in a cylinder with a movable piston.

Assumptions.

The piston moves without friction.
The pressure stays constant while the gas expands or is compressed.

Figure 8.3: Work done by the gas. The gas pushes the piston up through a distance

d

with force

F = P A

, so the volume increases by

Δ V = A d

Take a gas in a cylinder with a movable, frictionless piston of cross-sectional area $A$ (Fig. 8.3). In equilibrium the gas occupies volume $V$ and exerts pressure $P$ on the walls and on the piston. Pressure is force per unit area, so the force on the piston is
$P = \frac{F}{A} ⟹ F = P A$
The gas expands slowly by $Δ V$ , in order to stay in equilibrium. The piston rises through a small distance $d = Δ y$ . The work done by the gas is force times displacement.
$W = F Δ y$
Put $F = P A$ .
$W = P A Δ y$
The cylinder area times the distance moved by the piston is the change in volume of the gas, $A Δ y = Δ V$ .
$W = P Δ V$

Result

W = P Δ V (8.5)

$P$ is the external pressure (as opposed to the pressure of the gas in the system) and $Δ V$ is the change in volume of the gas: $Δ V = V_{final} - V_{initial}$ .

Graphical representation

Work done can be found from the area under a $P$ - $V$ graph. In Fig. 8.4 the area under the graph is $P Δ V$ , which is the work.

Work done at constant pressure

V_{1}

V_{2}

P

V

P

W = P Δ V

B

C

A

D

constant pressure

Figure 8.4: Area of the

P

V

graph equals the work done. At constant pressure the area under the line

B C

is the rectangle

A B C D

, of height

P

and width

Δ V = V_{2} - V_{1}

Example 8.2: pressure of argon in a cylinder

2 kg

sample of argon gas is heated to a temperature of

3 0^{\circ} C

. If the gas is contained in a cylinder of radius

1 m

and height

2 m

, what is the pressure of the gas?

Given

$m = 2 kg = 2000 g$
$T = 3 0^{\circ} C = 30 + 273 = 303 K$
$r = 1 m$
$h = 2 m$
$M = 40 g mo l^{- 1} (argon)$
$R = 8.314 J K^{- 1} mo l^{- 1}$

To find

pressure $P$

The gas fills the cylinder, so its volume is that of a cylinder, $π r^{2} h$ .
$V = π r^{2} h = 3.142 \times (1)^{2} \times 2 = 6.284 m^{3}$
Number of moles is mass divided by molar mass.
$n = \frac{mass of gas}{molar mass} = \frac{2000 g}{40 g mo l ^{- 1}} = 50 mol$
Use the general gas equation $P V = n R T$ and make $P$ the subject.
$P = \frac{n R T}{V}$
Substitute.
$P = \frac{50 \times 8.314 \times 303}{6.284} = 20044 Pa$

P \approx 2.0 \times 1 0^{4} Pa

Assignment 8.2

A sample of gas is compressed to one third of its initial volume at a constant pressure of

2.25 \times 1 0^{5} N m^{- 2}

. During the compression,

200 J

of work is done on the gas. Determine the final volume of the gas.

Given

$P = 2.25 \times 1 0^{5} Pa$
$V_{f} = \frac{1}{3} V_{i}$
$W_{e x t o n g a s} = 200 J$

To find

final volume $V_{f}$

Work done by the gas is $W = P Δ V$ . The gas is compressed, so work is done on it: the work done by the gas is $- 200 J$ .
$P Δ V = - 200 J$
Find the change in volume.
$Δ V = \frac{- 200}{2.25 \times 1 0 ^{5}} = - 8.89 \times 1 0^{- 4} m^{3}$
The volume falls from $V_{i}$ to $V_{i} /3$ , so the change is two thirds of $V_{i}$ .
$Δ V = V_{f} - V_{i} = \frac{V _{i}}{3} - V_{i} = - \frac{2}{3} V_{i}$
Solve for the initial volume.
$V_{i} = \frac{3}{2} \times 8.89 \times 1 0^{- 4} = 1.33 \times 1 0^{- 3} m^{3}$
Final volume is one third of this.
$V_{f} = \frac{V _{i}}{3} = \frac{1.33 \times 1 0 ^{- 3}}{3} = 4.4 \times 1 0^{- 4} m^{3}$

V_{f} \approx 4.4 \times 1 0^{- 4} m^{3}

8.2.2Difference between work done by the gas and on the gas

In thermodynamics the work done is given by W=P\Delta V. Its sign depends on whether the gas expands or contracts.

Positive work: work done by the gas

When a gas expands, it does work on the surroundings. The final volume is greater than the initial volume, so the change in volume is positive, \Delta V>0 (Fig. 8.5). The work done by the gas (the system) on the surroundings is therefore positive.

W=+P\Delta V

Figure 8.5: Work done by the system. The gas is heated and expands from V_A to V_B, lifting the load, so \Delta V>0 and the work W_{AB} done by the gas is positive.

Negative work: work done on the gas

When a gas contracts, the final volume is smaller than the initial volume, so \Delta V<0 (Fig. 8.6). The work done by the surroundings on the gas is then negative in the formula:

W=-P\Delta V

So when a gas contracts we say that the work done by the surroundings on the system is negative. A piston pushed into the cylinder does work on the gas.

Figure 8.6: Work done on the system. The piston is pushed into the cylinder, the volume falls from V_1 to V_2 and W=-P\Delta V.

Zero work

If the volume remains constant, the change in volume is zero (\Delta V=0), and the work done is zero.

W=P\Delta V=P(0)=0

Sign of the work done.
Process of the gas	Change in volume	Work W=P\Delta V	Who does the work
Expansion	\Delta V>0	Positive	The gas does work on the surroundings
Compression	\Delta V<0	Negative	The surroundings do work on the gas
Constant volume	\Delta V=0	Zero	No work is done

8.3The gas laws

It is a common observation that when the external conditions such as temperature and pressure are changed, the volume of a given quantity of any gas is affected. The gas laws give the relationships between the volume of a given amount of gas and variables such as temperature or pressure. Three laws follow: Boyle's law, Charles's law and Gay-Lussac's law.

The three gas laws.
Law	Kept constant	Relation	Proportionality
Boyle's law (8.3.1)	Temperature (and amount of gas)	PV=\text{constant}	V\propto 1/P
Charles's law (8.3.2)	Pressure (and amount of gas)	V/T=\text{constant}	V\propto T
Gay-Lussac's law (8.3.3)	Volume (and amount of gas)	P/T=\text{constant}	P\propto T

8.3.1Boyle's law

For an ideal gas system with a fixed quantity at constant temperature, the pressure and volume are related by Boyle's law. The relation can be derived from the ideal gas equation.

PV=nRT

8.6

If all the quantities on the right side of the equation (n, R and T) are constants, the right side is a constant and we can write

PV=\text{constant}

This shows that the product of pressure and volume remains constant. It can also be written as

V\propto\dfrac{1}{P}

As the pressure increases, the volume of the gas decreases, and vice versa. Graphically this is shown in Fig. 8.7.

Pressure against volume at constant temperature

Figure 8.7: Boyle's law. At constant temperature the product PV stays constant, so the graph of pressure against volume is a curve (a hyperbola) that falls as the volume increases.

Boyle's law: The volume of a given amount of a gas at constant temperature is inversely proportional to the pressure applied.

Worked example: Boyle's law for air in a syringe

Air of volume 60\,\mathrm{cm^{3}} at 1.0\times10^{5}\,\mathrm{Pa} is trapped in a syringe. The plunger is pushed in slowly, so the temperature stays constant, until the volume is 20\,\mathrm{cm^{3}}. Find the final pressure and the pressure at 40\,\mathrm{cm^{3}}.

Given

P_1 = 1.0\times10^{5}\,\mathrm{Pa}
V_1 = 60\,\mathrm{cm^{3}}
V_2 = 20\,\mathrm{cm^{3}}
V_B = 40\,\mathrm{cm^{3}}

To find

P_2 and P_B

The temperature and the amount of gas are constant, so Boyle's law applies: PV stays the same. Both volumes are in \mathrm{cm^{3}}, so the units cancel and no conversion is needed.
P_1V_1=P_2V_2
Final state.
P_2=\dfrac{P_1V_1}{V_2}=\dfrac{(1.0\times10^{5})(60)}{20}=3.0\times10^{5}\,\mathrm{Pa}
Intermediate state B.
P_B=\dfrac{(1.0\times10^{5})(60)}{40}=1.5\times10^{5}\,\mathrm{Pa}
The three points lie on one hyperbola P=60/V. Reducing the volume to one third raised the pressure to three times: pressure is inversely proportional to volume.

Boyle's law for the syringe

Isotherm PV=60 (in units of 10^{5}\,\mathrm{Pa}\,\mathrm{cm^{3}}). Each time the volume is cut to one third, the pressure rises to three times.

P_2=3.0\times10^{5}\,\mathrm{Pa},\qquad P_B=1.5\times10^{5}\,\mathrm{Pa}

8.3.2Charles's law

If the pressure of a given amount of gas is kept constant, there is a relation between the volume of the gas and its temperature, known as Charles's law.

Charles's law: The volume of a given amount of a gas at constant pressure is directly proportional to the absolute temperature.

The law can then be expressed as

V\propto T\qquad\text{or}\qquad\dfrac{V}{T}=\text{constant}

Graphically it is shown in Fig. 8.8.

Volume against temperature at constant pressure

Figure 8.8: Charles's law. The graph of volume against temperature is a straight line. Extended backwards (dashed) it meets the temperature axis at -273\,^\circ\mathrm{C}, where the volume would be zero.

The graph also shows that at a temperature of -273\,^\circ\mathrm{C} the volume of all gases reduces to zero, which shows that a gas cannot exist at such a low temperature.

8.3.3Gay-Lussac's law

To find the relation between the pressure and temperature of a given amount of gas at constant volume, we refer to Amontons's law, also called Gay-Lussac's law.

Gay-Lussac's law: The applied pressure of a given amount of a gas at constant volume is directly proportional to the absolute temperature.

Mathematically:

P\propto T\qquad\text{or}\qquad\dfrac{P}{T}=\text{constant}

Graphically it is shown in Fig. 8.9. For all gas laws the temperature is expressed in the kelvin scale.

Pressure against temperature at constant volume

Figure 8.9: Gay-Lussac's law. The graph of pressure against absolute temperature is a straight line through the origin.

8.4First law of thermodynamicsOften examined

Whenever heat is added to a system, the internal energy of the system increases because of the rise in its temperature. At the same time, if the system is allowed to expand, it may do work on its surroundings. The first law of thermodynamics is based on the law of conservation of energy applied to thermodynamic processes.

First law of thermodynamics: When heat (Q) is added to a system, it increases the internal energy (\Delta U) of the system and also does work (W) by the system on the environment.

Mathematically:

Q=\Delta U+W

8.7

Similarly, when heat is removed from the system, its internal energy decreases, and work is done on the system by compression of a gas (the system).

The change in internal energy \Delta U=U_f-U_i depends only on the initial and final states of the system. It is independent of the path followed. That is why we use the change in internal energy, in spite of the absolute value of the internal energy.

Sign conventions for the first law.
Quantity	Positive when	Negative when
Q	Heat is added to the system	Heat is removed from the system
W	The gas expands (work done by the system)	The gas is compressed (work done on the system)
\Delta U	The temperature rises	The temperature falls

Examples of the first law

The first law of thermodynamics dictates the behaviour of heat energy. The book gives these examples:

Melting of ice cubes is an example of the first law of thermodynamics.
Sweating in a crowded room: heat from a person's body transfers to sweat, which takes away heat from the body but warms the room by increasing the temperature of the atmosphere in the room.
Filling the air in the tube of a car wheel: the mechanical work of pumping the gas increases the internal energy of the system.
Human metabolism: when we do work, our internal energy decreases and the body temperature falls. To maintain internal energy we require food, which provides energy to our bodies.

Worked example: signs of Q, W and \Delta U in the first law

(a) A gas absorbs 500\,\mathrm{J} of heat and does 200\,\mathrm{J} of work on its surroundings. Find \Delta U. (b) The gas is then compressed: 120\,\mathrm{J} of work is done on it while it loses 50\,\mathrm{J} of heat. Find \Delta U for this step and say whether its temperature rises or falls.

Use Q=\Delta U+W, so \Delta U=Q-W. Heat added to the gas is positive, work done by the gas is positive.
(a) Q=+500\,\mathrm{J} and W=+200\,\mathrm{J}.
\Delta U=Q-W=500-200=+300\,\mathrm{J}
Part of the heat went out as work and the rest, 300\,\mathrm{J}, raised the internal energy, so the temperature rises.
(b) Heat is removed, so Q=-50\,\mathrm{J}. Work is done on the gas, so W=-120\,\mathrm{J} (negative).
\Delta U=Q-W=(-50)-(-120)=+70\,\mathrm{J}
The internal energy increases, so the temperature rises even though heat left the gas: the work done on it more than made up for the heat lost.

Case	Q	W	ΔU
(a) expands, heated	+500 J	+200 J	+300 J
(b) compressed, cooled	−50 J	−120 J	+70 J

\Delta U=+300\,\mathrm{J}\ \text{(a)},\qquad \Delta U=+70\,\mathrm{J}\ \text{(b), temperature rises}

8.4.1Isothermal processOften examined

Isothermal process: A thermodynamic process in which the temperature remains constant throughout the process.

This process is a condition for the application of Boyle's law, where the product of pressure and volume remains constant. Consider an ideal gas, as in Fig. 8.10, initially at pressure P_1 and volume V_1. After heating the system, the movable piston moves outward, increasing the volume to V_2 and decreasing the pressure to P_2, while the temperature of the gas remains constant. So the product of pressure and volume is the same for the initial and final values. As the gas expands, work is done by the gas; if the gas is compressed, work is done on the gas.

The internal energy of the system remains constant because the temperature is constant, so \Delta U=0. For an isothermal process the first law becomes

Q=0+W\qquad\text{or}\qquad Q=W

8.8

This shows that for a gas to expand isothermally, some amount of heat Q has to be given to the system to do the work. As heat takes time to move from one place to another, an isothermal process should be carried out slowly to keep the temperature constant.

Isotherm: The curve representing an isothermal process.

The isotherm is shown in Fig. 8.10, and the shaded area represents the work done.

Isothermal process

Figure 8.10: Isothermal process. The curve is the isotherm (T_2=T_1). The shaded area under it between V_1 and V_2 is the work W done by the gas.

The piston arrangement of Fig. 8.10. Heat Q enters, the gas expands from V_1 to V_2 at constant temperature and does work W=Q.

Examples of isothermal processes

Melting of ice at zero degrees centigrade.
Changing the states of different liquids through the processes of melting and evaporation.
The reaction in a heat pump.

8.4.2Adiabatic processOften examined

If we do not supply or remove heat from the system, can the system still perform work? Yes: when a system expands without addition of heat, it still does work.

Adiabatic process: A process in which no heat or mass transfers between a thermodynamic system and its surroundings.

In an adiabatic process the system transfers energy only to the surroundings, as work. So, with Q=0, the first law Q=\Delta U+W becomes

W=-\Delta U

This shows that if no heat is supplied to the system and the system still expands and performs work, the work is done at the expense of its internal energy.

If the gas expands and does work, its temperature and internal energy decrease.
If the gas is compressed adiabatically, its temperature and internal energy increase.
In an adiabatic process the gas expands or compresses rapidly, so there is no time for heat to enter or leave.

Adiabat: The curve representing an adiabatic process.

The adiabat is shown in Fig. 8.11. In an adiabatic process, Boyle's law takes the form

PV^{\gamma}=\text{constant}

Adiabatic process

Figure 8.11: Adiabatic process. The adiabat (coloured) falls from the isotherm at T_1 to the isotherm at T_2, with T_1>T_2. The shaded area is the work W done by the gas, at the expense of its internal energy.

The piston arrangement of Fig. 8.11. No heat enters (Q=0); the gas expands from V_1 to V_2 and does work W=-\Delta U.

Examples of adiabatic processes

The rapid compression and expansion of air as a sound wave passes through it.
The vertical flow of air in the atmosphere: air expands and cools as it rises, and contracts and gets warmer as it descends.
Expansion and contraction of a gas cloud in interstellar regions.

Worked example: expansion by an adiabat and by an isotherm

A gas in an insulated cylinder expands and does 150\,\mathrm{J} of work. (a) Find Q and \Delta U. (b) The gas starts at 2\,\mathrm{L} and 3\times10^{5}\,\mathrm{Pa} and expands to 4\,\mathrm{L} once along an isotherm and once along an adiabat. Find the final pressure on the isotherm, compare it with the adiabat and say which final state is cooler.

(a) The cylinder is insulated, so no heat enters or leaves: Q=0. The first law gives W=-\Delta U.
\Delta U=-W=-150\,\mathrm{J}
The internal energy falls, so the gas cools. The work was done at the expense of its internal energy.
(b) Isotherm: PV is constant.
P=\dfrac{(3\times10^{5})(2)}{4}=1.5\times10^{5}\,\mathrm{Pa}
Adiabat: the adiabat is steeper than the isotherm (Fig. 8.11), so at the same volume of 4\,\mathrm{L} it has fallen to a lower pressure than the isotherm: P_{adi}<1.5\times10^{5}\,\mathrm{Pa}.
Since PV=nRT, a smaller P at the same V means a lower temperature. The adiabat ends on a lower isotherm, so it is cooler: the gas did work at the expense of its internal energy. The isotherm stays at the starting temperature because heat flows in.

Isotherm and adiabat from the same state

Both curves start at A (2\,\mathrm{L}) and expand to 4\,\mathrm{L}. The adiabat (coloured) falls more steeply than the isotherm (dark), so it ends at a lower pressure and a lower temperature. The curves are drawn schematically.

\begin{aligned}Q&=0\\ \Delta U&=-150\,\mathrm{J}\\ P_{iso}&=1.5\times10^{5}\,\mathrm{Pa}\\ P_{adi}&<P_{iso}\ \text{(adiabat ends cooler)}\end{aligned}

8.4.3Isochoric processOften examined

When you heat a gas in a closed container, what will happen? There are thermodynamic processes in which the system does not change its volume. Such a process is called an isochoric process.

Isochoric (constant-volume) process: A thermodynamic process in which the volume of the system remains constant.

For example, a gas is enclosed in a thermally conducting but fixed container. When the gas is heated, its pressure increases from P_1 to P_2 and its temperature increases from T_1 to T_2, so its internal energy increases, as in Fig. 8.12. Work done is W=P\Delta V. Here the volume is constant (\Delta V=0), so

W=0

The first law of thermodynamics then reduces to

Q=\Delta U

This means that in an isochoric process all the heat given to the system is used in increasing its internal energy. If heat is removed from the system in an isochoric process, both the temperature and the pressure of the system decrease. The P-V graph of an isochoric process is a vertical line, called an isochor.

Isochoric process

Figure 8.12: Iso-choric process. At constant volume the P-V graph is a vertical line, the isochor (coloured), from state 1 on the isotherm at T_1 up to state 2 on the isotherm at T_2, with T_2>T_1.

The piston arrangement of Fig. 8.12. The piston is fixed, so V_2=V_1 and W=0; the heat Q supplied raises the internal energy.

Examples of isochoric processes

Cooking food in a pressure cooker: a closed container where, when heat is supplied, the temperature and pressure increase.
When a gasoline-air mixture is burnt in the engine of a car, it increases the pressure and temperature of the gas, but the volume of the gas remains constant.

8.4.4Isobaric processOften examined

The term isobaric comes from the Greek words "iso" meaning same and "baros" meaning pressure. In an isobaric process, when heat is transferred to the system, some work is done and there is also a change in the internal energy of the system. It is the process in which no quantity in the first law of thermodynamics becomes zero.

Isobaric (constant-pressure) process: A thermodynamic process in which the pressure of the system remains constant.

When a gas expands slowly because heat is supplied while the pressure is kept constant, it is an example of an isobaric process.

Derivation First law for an isobaric process

Aim. Write the first law for a gas heated at constant pressure.

Take a gas in a container of cross-sectional area A with initial pressure, temperature and volume P, T_1 and V_1 (Fig. 8.13). A quantity of heat Q changes the temperature and volume to T_2 and V_2, while the pressure stays constant. The gas expands, so it does work and its internal energy also increases. The force on the piston is
F=PA
If the gas moves the piston through a displacement \Delta y, the work done by the gas is force times displacement.
W=PA\,\Delta y=P\,\Delta V
Put this work in the first law Q=\Delta U+W.
Q=\Delta U+P\,\Delta V

ResultQ=\Delta U+P\Delta V

The heat supplied is shared between increasing the internal energy and doing the work P\Delta V. The P-V graph of an isobaric process is a horizontal line called an isobar, and the area under it is the work done.

Isobaric process

Figure 8.13: Iso-baric process. At constant pressure the P-V graph is a horizontal line, the isobar (coloured). The shaded area under it is the work W=P(V_2-V_1) done by the gas.

The piston arrangement of Fig. 8.13. Heat Q enters, the gas expands from V_1 to V_2 at constant pressure and does work W=P(V_2-V_1).

Examples of isobaric processes

Boiling of water to steam in an open container.
A solid moth ball (made of solid material that slowly changes to gas at room temperature and becomes fumes) sublimating into an insect repelling gas in your linen closet.
Ice formation in nature, in which the volume changes to compensate for energy removal while the pressure remains constant.

Example 8.3: first law applied to a gas

The internal energy of a gas decreases by 200\,\mathrm{J}, while 50\,\mathrm{J} of work is done by the gas. How much energy is transferred as heat? Is the work done positive or negative?

Given

\Delta U = -200\,\mathrm{J}\ \text{(a decrease)}
W = +50\,\mathrm{J}\ \text{(done by the gas)}

To find

heat Q

Write the first law of thermodynamics.
Q=\Delta U+W
The internal energy decreases, so \Delta U is negative. The gas does the work, so W is positive. Substitute.
Q=-200\,\mathrm{J}+50\,\mathrm{J}=-150\,\mathrm{J}
T
h
e

Q=-150\,\mathrm{J}\ (\text{heat given out})

Assignment 8.3

A cup is filled with hot tea and a spoon is used for stirring. Initially the tea has an internal energy of 200\,\mathrm{kJ}. On cooling, it loses 150\,\mathrm{kJ} of heat. The spoon does 25\,\mathrm{kJ} of work while stirring. Find the change in the internal energy of the tea and the final internal energy of the tea.

Given

U_i = 200\,\mathrm{kJ}
Q = -150\,\mathrm{kJ}\ \text{(heat lost)}
W = -25\,\mathrm{kJ}\ \text{(work done on the tea)}

To find

\Delta U and the final internal energy U_f

The tea loses heat, so Q is negative. The spoon does work on the tea, so the work done by the tea is negative.
Q=-150\,\mathrm{kJ},\qquad W=-25\,\mathrm{kJ}
Use the first law Q=\Delta U+W and make \Delta U the subject.
\Delta U=Q-W=-150-(-25)=-125\,\mathrm{kJ}
Final internal energy is the initial value plus the change.
U_f=U_i+\Delta U=200-125=75\,\mathrm{kJ}

\Delta U=-125\,\mathrm{kJ},\qquad U_f=75\,\mathrm{kJ}

The four processes at a glance

The first law applied to the four processes.
Process	What stays constant	W	First law becomes	P-V curve
Isothermal	Temperature (\Delta U=0)	W	Q=W	Isotherm
Adiabatic	Heat (Q=0)	W	W=-\Delta U	Adiabat
Isochoric	Volume (\Delta V=0)	0	Q=\Delta U	Isochor (vertical line)
Isobaric	Pressure	P\Delta V	Q=\Delta U+P\Delta V	Isobar (horizontal line)

8.5Heat engineOften examined

Heat and energy are two distinct quantities but they are closely related: heat can be converted into work and vice versa.

Heat engine: A device that converts heat energy into mechanical work.

A petrol engine and a steam engine are examples of heat engines. The earliest heat engine was a steam engine. Every heat engine must have three parts: a hot reservoir, a cold reservoir and a working substance. The working principle of a heat engine is the second law of thermodynamics, by which we can extract useful work during the transfer of heat from a hotter body to a colder one.

Figure 8.14: Heat engine. The engine takes heat Q_1 from the hot reservoir at T_1, does work W and rejects heat Q_2 to the cold reservoir at T_2.

Hot reservoir: The source of heat that is to be converted into useful work. Its temperature must be maintained at a higher level, denoted by T_1 in Fig. 8.14. It supplies a quantity of heat Q_1 flowing towards low temperature.
Cold reservoir (heat sink): A necessary part of a heat engine. The difference between the temperatures of the two reservoirs is responsible for the heat flow. The heat sink absorbs heat Q_2 at temperature T_2 from the heat source. The cold reservoir is essential, as heat can only flow when there is a temperature difference.
Working substance: The main part of the engine, which practically converts heat into useful mechanical work. In most heat engines it is a gas. It performs work due to the heat flowing from the heat source to the heat sink.

The work done by a heat engine is given by

W=Q_1-Q_2

For continuous operation, the heat engine is made cyclic. Leonard Sadi Carnot invented the first heat engine, which was a type of hot air engine.

Thermodynamics and the auto-engine

A petrol engine of a car extracts heat from burning fuel and converts some of its energy into mechanical work. The engine expels the rest of the energy to the atmosphere. A petrol engine roughly has an efficiency of about 25% to 30%. Motorbikes normally have two cylinders, whereas cars have four cylinders on the same crankshaft, timed to fire in turn in succession for smooth running of the car. It has four steps of working.

Intake stroke: The piston moves downward and the petrol-air mixture is drawn through the inlet valve, Fig. (a).
Compression stroke: The piston moves upward, the inlet valve is closed and the petrol-air mixture compresses adiabatically, Fig. (b).
Power stroke: A spark ignites the mixture, causing a rapid increase in temperature and pressure. It expands adiabatically and so forces the piston to move downwards. This stroke delivers the power to the engine and is called the power stroke, Fig. (c).
Exhaust stroke: The piston moves upward, the outlet valve opens and the residual gases are expelled into the atmosphere, Fig. (d).

The four strokes of a petrol engine: (a) intake, (b) compression, (c) power and (d) exhaust. The arrow beside each piston shows its direction of motion; in the power stroke the force is F=PA.

8.6Reversible and irreversible processes

In thermodynamics a process occurs when energy is exchanged between the system and its surroundings. There are mainly two types of processes: reversible and irreversible.

Reversible process

Reversible process: A process which can be retraced in exactly reverse order without causing any change to its surroundings.

For example, if heat is absorbed in a direct process, it will be evolved in the reverse process. If work is done by the system in the direct process, it will be done on the system in the reverse process. In reality no actual change is exactly reversible, but some changes can be taken as reversible if done slowly. Some examples of reversible processes:

The transformation between water and ice.
Liquefaction and evaporation of a substance performed slowly.
Melting, freezing, boiling and condensing are reversible processes.

Irreversible process

Irreversible process: A process which cannot be retraced in exactly the reverse order without causing a change in its surroundings.

Any process involving friction or dissipation of energy.
Decaying of animals and plants.
Convection, conduction and radiation are irreversible processes.

Reversible and irreversible processes.
	Reversible process	Irreversible process
Retracing	Can be retraced in exactly reverse order	Cannot be retraced in exactly reverse order
Effect on surroundings	No change in the surroundings	A change is caused in the surroundings
Examples	Water and ice, slow liquefaction and evaporation, melting, freezing, boiling, condensing	Friction or dissipation of energy, decay, convection, conduction and radiation

Cyclic process

Cyclic process: A succession of events which bring the system back to its initial condition.

For example, an event A brings the system from condition 1 to condition 2, as in Fig. 8.15. Then another event B brings the system back from condition 2 to condition 1. This is a cyclic process.

A cyclic process

Figure 8.15: A cyclic process. Event A takes the system from condition 1 to condition 2, and event B brings it back from 2 to 1.

Heat engines are normally operated in cycles. In a reversible cycle all the changes are reversible, but this is an idealisation. We can take some real processes as close to reversible ones by ignoring minute changes.

8.7Second law of thermodynamicsOften examined

The first law of thermodynamics deals with the conversion of heat into work and vice versa. It is simply the law of conservation of energy. However, it does not tell us the conditions under which this conversion takes place. To find this condition we need another law, the second law of thermodynamics. This law deals with the circumstances under which heat can be converted into work and also with the direction of heat flow. It can be stated in two different ways.

Lord Kelvin's statement

Kelvin's statement: It is impossible for a heat engine to convert heat taken from a single reservoir entirely into work without leaving any change in the working system.

It is clear that in any heat engine, a single heat reservoir, no matter how much energy it contains, cannot be made to perform any work. This is why the heat content of our atmosphere cannot be utilised for work, as it is a single source and there is no sink relative to it.

Rudolf Clausius's statement

Clausius's statement: It is impossible to cause heat to flow from a cold body to a hot body without the expenditure of work.

The second law of thermodynamics explains that both a hot and a cold reservoir are essential for a working substance to do work. As in Fig. 8.14, a heat source Q_1 rejects heat towards a heat sink Q_2 at lower temperature, while some of the heat is converted into mechanical work W, which can be given as

W=Q_h-Q_c

Q_h is the heat from the hot reservoir (Q_1) and Q_c is the heat rejected to the cold reservoir (Q_2).

In a cyclic process, the substance eventually returns to its initial state, so the change in internal energy is equal to zero, \Delta U=0.

8.8Carnot's engineOften examined

A Carnot heat engine is a theoretical engine that operates on the principle of the Carnot cycle. This concept was introduced by Sadi Carnot in 1824. A heat engine operating in an ideal reversible cycle between a heat source and a heat sink would be the most efficient engine. For the operation of the Carnot engine, we consider the Carnot cycle using an ideal gas as the working substance.

The working principle of Carnot's engine is the isothermal and adiabatic processes. The engine works in a cycle of isothermal expansion, adiabatic expansion, isothermal compression and finally adiabatic compression to reach the initial state, while taking some useful work as output. This working process is shown in the P-V diagram of Fig. 8.16 and can be explained in four steps.

P-V diagram of the Carnot cycle

Figure 8.16: P-V diagram for the Carnot cycle a\to b\to c\to d\to a (small arrows). ab and cd are isotherms (coloured); bc and da are adiabats. The area enclosed by abcda is the work done in one cycle.

Step 1: isothermal expansion (ab)

In the first step the gas is allowed to expand isothermally at pressure P_1 and volume V_1. The temperature of the gas remains high, T_1. Absorbing heat Q_1 from the hot reservoir, the pressure of the gas decreases to P_2 and its volume increases to V_2. Work is done by the gas while the temperature remains constant. This is the isotherm ab in Fig. 8.16 and is shown mechanically in Fig. 8.17 (A).

Step 2: adiabatic expansion (bc)

The cylinder is insulated so that the gas neither absorbs nor releases heat, Fig. 8.17 (B). The gas expands adiabatically and its temperature drops from the high temperature T_1 to a low temperature T_2. In the adiabat bc of Fig. 8.16 the pressure falls further to P_3 and the volume increases to V_3. The piston moves outward by doing work.

Step 3: isothermal compression (cd)

The gas is kept at the low temperature T_2 and it is compressed by rejecting heat Q_2, Fig. 8.17 (C). The pressure increases to P_4 and the volume decreases to V_4, so work is done on the gas. This is the isotherm cd of Fig. 8.16.

Step 4: adiabatic compression (da)

Finally, the gas is again placed in an insulated cylinder, so that no heat enters or leaves. The gas is compressed adiabatically and its temperature rises from T_2 to T_1, Fig. 8.17 (D). The pressure increases to P_1 and the volume decreases to V_1. This is the curve da of Fig. 8.16.

Figure 8.17: Working of a Carnot engine. (A) Isothermal expansion at T_1 with heat Q_1 absorbed. (B) Adiabatic expansion from T_1 to T_2 in an insulated cylinder. (C) Isothermal compression at T_2 with heat Q_2 rejected. (D) Adiabatic compression from T_2 to T_1. The arrow beside the piston gives its direction.

The four steps of the Carnot cycle.
Step	Curve	Process	Heat	Temperature
1	ab	Isothermal expansion	Q_1 absorbed from the hot reservoir	T_1 constant
2	bc	Adiabatic expansion	None (Q=0)	Falls from T_1 to T_2
3	cd	Isothermal compression	Q_2 rejected to the cold reservoir	T_2 constant
4	da	Adiabatic compression	None (Q=0)	Rises from T_2 to T_1

Thermal and mechanical equilibrium is maintained throughout the process to make it perfectly reversible. As the working substance returns to its initial state, there is no change in internal energy, \Delta U=0.

8.8.1Efficiency of Carnot engineOften examined

Derivation Efficiency of a Carnot engine

Aim. Find the efficiency of a Carnot engine in terms of the temperatures of the hot and cold reservoirs.

During one cycle the total work done by the gas is found from the area enclosed by the path abcda in Fig. 8.16. It can also be found from the difference between the heat Q_1 absorbed by the gas during isothermal expansion (input) and the heat Q_2 rejected by the gas during isothermal compression.
W=Q_1-Q_2\qquad(8.8)
This follows from the first law, Q=\Delta U+W, with \Delta U=0 for a complete cycle. The net heat taken in is Q=Q_1-Q_2, so
W=Q=Q_1-Q_2
Efficiency is the useful output divided by the input heat.
\eta=\dfrac{\text{output (work)}}{\text{input (heat)}}=\dfrac{W}{Q_1}
Put W=Q_1-Q_2 and split the fraction.
\eta=\dfrac{Q_1-Q_2}{Q_1}=1-\dfrac{Q_2}{Q_1}
The heat Q is proportional to the absolute temperature T, so Q_2/Q_1=T_2/T_1.
\eta=1-\dfrac{T_2}{T_1}

Result\eta=\left(1-\dfrac{T_2}{T_1}\right)\qquad(8.9)

Efficiency is usually expressed as a percentage, so \text{Percentage efficiency}=\eta\times100\%=\left(1-\dfrac{T_2}{T_1}\right)\times100\%.

Conclusion. The efficiency of a Carnot engine is independent of the nature of the working substance. It depends only on the temperatures of the hot and cold reservoirs. For the efficiency to be 100% the ratio T_2/T_1 should be zero. This happens only if T_2=0 or T_1=\infty, which is impossible, because we do not have a reservoir at absolute zero or at infinite temperature. Therefore it is impossible for a real engine to have an efficiency of 100%.

Worked example: Carnot efficiency: raise the source or lower the sink

A Carnot engine works between 400\,\mathrm{K} and 300\,\mathrm{K}. Find its efficiency. Which change gives the larger efficiency: raising the source temperature by 100\,\mathrm{K} or lowering the sink temperature by 100\,\mathrm{K}? If it takes 2500\,\mathrm{J} from a 500\,\mathrm{K} source with a 300\,\mathrm{K} sink, how much work does it do?

Efficiency of a Carnot engine depends only on the two Kelvin temperatures.
\eta=1-\dfrac{T_2}{T_1}=1-\dfrac{300}{400}=0.25=25\%
Raise the source to 500\,\mathrm{K}.
\eta_1=1-\dfrac{300}{500}=0.40=40\%
Lower the sink to 200\,\mathrm{K}.
\eta_2=1-\dfrac{200}{400}=0.50=50\%
Lowering the sink by 100\,\mathrm{K} helps more, because the ratio T_2/T_1 falls further (from 0.75 to 0.5 instead of to 0.6).
Work from the 500\,\mathrm{K} engine: \eta=0.40 and W=\eta Q_1.
W=0.40\times2500=1000\,\mathrm{J},\qquad Q_2=Q_1-W=1500\,\mathrm{J}

\eta=25\%;\ 40\%\ (\text{raise source})\ \text{or}\ 50\%\ (\text{lower sink});\ W=1000\,\mathrm{J}

8.8.2Carnot theorem

Carnot's theorem: No heat engine can be more efficient than a Carnot engine operating between the same two temperatures.

In most cases the cold reservoir is our atmosphere. To increase the efficiency of a heat engine, we have to increase the temperature of the hot reservoir. All real heat engines are less efficient than Carnot's engine because of heat losses.

Example 8.4: efficiency of a heat engine

Find the efficiency of a heat engine operating between 400^\circ\mathrm{C} and 30^\circ\mathrm{C}.

Given

T_1 = 400+273=673\,\mathrm{K}
T_2 = 30+273=303\,\mathrm{K}

To find

percentage efficiency

Use the efficiency of a Carnot engine in percentage.
\text{Percentage efficiency}=\left(1-\dfrac{T_2}{T_1}\right)\times100\%
Substitute the temperatures in kelvin.
\eta=\left(1-\dfrac{303}{673}\right)\times100\%=(1-0.45)\times100\%
Evaluate.
\eta=55\%

\eta=55\%

Assignment 8.4

A heat engine is working with a thermal efficiency of 40\%. It receives 4\,\mathrm{kJ} of heat from a furnace. What amount of waste heat does it reject?

Given

\eta = 40\%=0.40
Q_1 = 4\,\mathrm{kJ}

To find

waste heat Q_2

Efficiency is work output divided by heat input, so find the work done.
W=\eta Q_1=0.40\times4=1.6\,\mathrm{kJ}
The heat rejected is the heat taken in minus the work done (W=Q_1-Q_2).
Q_2=Q_1-W=4-1.6=2.4\,\mathrm{kJ}

Q_2=2.4\,\mathrm{kJ}

8.9RefrigeratorOften examined

We know that in nature heat flows from a hotter body towards a colder body. If we want to move heat from a colder body to a hotter body, we need to reverse the working of a heat engine, and work has to be done on the system.

Refrigerator: A device designed to remove heat from a region that is at a lower temperature than its surroundings. In other words, a device in which the working substance performs a cycle in a direction opposite to that of a heat engine.

The same device can be used to heat a space that is at a higher temperature than the surroundings. In this case the device is called a heat pump, as in Fig. 8.18. In refrigerators, heat flows from the lower temperature source to the higher temperature sink.

Figure 8.18: Schematic diagram of a heat pump. Work W is done on the working substance. It removes heat Q_2 from the cold reservoir at T_2 and delivers heat Q_1 to the hot reservoir at T_1. Compare Fig. 8.14, where all the arrows point the other way.

The refrigerator removes some amount of heat Q_2 from a cold reservoir at low temperature T_2. Some work W is performed by the compressor on the working substance. The quantity of heat Q_1 is rejected to a hot reservoir at higher temperature T_1. The working substance in this case is usually a gas called the refrigerant.

Coefficient of performance of a refrigerator

Coefficient of performance for cooling: The ratio of the amount of heat removed from the cold reservoir to the work required to do so.

\text{CP}_{\text{cooling}}=\dfrac{Q_2}{W}

Derivation Coefficient of performance in terms of heat and temperature

Aim. Express the coefficient of performance for cooling and for heating in terms of the heats and the temperatures.

The work done on the working substance is the difference of the two heats.
W=Q_1-Q_2
Put this W in the definition for cooling.
\text{CP}_{\text{cooling}}=\dfrac{Q_2}{Q_1-Q_2}\qquad(8.10)
Heat Q is proportional to temperature T, so Q_1\to T_1 and Q_2\to T_2.
\text{CP}_{\text{cooling}}=\dfrac{T_2}{T_1-T_2}
For heating, the coefficient of performance is the heat delivered to the hot reservoir divided by the work, Q_1/W. Put W=Q_1-Q_2.
\text{CP}_{\text{heating}}=\dfrac{Q_1}{W}=\dfrac{Q_1}{Q_1-Q_2}\qquad(8.11)
Again using Q\propto T:
\text{CP}_{\text{heating}}=\dfrac{T_1}{T_1-T_2}

Result\text{CP}_{\text{cooling}}=\dfrac{T_2}{T_1-T_2},\qquad \text{CP}_{\text{heating}}=\dfrac{T_1}{T_1-T_2}

The coefficient of performance is not limited to 1 and can be larger. The smaller the temperature difference T_1-T_2, the larger the heat moved per joule of work.

No cyclic device has ever been built that can extract heat Q_2 from a cold reservoir and reject it entirely into a hot reservoir without expenditure of work.

Example 8.5: heat delivered by a refrigerator

The temperature inside a refrigerator is 6^\circ\mathrm{C} and the room temperature is 37^\circ\mathrm{C}. How many joules of heat will be delivered to the room for each joule of electricity consumed by the refrigerator? Assume the refrigerator is ideal.

Given

T_2 = 6^\circ\mathrm{C}=6+273=279\,\mathrm{K}\ \text{(low temperature)}
T_1 = 37^\circ\mathrm{C}=37+273=310\,\mathrm{K}\ \text{(high temperature)}

To find

heat Q_1 delivered to the room for each joule of work

The coefficient of performance for cooling of the ideal refrigerator is
\text{CP}_{\text{cooling}}=\dfrac{T_2}{T_1-T_2}=\dfrac{279}{310-279}=9
The coefficient is also the heat removed per unit of work, so the heat removed from the inside is 9 times the work W.
\text{CP}_{\text{cooling}}=\dfrac{Q_2}{W}\ \Rightarrow\ Q_2=9W
The heat delivered to the room is the heat removed plus the work done by the pump.
Q_1=Q_2+W=9W+W=10W
For each joule of work (each joule of electricity consumed) the heat pumped out into the room is
Q_1=10\,\mathrm{J}

Q_1=10\,\mathrm{J}\ \text{for each joule of work}

Assignment 8.5

The temperature inside a refrigerator is t_2\,^\circ\mathrm{C} and the room temperature is t_1\,^\circ\mathrm{C}. Find the amount of heat delivered to the room for each joule of electricity consumed ideally.

Change to kelvin: T_2=t_2+273 and T_1=t_1+273. The coefficient of performance for cooling is
\text{CP}_{\text{cooling}}=\dfrac{T_2}{T_1-T_2}=\dfrac{t_2+273}{t_1-t_2}
The heat removed is Q_2=W\times\text{CP}_{\text{cooling}}, so the heat delivered to the room per unit work is
Q_1=Q_2+W=W\left(\text{CP}_{\text{cooling}}+1\right)=W\,\dfrac{T_1}{T_1-T_2}
For W=1\,\mathrm{J}:
Q_1=\dfrac{t_1+273}{t_1-t_2}\ \mathrm{J}

Q_1=\dfrac{t_1+273}{t_1-t_2}\,\mathrm{J}\ \text{per joule}

8.10EntropyOften examined

The concept of entropy was introduced into thermodynamics by Rudolf Clausius in 1856 to give a quantitative statement of the second law of thermodynamics. It introduced another state variable, along with temperature, pressure, volume and internal energy. Entropy can be defined in many ways.

Entropy: A thermodynamic quantity that represents the unavailability of a system's thermal energy for conversion into mechanical work.

It is simply the degree of disorder or randomness in a system. For the mathematical form of entropy, consider a system that undergoes a reversible process during which it absorbs an amount of heat \Delta Q at an absolute temperature T. The increase in the state variable called entropy S of the system is given by

\Delta S=\dfrac{\Delta Q}{T}

8.12

From this equation we find that entropy is the measure of a system's thermal energy, per unit temperature, that is unavailable for doing useful work. Like potential energy and internal energy, it is the change in entropy that is important rather than the absolute value. The unit of entropy is \mathrm{J\,K^{-1}}.

Examples of entropy changes

Burning of wood, as wood converts into ash, smoke and gases, releasing the stored energy within the wood.
Melting of ice, as the molecules of ice lose their order during melting (Fig. 8.19). If we place ice in a bucket for some time in a room, with the passage of time the ice converts into liquid water by absorbing heat. The molecules are ordered in ice while in water they move randomly. In all natural processes the entropy of the system as well as of the universe is increasing, and hence the universe is going from an order to a disorder state with time.

Figure 8.19: From order to disorder. The molecules of ice are held in an ordered arrangement; in water they move randomly. Melting increases the entropy, \Delta S>0.

8.10.1Sign convention for entropy

During a change the sign of the entropy change is important. It depends on the direction of heat flow, that is, on whether heat is added to the system or removed from it.

The change in entropy is positive if heat is added to the system.
The change in entropy is negative if heat is removed from the system.

Derivation Entropy change when heat flows from a hot body to a cold body

Aim. Show that the net entropy change is positive when heat flows from a hotter body to a colder body.

Heat Q flows from a heat source at temperature T_1 to a heat sink at temperature T_2, with T_1>T_2.
The source loses heat, so its entropy change is negative.
\Delta S_{\text{source}}=-\dfrac{Q}{T_1}
The sink gains heat, so its entropy change is positive.
\Delta S_{\text{sink}}=+\dfrac{Q}{T_2}
Add the two changes to get the net change.
\Delta S=\dfrac{Q}{T_2}-\dfrac{Q}{T_1}
Since T_1>T_2, the quantity Q/T_2 is larger than Q/T_1, so the net change is positive.
\Delta S>0

Result\Delta S=\dfrac{Q}{T_2}-\dfrac{Q}{T_1}>0

In all natural processes in which heat flows from a hotter body to a colder body there is a net increase in entropy.

We can define the second law of thermodynamics in terms of entropy as follows.

Second law in terms of entropy: If a system undergoes a natural process, it will proceed in the direction that causes the entropy of the system plus the environment to increase.

It is observed that all natural processes tend to proceed towards a state of greater disorder. There is a relation between entropy and molecular disorder, as shown in Fig. 8.20.

Figure 8.20: Randomness of molecules. From left to right the molecules become less random: the disorder and the entropy fall. From right to left, the natural direction, they spread out and the entropy rises.

Arrow of time

Entropy requires a particular direction for time, known as the "Arrow of Time". As one goes forward in time, the entropy of an isolated system increases, hence the measurement of entropy is a way of distinguishing the past from the future.

Worked example: entropy change when heat flows from hot to cold

A quantity of heat Q=600\,\mathrm{J} flows from a body at 400\,\mathrm{K} to a body at 300\,\mathrm{K}. Find the change of entropy of each body and of the two together. What would a total change of -0.5\,\mathrm{J\,K^{-1}} mean?

Given

Q = 600\,\mathrm{J}
T_1 = 400\,\mathrm{K}
T_2 = 300\,\mathrm{K}

To find

\Delta S_1, \Delta S_2, total

The hot body loses heat, so its entropy change is negative.
\Delta S_1=-\dfrac{Q}{T_1}=-\dfrac{600}{400}=-1.5\,\mathrm{J\,K^{-1}}
The cold body gains the same heat at a lower temperature, so its entropy change is positive and larger in size.
\Delta S_2=+\dfrac{Q}{T_2}=+\dfrac{600}{300}=+2.0\,\mathrm{J\,K^{-1}}
Add the two.
\Delta S=\dfrac{Q}{T_2}-\dfrac{Q}{T_1}=2.0-1.5=+0.5\,\mathrm{J\,K^{-1}}
The total entropy increases, as the second law says for a natural process. A total of -0.5\,\mathrm{J\,K^{-1}} would be heat flowing from the cold body to the hot body by itself, which never happens.

Entropy change when heat flows from a hot body to a cold body. The cold body gains more entropy than the hot body loses, so the total increases.

\Delta S_1=-1.5,\ \Delta S_2=+2.0,\ \Delta S_{\text{total}}=+0.5\ \ \mathrm{J\,K^{-1}}

8.10.2Entropy and energyOften examined

In nature only those processes are more likely to occur in which the entropy of the system increases or remains constant.

For all reversible processes, the entropy of the system remains constant.
The entropy increases for an irreversible process.

Every time entropy increases, the opportunity to convert some heat into work is lost.

Entropy and energy: An increase in entropy means the degradation of energy.

For example, in a car engine the burning of fuel produces some work while the remaining heat energy is released into the atmosphere. The heat energy released into the atmosphere is no longer available for work. Before the entropy change, all the heat from this process was available for work, but after the process a major part of the heat energy is lost into the atmosphere.

Even in reverse processes, such as in a refrigerator or an air conditioner, the temperature of one system decreases at the expense of an increase in the temperature of another system. When these systems are considered together as the universe, the entropy of the universe always increases.

Example 8.6: entropy change on melting ice

Calculate the entropy change when 700\,\mathrm{g} of ice melts at 0^\circ\mathrm{C}. The latent heat of fusion for ice is L_f=3.36\times10^{5}\,\mathrm{J\,kg^{-1}}. Also discuss the sign of the entropy change.

Given

m = 700\,\mathrm{g}=0.7\,\mathrm{kg}
T = 0^\circ\mathrm{C}=273\,\mathrm{K}
L_f = 3.36\times10^{5}\,\mathrm{J\,kg^{-1}}

To find

entropy change \Delta S

The heat absorbed by the melting ice is the mass times the latent heat of fusion.
\Delta Q=mL_f
Entropy change is heat divided by the (constant) absolute temperature.
\Delta S=\dfrac{\Delta Q}{T}=\dfrac{mL_f}{T}
Substitute.
\Delta S=\dfrac{0.7\,\mathrm{kg}\times336000\,\mathrm{J\,kg^{-1}}}{273\,\mathrm{K}}=861.5\,\mathrm{J\,K^{-1}}
T
h
e

\Delta S=861.5\,\mathrm{J\,K^{-1}}\ (\text{positive})

Assignment 8.6

A body at 150^\circ\mathrm{C} undergoes a reversible isothermal process. The heat energy removed in the process is 7373\,\mathrm{J}. Find the entropy change of the body.

Given

T = 150+273=423\,\mathrm{K}
Q = -7373\,\mathrm{J}\ \text{(heat removed)}

To find

entropy change \Delta S

Heat is removed from the body, so \Delta Q is negative. Use \Delta S=\Delta Q/T at the constant temperature.
\Delta S=\dfrac{\Delta Q}{T}=\dfrac{-7373}{423}
Divide.
\Delta S=-17.4\,\mathrm{J\,K^{-1}}
The negative sign shows that the entropy of the body decreases when heat is removed.

\Delta S\approx-17.4\,\mathrm{J\,K^{-1}}

Formula sheet

Change in internal energy

\Delta U=U_2-U_1

U_1, U_2: initial and final internal energy

Unit: \mathrm{J}

Average kinetic energy of a gas molecule

\left\langle\tfrac12 mv^{2}\right\rangle=\tfrac32\,kT

k: Boltzmann\'s constant; T: absolute temperature (8.1)

Unit: \mathrm{J}

Ideal gas equation

PV=nRT

P: pressure; V: volume; n: number of moles; R=8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}; T: absolute temperature (8.2)

Unit: \mathrm{Pa\,m^{3}}=\mathrm{J}

Number of moles

n=\dfrac{N}{N_A}=\dfrac{m}{M}

N: number of molecules; N_A: Avogadro\'s number; m: mass of gas; M: molar mass

Unit: \mathrm{mol}

Ideal gas equation in molecules

PV=NkT

N: number of molecules; k: Boltzmann\'s constant (8.3)

Unit: \mathrm{J}

Boltzmann constant

k=\dfrac{R}{N_A}=1.38\times10^{-23}\,\mathrm{J\,K^{-1}}

R: universal gas constant; N_A: Avogadro\'s number

Unit: \mathrm{J\,K^{-1}}

Density of a gas

\rho=\dfrac{PM}{RT}

P: pressure; M: molar mass; T: absolute temperature (8.4)

Unit: \mathrm{kg\,m^{-3}}

Work done by a gas

W=P\Delta V

P: external pressure; \Delta V=V_{\text{final}}-V_{\text{initial}} (8.5)

Unit: \mathrm{J}

Boyle's law

PV=\text{constant}\quad\left(V\propto\dfrac1P\right)

constant temperature and amount of gas

Unit: \mathrm{J}

Charles's law

\dfrac{V}{T}=\text{constant}

constant pressure; T in kelvin

Unit: \mathrm{m^{3}\,K^{-1}}

Gay-Lussac's law

\dfrac{P}{T}=\text{constant}

constant volume; T in kelvin

Unit: \mathrm{Pa\,K^{-1}}

First law of thermodynamics

Q=\Delta U+W

Q: heat added; \Delta U: change in internal energy; W: work done by the system (8.7)

Unit: \mathrm{J}

Isothermal process

\Delta U=0,\qquad Q=W

temperature constant (8.8)

Unit: \mathrm{J}

Adiabatic process

Q=0,\qquad W=-\Delta U,\qquad PV^{\gamma}=\text{constant}

no heat transfer; \gamma is a constant of the gas

Unit: \mathrm{J}

Isochoric process

W=0,\qquad Q=\Delta U

volume constant

Unit: \mathrm{J}

Isobaric process

Q=\Delta U+P\Delta V

pressure constant

Unit: \mathrm{J}

Work done by a heat engine

W=Q_1-Q_2

Q_1: heat taken from the hot reservoir; Q_2: heat rejected to the cold reservoir

Unit: \mathrm{J}

Efficiency of a heat engine

\eta=\dfrac{W}{Q_1}=\dfrac{Q_1-Q_2}{Q_1}=1-\dfrac{Q_2}{Q_1}

W: work output; Q_1: heat input

no unit (a ratio, or a percentage when multiplied by 100\%)

Efficiency of a Carnot engine

\eta=1-\dfrac{T_2}{T_1}

T_1: hot reservoir; T_2: cold reservoir (kelvin) (8.9)

no unit (multiply by 100\% for a percentage)

Coefficient of performance for cooling

\text{CP}_{\text{cooling}}=\dfrac{Q_2}{W}=\dfrac{Q_2}{Q_1-Q_2}=\dfrac{T_2}{T_1-T_2}

Q_2: heat removed from the cold reservoir; W: work done on the working substance (8.10)

no unit

Coefficient of performance for heating

\text{CP}_{\text{heating}}=\dfrac{Q_1}{W}=\dfrac{Q_1}{Q_1-Q_2}=\dfrac{T_1}{T_1-T_2}

Q_1: heat delivered to the hot reservoir (8.11)

no unit

Change of entropy

\Delta S=\dfrac{\Delta Q}{T}

\Delta Q: heat absorbed in a reversible process at absolute temperature T (8.12)

Unit: \mathrm{J\,K^{-1}}

Entropy change in heat flow

\Delta S=\dfrac{Q}{T_2}-\dfrac{Q}{T_1}

heat Q flows from a source at T_1 to a sink at T_2; T_1>T_2

Unit: \mathrm{J\,K^{-1}}

Entropy change on melting

\Delta S=\dfrac{mL_f}{T}

m: mass; L_f: latent heat of fusion; T: melting point (kelvin)

Unit: \mathrm{J\,K^{-1}}

Key terms

Thermodynamics: The branch of physics in which we study the transformation of heat into other forms of energy.
Thermal energy: The energy possessed by a system due to the movement of its particles.
Heat: The form of energy associated with the kinetic energies of all the molecules in a substance; its unit is the joule.
Temperature: The degree of hotness and coldness of an object; a measure of the average kinetic energy of the molecules. Its unit is the kelvin.
Thermal equilibrium: The state of two systems at the same temperature; no net heat flows between them.
System: The matter or space region that is being studied, for example a gas in a cylinder.
Surroundings: Everything outside the system.
Boundary of the system: The surface that separates a system from its surroundings; all energy changes take place through it.
Closed system: A system in which heat energy can enter or leave but mass cannot.
Open system: A system in which both heat energy and mass can enter or leave.
Isolated system: A system in which there is no transfer of mass and heat energy across its boundary.
Diathermic substance: A substance that allows heat to pass through it.
Internal energy: The sum of all the forms of molecular energies (kinetic and potential) of a substance. It is a function of state.
Ideal gas equation: The equation of state PV=nRT=NkT relating pressure, volume, quantity of gas and temperature.
Boltzmann constant: k=R/N_A=1.38\times10^{-23}\,\mathrm{J\,K^{-1}}, the gas constant per molecule.
Kinetic theory of gases: The theory that a gas is made of a very large number of tiny, hard spheres (molecules) in random motion, colliding elastically with one another and with the walls, with no forces between collisions.
Work in thermodynamics: The product of pressure and the change in volume of the system, W=P\Delta V.
Boyle's law: The volume of a given amount of a gas at constant temperature is inversely proportional to the pressure applied.
Charles's law: The volume of a given amount of a gas at constant pressure is directly proportional to the absolute temperature.
Gay-Lussac's law: The applied pressure of a given amount of a gas at constant volume is directly proportional to the absolute temperature.
First law of thermodynamics: When heat is added to a system, it increases the internal energy of the system and also does work by the system on the environment: Q=\Delta U+W.
Isothermal process: A thermodynamic process in which the temperature remains constant throughout the process; the curve is called an isotherm.
Adiabatic process: A process in which no heat or mass transfers between a thermodynamic system and its surroundings; the curve is called an adiabat.
Isochoric process: A thermodynamic process in which the volume of the system remains constant; the curve is called an isochor.
Isobaric process: A thermodynamic process in which the pressure of the system remains constant; the curve is called an isobar.
Heat engine: A device that converts heat energy into mechanical work.
Hot reservoir: The heat source of an engine, kept at a higher temperature T_1, from which heat Q_1 is taken.
Cold reservoir (heat sink): The reservoir at the lower temperature T_2 that absorbs the rejected heat Q_2.
Working substance: The substance (usually a gas) in a heat engine that practically converts heat into useful work.
Reversible process: A process which can be retraced in exactly reverse order without causing any change to its surroundings.
Irreversible process: A process which cannot be retraced in exactly the reverse order without causing a change in its surroundings.
Cyclic process: A succession of events which bring the system back to its initial condition.
Second law of thermodynamics: Kelvin: it is impossible for a heat engine to convert heat taken from a single reservoir entirely into work without leaving any change in the working system. Clausius: it is impossible to cause heat to flow from a cold body to a hot body without the expenditure of work.
Carnot engine: A theoretical engine working in an ideal reversible cycle of isothermal expansion, adiabatic expansion, isothermal compression and adiabatic compression between a heat source and a heat sink.
Carnot's theorem: No heat engine can be more efficient than a Carnot engine operating between the same two temperatures.
Refrigerator: A device in which the working substance performs a cycle in a direction opposite to that of a heat engine, removing heat from a region at a lower temperature than its surroundings.
Heat pump: A refrigerator-type device used to heat a space that is at a higher temperature than the surroundings.
Coefficient of performance (cooling): The ratio of the amount of heat removed from the cold reservoir to the work required to do so, Q_2/W.
Entropy: A thermodynamic quantity that represents the unavailability of a system's thermal energy for conversion into mechanical work; the degree of disorder or randomness of a system, \Delta S=\Delta Q/T.
Arrow of time: The particular direction of time in which the entropy of an isolated system increases.

Multiple choice questionsp. 150–151

A real gas can be approximated as an ideal gas at:

Alow densityAnswer
Bhigh pressure
Chigh density
Dlow temperature

Why

An ideal gas has negligible molecular volume and no intermolecular forces. Both conditions hold when the molecules are far apart, which is the case at low density (and high temperature).

High pressure (B), high density (C) and low temperature (D) all push the molecules together or slow them down, so attractions and molecular size matter and the gas deviates from ideal behaviour.

A thermos flask containing milk as a system is shaken vigorously, and the temperature of the milk rises. This process is called:

Aisochoric
Bisothermal
Cisobaric
DadiabaticAnswer

Why

A thermos flask is insulated, so no heat enters or leaves the milk: Q=0. The temperature rises because work is done on the milk by shaking it, which increases its internal energy. A process with no heat transfer is adiabatic.

Isothermal (B) needs constant temperature, which is not the case here. Isochoric (A) and isobaric (C) describe constant volume and constant pressure, and the question says nothing about either.

Considering your metabolism as a system, during fasting in Ramadan, it is a/an ______ process.

Aisochoric
BisothermalAnswer
Cisobaric
Dadiabatic

Why

The body keeps its temperature nearly constant at about 37\,^\circ\mathrm{C} while fasting, so the metabolic process takes place at constant temperature: it is isothermal.

Adiabatic (D) is wrong because the body still exchanges heat with its surroundings. Isochoric (A) and isobaric (C) would need constant volume or pressure, which the body does not maintain as a rule.

Note. The book prints no key for this question. The key B is worked out from the constant body temperature, which is the reading the book's text on isothermal processes supports.

Change in internal energy of a gas kept in a rigid container when Q J of energy is added to it is:

A\dfrac{Q}{2}
B\dfrac{Q}{3}
CQAnswer
D2Q

Why

A rigid container keeps the volume constant, so the gas does no work: W=0. The first law gives \Delta U=Q-W=Q.

Options A, B and D would need the energy to be shared or doubled, which the first law does not allow when no work is done.

The volume of an ideal gas increases from 5\,\mathrm{m^3} to 20\,\mathrm{m^3} under a constant pressure of 6\times10^{5}\,\mathrm{Pa}. The work done by the gas is:

A5\times10^{5}\,\mathrm{J}
B6\times10^{6}\,\mathrm{J}
C9\times10^{6}\,\mathrm{J}Answer
D9\times10^{7}\,\mathrm{J}

Why

At constant pressure W=P\,\Delta V=6\times10^{5}\times(20-5)=6\times10^{5}\times15=9\times10^{6}\,\mathrm{J}.

B equals P multiplied by 10\,\mathrm{m^3}, which is not the change in volume. D uses the wrong power of ten.

Which process involves an increase in entropy?

Acrystallization
BsublimationAnswer
Cfreezing
Dcondensation

Why

Sublimation turns a solid straight into a gas. The molecules move from a fixed, ordered arrangement into a gas where they move freely and randomly, so the disorder and the entropy increase.

Crystallization, freezing and condensation all make the molecules more ordered (liquid or gas to solid, or gas to liquid), so their entropy decreases.

A Carnot engine working between 300\,\mathrm{K} and 600\,\mathrm{K} has work output 800\,\mathrm{J/cycle}. The heat supplied is:

A1000\,\mathrm{J/cycle}
B1600\,\mathrm{J/cycle}Answer
C1800\,\mathrm{J/cycle}
D2000\,\mathrm{J/cycle}

Why

Efficiency: \eta=1-\dfrac{T_2}{T_1}=1-\dfrac{300}{600}=0.5. Heat supplied: Q_1=\dfrac{W}{\eta}=\dfrac{800}{0.5}=1600\,\mathrm{J}.

A would need an efficiency of 0.8, which is not this engine's efficiency. C and D come from wrong efficiencies as well.

A Carnot engine working between 0\,^\circ\mathrm{C} and 200\,^\circ\mathrm{C} has efficiency \eta_1. The same engine working between 0\,^\circ\mathrm{C} and -200\,^\circ\mathrm{C} has efficiency \eta_2. The ratio \eta_1/\eta_2 is:

A0.577Answer
B0.638
C0.733
D0.85

Why

First case: T_1=473\,\mathrm{K}, T_2=273\,\mathrm{K}, so \eta_1=1-\dfrac{273}{473}=\dfrac{200}{473}.

Second case: T_1=273\,\mathrm{K}, T_2=73\,\mathrm{K}, so \eta_2=1-\dfrac{73}{273}=\dfrac{200}{273}. The ratio is \dfrac{\eta_1}{\eta_2}=\dfrac{273}{473}=0.577.

Maximum work is obtained by a process called:

Aisochoric
Bisothermal
CisobaricAnswer
Dadiabatic

Why

Work is the area under the P-V curve. Starting from the same state and expanding to the same final volume, the pressure stays at its highest value in an isobaric process, so the area P\,\Delta V is the largest. The isothermal curve falls below it and the adiabatic curve falls further still.

Isochoric (A) gives no work at all because \Delta V=0.

Note. The book prints no key for this question. The question is open to the reading 'maximum work' from one initial state to one final volume, and isobaric is the answer under that reading. Under a reading that ranks the processes in some other way, isothermal (B) could be intended.

The coefficient of performance of a refrigerator operating at room temperature is 5. The temperature inside the refrigerator is:

A150\,\mathrm{K}
B200\,\mathrm{K}
C250\,\mathrm{K}Answer
D300\,\mathrm{K}

Why

Coefficient of performance: \mathrm{CP}_{\text{cooling}}=\dfrac{T_2}{T_1-T_2}. With room temperature T_1=300\,\mathrm{K} and \mathrm{CP}=5: 5(300-T_2)=T_2, so T_2=250\,\mathrm{K}.

Check: \dfrac{250}{300-250}=5. A gives \dfrac{150}{150}=1 and B gives \dfrac{200}{100}=2, so both are wrong.

112026 Annual I, Set B, MCQ 8

If the temperature of the heat source is doubled than before, the efficiency increases by:

A1.5 timesAnswer
B2 times
C2.5 times
D3 times

Why

Take the cold reservoir at 300\,\mathrm{K} and the source at 600\,\mathrm{K}. Then \eta=1-\dfrac{300}{600}=0.50.

Doubling the source to 1200\,\mathrm{K} gives \eta=1-\dfrac{300}{1200}=0.75. The efficiency rises by a factor of \dfrac{0.75}{0.50}=1.5.

Note. The book prints no key. The factor depends on the cold reservoir temperature; with the room temperature of 300 K used elsewhere in the exercise it is 1.5.

Thermodynamics is that branch of physics in which we study:

ANo heat transfers to the environment.
BNeither heat nor any mass are transferred to the environment.Answer
CNo dissipated energy and heat transferred to the environment.
DNo mass transfers to the environment.

Why

Thermodynamics studies systems through their boundaries, and the simplest case treated is the isolated system, across whose boundary neither heat nor mass passes. Option B states this.

A and D each leave out one of heat or mass, while C mixes in dissipated energy, which is not part of the definition.

Note. The book prints no key, and the wording is loose. B is the option that matches the book's isolated-system idea; this is a judgement, not a printed answer.

If two objects are in thermal equilibrium with each other, they cannot:

Abe moving.
Bbe undergoing an elastic collision.
Chave different pressures.
Dbe at different temperatures.Answer

Why

Thermal equilibrium means the two objects are at the same temperature, so no net heat flows between them. This is the one thing they cannot do.

Their molecules can be moving or colliding, and gases in thermal equilibrium can still have different pressures if they are not in mechanical equilibrium, so A, B and C are possible.

What is the change in internal energy in the figure?

AzeroAnswer
B20\,\mathrm{J}
C40\,\mathrm{J}
D160\,\mathrm{J}

Why

The curve in the figure is marked T=\text{constant}, so the process runs along an isotherm. The start and end temperatures are equal, so \Delta U=0 whatever the path.

Options B, C and D are numbers that would need a temperature change, and the figure gives none.

Note. The book gives the figure without a key. The figure is redrawn in the book at a small scale; the curve is marked T = constant, which is the basis of the answer.

The change in internal energy is independent of the path taken, similar to:

AK.E.
Bwave energy
Cgravitational P.E.Answer
Dsolar energy

Why

Internal energy is a function of state: only the initial and final states decide \Delta U. The book compares this with gravitational potential energy, which also depends only on position and not on the path.

Kinetic energy, wave energy and solar energy are not chosen because the book's comparison is with gravitational potential energy.

Boltzmann constant k is equal to:

A\dfrac{N_A}{R}
B\dfrac{R}{N_A}Answer
C\dfrac{R}{N_o}
D\dfrac{N_o}{R}

Why

Writing the ideal gas equation for N molecules gives PV=NkT, where k=\dfrac{R}{N_A}. Here R is the universal gas constant and N_A is Avogadro's number.

A and D are reciprocals of the wrong ratio, so they do not have the units of k (J K^{-1} per molecule). C has the right form, but the book's own expression is R/N_A.

Note. Option C is printed in the book with N_o, which is the Avogadro symbol written differently (a misprint for N_A). If read literally it would also equal k, so the key follows the book's notation k=R/N_A.

Two spheres of the same size are made of the same material, but one is hollow and the other is solid. They are heated to the same temperature. Then:

ABoth spheres will expand equally.Answer
BThe hollow sphere will expand more than the solid one.
CThe solid sphere will expand more than the hollow one.
DNo conclusion can be drawn about their relative expansions unless the nature of the material is known.

Why

Expansion of a solid depends on its material and the rise in temperature. Both spheres are made of the same material, have the same outside size and get the same temperature rise, so their outside dimensions grow by the same amount.

The hollow cavity also expands, as if it were filled with the same material, so the hollow sphere does not expand more. The material is known (it is the same for both), so D is wrong.

Note. The book prints no key. A is the answer for equal outside size and the same material; the hollow cavity expands with the shell.

Expansion during heating:

Aoccurs at the same rate for all materials.
Bincreases the weight of a material.
Coccurs at different rates for different materials.Answer
Dincreases the density of material.

Why

Each material has its own coefficient of expansion, so the same rise in temperature gives a different expansion for different materials.

A is false for the reason above. B and D are false because heating does not change the mass (weight) of a material, and expansion lowers its density rather than raising it.

Short questionsp. 152

13 marks

Why does the gas pressure in the tyres rise when a car runs on a road through some distance?

Solution

Friction between the tyres and the road, and the flexing of the rubber, do work on the tyre and air. This work is converted into internal energy, so the temperature of the air in the tyre rises.

The tyre is almost rigid, so the volume of the air stays nearly constant. From the gas law P\propto T at constant volume, a rise in temperature gives a rise in pressure: \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}.

23 marks

When the internal energy of a system rises, can its temperature remain the same?

Solution

Yes. The internal energy includes the potential energy of the molecules as well as their kinetic energy. When a substance melts or boils, heat is used to break the bonds between particles, which raises the potential energy and so the internal energy, while the temperature stays constant.

For an ideal gas the internal energy depends only on temperature, so there the answer is no. The answer is yes for a substance changing state, and the book's example is melting ice at 0\,^\circ\mathrm{C}.

33 marks

Can the temperature of an isolated system change? Explain.

Solution

No, not for the system as a whole. An isolated system has no exchange of heat or mass with its surroundings, and no work is done on it, so Q=0 and W=0. The first law gives \Delta U=0, and the internal energy of a given system depends on its temperature, so the temperature stays the same.

Parts of the isolated system can still exchange energy with each other, but the total internal energy is fixed.

43 marks

The internal energy of a system is increased as a result of some process. How can one tell that the increase was due to orderly macroscopic work W or due to the flow of disorderly microscopic energy Q?

Solution

Work is an orderly transfer of energy: a force moves through a displacement, as when a piston is pushed or a stirrer is turned. It can be measured from W=F\,\Delta y or W=P\,\Delta V and it shows up as a visible movement of the whole system.

Heat is a disorderly transfer of energy at the microscopic level, driven by a temperature difference between the system and its surroundings, as when a hot body is placed in contact with a cold one. If the system is thermally insulated and only moved mechanically, the change is due to work alone, with Q=0.

53 marks

At what temperature (in ^\circ\mathrm{C}) does the volume of a gas at 0\,^\circ\mathrm{C} become double, while the pressure remains constant?

Solution

At constant pressure V\propto T (Kelvin), so \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}. Here T_1=273\,\mathrm{K} and V_2=2V_1, so T_2=2\times273=546\,\mathrm{K}.

T_2=546\,\mathrm{K}=546-273=273\,^\circ\mathrm{C}

The gas must be heated to 273\,^\circ\mathrm{C} for its volume to double.

273\,^\circ\mathrm{C}

63 marks

Should the internal energy of a system necessarily increase if heat is added to it?

Solution

No. By the first law, \Delta U=Q-W. Part or all of the heat can leave the system as work done by it. For example, in an isothermal expansion of an ideal gas, Q=W and \Delta U=0, so the internal energy does not change even though heat is added.

Heat added raises the internal energy only if it is not all used up in doing work on the surroundings.

73 marks

Why is the slope of an adiabatic curve steeper than the slope of an isothermal curve?

Solution

For an isotherm PV=\text{constant}, so \dfrac{dP}{dV}=-\dfrac{P}{V}. For an adiabat PV^{\gamma}=\text{constant} with \gamma>1, so \dfrac{dP}{dV}=-\gamma\dfrac{P}{V}, which is \gamma times steeper.

On an adiabat the temperature changes as well. When the gas is compressed, no heat escapes, so its temperature rises and the pressure rises faster than on an isotherm, where the temperature is held constant.

83 marks

Why do isothermal processes typically take place slowly and adiabatic processes fast?

Solution

An isothermal process must keep the temperature constant, so heat has to flow into or out of the gas to balance the work done. Heat flows slowly through the walls, so the process must be slow.

An adiabatic process has no heat flow (Q=0), so nothing has to be waited for: the gas expands or compresses rapidly, as in the passage of a sound wave through air.

93 marks

Can a process be both adiabatic and isothermal? Explain.

Solution

Only in the trivial case where nothing changes. Adiabatic means Q=0 and isothermal means \Delta T=0, so \Delta U=0. The first law then gives W=Q-\Delta U=0: no work is done and the volume does not change.

A non-trivial expansion or compression cannot be both, because an adiabatic change does work at the expense of internal energy and so lowers the temperature, while an isothermal change needs heat to flow in or out.

103 marks

Does the entropy of a Carnot engine increase for each cycle?

Solution

No. The Carnot cycle is reversible, so the total entropy change over one complete cycle is zero. The working gas returns to its initial state, so its entropy is the same at the end of each cycle as at the start.

The book states that for all reversible processes the entropy remains constant.

113 marks

Does the entropy of a system increase or decrease due to friction?

Solution

It increases. Friction is an irreversible process: it converts organised mechanical energy into disorderly thermal energy in the surfaces and the surroundings. The disorder grows, so the entropy increases.

This is an example of the rule that an increase in entropy means the degradation of energy.

123 marks

What are the similarities and dissimilarities in the working of a refrigerator and an air conditioner?

Solution

Similarities: both are heat pumps, which means heat engines run in reverse. Both use work from a compressor to move heat from a cooler region to a hotter one, through a working gas called the refrigerant that goes round a cycle.

Dissimilarities: a refrigerator removes heat from the enclosed interior of a box and discharges it into the room, while an air conditioner removes heat from a room and discharges it to the outside air. The refrigerator's cold reservoir is the small interior, while the air conditioner's cold reservoir is the room.

133 marks

When a coffee-filled thermos is vigorously shaken: (i) is the work done on the coffee? (ii) Does the internal energy of the coffee increase? (iii) Does the temperature of the coffee rise? (iv) Does the coffee get heat from outside?

Solution

(i) Yes. Shaking is a mechanical action on the coffee: work is done on it by the hand through the flask.

(ii) Yes. Work done on the coffee increases its internal energy, \Delta U=W on the system, since Q=0.

(iii) Yes. Internal energy is linked to temperature for the coffee, so the temperature rises.

(iv) No. A thermos is insulated, so no heat enters from outside: the energy comes in as work.

(i) Is the work done on the coffee?

(i) Yes. Shaking is a mechanical action on the coffee: work is done on it by the hand through the flask.

(ii) Does the internal energy increase?

(ii) Yes. Work done on the coffee increases its internal energy, \Delta U=W on the system, since Q=0.

(iii) Does the temperature rise?

(iii) Yes. Internal energy is linked to temperature for the coffee, so the temperature rises.

(iv) Does the coffee get heat from outside?

(iv) No. A thermos is insulated, so no heat enters from outside: the energy comes in as work.

Comprehensive questionsp. 152–153

8.1

Define internal energy of a system and find its relation with absolute temperature.

Solution

Internal energy

The internal energy U of a system is the sum of all the microscopic energies of its molecules: their random kinetic energies and the potential energies of the forces between them. It is a state function, so only its change \Delta U between two states has a meaning.

Relation with absolute temperature

For an ideal gas there are no intermolecular forces, so its internal energy is only the translational kinetic energy of the molecules. This is the case treated in the book.

Derivation Internal energy of an ideal gas

Aim. Show that the internal energy of an ideal gas is proportional to the absolute temperature T.

Average translational kinetic energy of one molecule from kinetic theory:
\left\langle \tfrac12 mv^{2}\right\rangle=\tfrac32\,kT
The internal energy is the sum of this over all N molecules, because the molecules are identical and have the same average energy.
U=N\left\langle \tfrac12 mv^{2}\right\rangle
Substitute the average kinetic energy.
U=\tfrac32\,NkT
Write k=\dfrac{R}{N_A} and N=nN_A, where n is the number of moles. Then Nk=nN_A\dfrac{R}{N_A}=nR.
U=\tfrac32\,nRT

ResultU=\tfrac32\,nRT

So the internal energy of an ideal gas is directly proportional to its absolute temperature: U\propto T. A rise in temperature raises the average kinetic energy of the molecules and so the internal energy. The temperature must be taken in kelvin, because the absolute scale is the one in which U is proportional to T and it is zero at zero kelvin.

In a process \Delta U=\tfrac32\,nR\,\Delta T, which is why the change in internal energy is what matters in thermodynamics, not its absolute value.

8.2

Explain the gas laws.

Solution

The gas laws give the relation between the volume, pressure and temperature of a fixed quantity of gas. Each law holds with one quantity constant, and they combine into the ideal gas equation.

Boyle's law (constant temperature)

For a fixed amount of gas at constant temperature, the volume is inversely proportional to the pressure: PV=\text{constant}, so V\propto\dfrac{1}{P}. The isotherm is a hyperbola, and a higher temperature gives a curve further from the axes.

Boyle's law

Figure 8.7: Boyle's law. The curves PV=\text{constant} are isotherms; the curve for the higher temperature T_2 lies further from the origin.

Charles's law (constant pressure)

For a fixed amount of gas at constant pressure, the volume is directly proportional to the absolute temperature: \dfrac{V}{T}=\text{constant}. The line extended backwards meets zero volume at -273\,^\circ\mathrm{C}, which is 0\,\mathrm{K}.

Charles's law

Figure 8.8: Charles's law. Volume is a straight line through the origin when plotted against absolute temperature.

Gay-Lussac's law (constant volume)

For a fixed amount of gas at constant volume, the pressure is directly proportional to the absolute temperature: \dfrac{P}{T}=\text{constant} (Figure 8.9 of the book).

Summary and the ideal gas equation

Law	Held constant	Relation
Boyle	T	PV=\text{constant}
Charles	P	\dfrac{V}{T}=\text{constant}
Gay-Lussac	V	\dfrac{P}{T}=\text{constant}

Combining the three laws for a fixed amount of gas gives \dfrac{PV}{T}=\text{constant}. Using n moles and the universal gas constant R=8.314\,\mathrm{J\,mol^{-1}\,K^{-1}} gives the ideal gas equation:

PV=nRT

8.32026 Annual I, Set D, Q.2 (vi) OR

Explain first law of thermodynamics.

Solution

Statement

The first law of thermodynamics is the law of conservation of energy applied to a thermodynamic process. It states that when heat Q is added to a system, the heat increases the internal energy \Delta U of the system and the system does work W on its environment.

Q=\Delta U+W

(8.7)

Sign conventions

Q>0: heat is added to the system; Q<0: heat is removed from it.
\Delta U>0: the temperature of the system rises; \Delta U<0: it falls.
W>0: the system does work on the surroundings (it expands); W<0: work is done on the system (it is compressed).

Work done by a gas

The first law needs the work done by the gas. For a gas in a cylinder with a frictionless piston of area A, the work done when the gas expands is derived below.

Derivation Pressure-volume work

Aim. Show that the work done by a gas at constant external pressure P is W=P\,\Delta V.

Assumptions.

The piston is frictionless and moves slowly, so the gas pressure is the same as the external pressure P.
The cylinder has a uniform cross-section A.

Figure 8.3: Work done by the gas. The gas pushes the piston up through \Delta y, so it does work P\,\Delta V.

The gas pushes on the piston with force F=PA, because pressure is force per unit area.
F=PA
The piston moves up through a small displacement \Delta y while the pressure stays at P.
Work done by the gas on the piston:
W=F\,\Delta y
Substitute F=PA.
W=PA\,\Delta y
The volume swept by the piston is A\,\Delta y=\Delta V, the change in the volume of the gas.
W=P\,\Delta V

ResultW=P\,\Delta V

Where \Delta V=V_{\text{final}}-V_{\text{initial}}. The work is positive when the gas expands (\Delta V>0) and negative when it is compressed (\Delta V<0). On a P-V graph it is the area under the curve.

Application to processes

Applying the first law to the special processes of the unit gives the results used in parts (a) to (d) of question 8.4. For example, if \Delta U=0 then Q=W, and if W=0 then Q=\Delta U.

8.4

Define and explain: (a) isothermal process (b) isobaric process (c) adiabatic process (d) isochoric process.

Solution

Each process keeps one quantity fixed. The four curves below share one set of axes so the difference between them can be seen. The book shows each process separately in Figures 8.10 to 8.13.

Four processes on one P-V diagram

The isotherm falls more slowly than the adiabat, so the adiabat is steeper. The isobar is horizontal and the isochor vertical.

(a) Define and explain the isothermal process.

An isothermal process is a thermodynamic process in which the temperature stays constant throughout. The gas exchanges heat with its surroundings slowly, so that its temperature does not change.

For an ideal gas at constant temperature Boyle's law holds, PV=\text{constant}, so the curve is an isotherm. Since \Delta U=0, the first law gives Q=W: all the heat supplied is used as work.

\Delta U=0\quad\Rightarrow\quad Q=W

Example in the book: melting of ice at 0\,^\circ\mathrm{C} is an isothermal process.

(b) Define and explain the isobaric process.

An isobaric process is a thermodynamic process in which the pressure of the system stays constant. The work is W=P\,\Delta V, so the first law becomes:

Q=\Delta U+P\,\Delta V

On a P-V graph the isobar is a horizontal line and the area under it is the work done. Example in the book: boiling water to steam in an open container.

An adiabatic process is a thermodynamic process in which no heat or mass transfers between the system and its surroundings: Q=0. The first law then gives W=-\Delta U. The work is done at the expense of the internal energy.

Q=0\quad\Rightarrow\quad W=-\Delta U

So an adiabatic expansion cools the gas and an adiabatic compression warms it. For an ideal gas the adiabat is PV^{\gamma}=\text{constant}, which is steeper than the isotherm. Example in the book: the rapid compression and expansion of air as a sound wave passes through it.

(d) Define and explain the isochoric process.

An isochoric process (constant-volume process) is a thermodynamic process in which the volume of the system stays constant. Since \Delta V=0, the work is zero and the first law reduces to:

W=0\quad\Rightarrow\quad Q=\Delta U

All the heat supplied goes into the internal energy, so the pressure and temperature both rise. On a P-V graph the isochor is a vertical line. Example in the book: cooking in a closed pressure cooker.

8.5

Explain the working of a heat engine.

Solution

Definition

A heat engine is a device that converts heat energy into mechanical work. Petrol engines and steam engines are examples. The working principle is the second law of thermodynamics: useful work can be extracted when heat is transferred from a hotter body to a colder one.

Parts of a heat engine

Hot reservoir (heat source) at a higher temperature T_1. It supplies heat Q_1 to the engine.
Cold reservoir (heat sink) at a lower temperature T_2. It absorbs heat Q_2 from the engine. Heat flows only when there is a temperature difference, so a cold reservoir is essential.
Working substance, usually a gas, which takes in heat, expands and does the mechanical work, and returns to its initial state each cycle.

Working and the work relation

Derivation Work output of a heat engine

Aim. Show that the work done by a heat engine in one cycle is W=Q_1-Q_2.

Figure 8.14: Heat engine. Heat Q_1 flows from the hot reservoir into the engine, work W comes out and heat Q_2 goes to the cold reservoir.

The engine works in cycles, so after one complete cycle the working substance returns to its initial state. The internal energy is therefore the same as at the start.
\Delta U=0
Apply the first law to one complete cycle: the net heat taken in equals the work done plus the change in internal energy.
Q_{\text{net}}=\Delta U+W
The net heat taken in is Q_1 from the hot reservoir minus Q_2 given to the cold reservoir.
Q_{\text{net}}=Q_1-Q_2
With \Delta U=0 the work is the net heat.
W=Q_1-Q_2

ResultW=Q_1-Q_2

The work output equals the heat taken from the hot reservoir minus the heat rejected to the cold reservoir.

The efficiency of the engine is the fraction of the heat taken in that is converted into work:

\eta=\dfrac{W}{Q_1}=1-\dfrac{Q_2}{Q_1}

8.62025 Annual I, Set B, Q.2 (xiii) OR2025 Annual I, Set D, Q.2 (xi)

Define and explain second law of thermodynamics.

Solution

Statement

The second law of thermodynamics deals with the conditions under which heat can be converted into work and with the direction in which heat flows. The first law says energy is conserved; the second law says which energy transfers can actually happen. It is stated in two equivalent forms.

Lord Kelvin's statement

It is impossible for a heat engine to convert heat taken from a single reservoir entirely into work without leaving any change in the working system.

A single reservoir, however much energy it holds, cannot make an engine do work on its own. A cold reservoir is always needed, which is why the heat content of the atmosphere cannot be used as a single source.

Rudolf Clausius's statement

It is impossible to cause heat to flow from a cold body to a hot body without the expenditure of work. Heat flows naturally only from hot to cold, and a refrigerator needs work to reverse this flow.

Meaning

Heat cannot be converted completely into work.
Heat flows spontaneously only from a hotter body to a colder one.
Every natural process moves towards a state of greater disorder, which is described by entropy (question 8.10).

8.72026 Annual I, Set D, Q.5

Define and explain Carnot's engine, also discuss the relation for the efficiency of Carnot's engine.

Solution

Carnot's engine

A Carnot engine is a theoretical heat engine that operates on the Carnot cycle, between a hot source and a cold sink, using a reversible cycle. It was introduced by Sadi Carnot in 1824. It is the most efficient engine that can work between two given temperatures.

The Carnot cycle

Isothermal expansion a\to b at T_1: the gas absorbs heat Q_1 from the hot reservoir, expands and does work.
Adiabatic expansion b\to c: the cylinder is insulated, the gas cools from T_1 to T_2 and does more work.
Isothermal compression c\to d at T_2: the gas rejects heat Q_2 to the cold reservoir and is compressed.
Adiabatic compression d\to a: the gas is compressed in an insulated cylinder, its temperature rises from T_2 back to T_1, and the cycle is complete.

P-V diagram of the Carnot cycle

Figure 8.16: P-V diagram for the Carnot cycle. Two isotherms at T_1 and T_2 are joined by two adiabats.

Efficiency of the Carnot engine

Derivation Efficiency of a Carnot engine

Aim. Express the efficiency of a Carnot engine in terms of the absolute temperatures T_1 (hot) and T_2 (cold).

Work from the first law over one cycle, as in question 8.5:
W=Q_1-Q_2
Efficiency is output work divided by input heat:
\eta=\dfrac{W}{Q_1}=1-\dfrac{Q_2}{Q_1}
For a reversible Carnot cycle the heat exchanged in each isotherm is proportional to the absolute temperature of that isotherm, so \dfrac{Q_2}{Q_1}=\dfrac{T_2}{T_1}.
\dfrac{Q_2}{Q_1}=\dfrac{T_2}{T_1}
Substitute into the efficiency.
\eta=1-\dfrac{T_2}{T_1}

Result\eta=1-\dfrac{T_2}{T_1}

As a percentage, \eta\,(\%)=\left(1-\dfrac{T_2}{T_1}\right)\times100\,\%. The efficiency depends only on the two reservoir temperatures, not on the working substance. It can reach 100\,\% only if T_2=0\,\mathrm{K} or T_1\to\infty, which is impossible, so no real engine is perfectly efficient.

Carnot's theorem

No heat engine can be more efficient than a Carnot engine working between the same two temperatures. Real engines are less efficient because of heat losses.

8.92026 Annual I, Set B, Q.52025 Annual I, Set B, Q.4

Explain the working of a refrigerator.

Solution

Definition

A refrigerator is a device that works in the reverse direction to a heat engine. It removes heat from a region at a lower temperature than its surroundings and gives it to a region at a higher temperature, and work must be done on the working substance to make this happen.

Working

The working substance, a gas called the refrigerant, takes in heat Q_2 from the cold reservoir at T_2 (the inside of the refrigerator).
A compressor does work W on the refrigerant, raising its pressure and temperature.
The refrigerant gives out heat Q_1 to the hot reservoir at T_1 (the room), where Q_1=Q_2+W.

Figure 8.18: Schematic diagram of a heat pump. A refrigerator works the same way: work W moves heat Q_2 from the cold reservoir to the hot reservoir.

Coefficient of performance

Derivation Coefficient of performance for cooling

Aim. Express the coefficient of performance for cooling, \mathrm{CP}_{\text{cooling}}, in terms of the temperatures.

Definition: heat removed from the cold reservoir divided by the work needed to remove it.
\mathrm{CP}_{\text{cooling}}=\dfrac{Q_2}{W}
From the energy balance W=Q_1-Q_2.
\mathrm{CP}_{\text{cooling}}=\dfrac{Q_2}{Q_1-Q_2}
Heat is proportional to absolute temperature, so \dfrac{Q_2}{Q_1}=\dfrac{T_2}{T_1}. Divide the top and bottom by Q_1, or use Q\propto T directly.
\mathrm{CP}_{\text{cooling}}=\dfrac{T_2}{T_1-T_2}

Result\mathrm{CP}_{\text{cooling}}=\dfrac{T_2}{T_1-T_2}

The closer the inside temperature T_2 is to the room temperature T_1, the larger the coefficient of performance and the less work is needed.

For heating, the coefficient of performance is \mathrm{CP}_{\text{heating}}=\dfrac{Q_1}{W}=\dfrac{T_1}{T_1-T_2}. No cyclic device can extract heat from a cold reservoir and deliver it entirely to a hot one without work.

Note. The book's comprehensive list runs from 8.7 to 8.9 with no question numbered 8.8. This answer keeps the book's numbering.

8.10

What is entropy? Explain.

Solution

Definition

Entropy S is a thermodynamic quantity that measures the unavailability of a system's thermal energy for conversion into mechanical work. It is also the degree of disorder or randomness in a system.

Clausius introduced entropy in 1856 as a quantitative statement of the second law. For a reversible process in which a system absorbs heat \Delta Q at absolute temperature T, the increase in entropy is:

\Delta S=\dfrac{\Delta Q}{T}

(8.12)

Entropy is measured in \mathrm{J\,K^{-1}}. Only the change in entropy matters, as with potential energy and internal energy.

Sign convention

The change in entropy is positive when heat is added to the system.
The change in entropy is negative when heat is removed from the system.

Derivation Net entropy change when heat flows between two reservoirs

Aim. Find the net change in entropy when heat Q flows from a hot source at T_1 to a cold sink at T_2, where T_1>T_2.

The source loses heat Q, so its entropy change is negative.
\Delta S_{\text{source}}=-\dfrac{Q}{T_1}
The sink gains heat Q, so its entropy change is positive.
\Delta S_{\text{sink}}=+\dfrac{Q}{T_2}
Add the two changes for the net entropy change.
\Delta S_{\text{net}}=\dfrac{Q}{T_2}-\dfrac{Q}{T_1}
Since T_1>T_2, \dfrac{1}{T_2}>\dfrac{1}{T_1}, so the net change is positive.
\Delta S_{\text{net}}>0

Result\Delta S_{\text{net}}>0

The total entropy increases whenever heat flows from a hotter body to a colder body. Natural processes increase the total entropy of the system and its surroundings. This is the second law stated in terms of entropy.

8.112026 Annual I, Set B, Q.52025 Annual I, Set D, Q.2 (i) OR

Explain that entropy is the degradation of energy.

Solution

Entropy is said to measure the degradation of energy: as entropy increases, less of the energy in the system is available to do useful work. In a natural process the energy is not lost, but its ability to do work is reduced.

Derivation Energy made unavailable for work by an irreversible flow of heat

Aim. Show that the work lost when heat flows directly from T_1 to T_2 equals T_2\,\Delta S.

Heat Q flows directly from the hot body at T_1 to the cold body at T_2, with no work done. The energy is not available for work now.
Had the same heat been used by a Carnot engine between T_1 and T_2, the maximum work obtainable would have been:
W_{\max}=Q\left(1-\dfrac{T_2}{T_1}\right)
Expand the maximum work.
W_{\max}=Q-\dfrac{QT_2}{T_1}
Write it in terms of entropy, using \Delta S=\dfrac{Q}{T_2}-\dfrac{Q}{T_1} from question 8.10.
T_2\,\Delta S=Q-\dfrac{QT_2}{T_1}
Compare the two results: the work lost is equal to T_2\,\Delta S.
W_{\text{lost}}=T_2\,\Delta S

ResultW_{\text{lost}}=T_2\,\Delta S

Each joule of entropy increase at temperature T_2 removes that much energy from the possibility of doing work. Entropy keeps increasing in natural processes, so the energy available for doing work keeps falling. The total energy is conserved, but it becomes less useful: it is degraded.

In a reversible process the entropy is constant and no energy is degraded. Natural processes are irreversible, so energy is always being degraded.

Numerical problemsp. 153

8.1

Calculate the number of moles of air in an inflated balloon of radius 10\,\mathrm{cm} and pressure 180\,\mathrm{kPa} at room temperature.

Solution

Given

r = 0.10\,\mathrm{m} radius of the balloon (10 cm)
P = 180\,\mathrm{kPa}=1.8\times10^{5}\,\mathrm{Pa}
T = 298\,\mathrm{K} room temperature, 25 ^\circC
R = 8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}

To find

n (number of moles)

The balloon is a sphere, so its volume is:
V=\dfrac{4}{3}\pi r^{3}=\dfrac{4}{3}\pi(0.10)^{3}=4.19\times10^{-3}\,\mathrm{m^{3}}
Use the ideal gas equation PV=nRT and make n the subject.
n=\dfrac{PV}{RT}
Substitute the values.
n=\dfrac{1.8\times10^{5}\times4.19\times10^{-3}}{8.314\times298}=\dfrac{754}{2478}
Evaluate.
n=0.304\,\mathrm{mol}

n\approx0.30\,\mathrm{mol}

Book answer. 0.3 moles

Note. Taking room temperature as 300\,\mathrm{K} gives the same 0.30\,\mathrm{mol}. The temperature used here is 298\,\mathrm{K} (25 ^\circC), the value the book uses in Example 8.1.

8.2

A heat engine operates between the freezing and boiling points of water. Find its efficiency.

Solution

Given

T_1 = 373\,\mathrm{K} boiling point, 100 ^\circC
T_2 = 273\,\mathrm{K} freezing point, 0 ^\circC

To find

\eta

The efficiency of a heat engine working between two reservoirs is at most the Carnot value:
\eta=1-\dfrac{T_2}{T_1}
Substitute.
\eta=1-\dfrac{273}{373}=\dfrac{100}{373}
Evaluate and express as a percentage.
\eta=0.268=26.8\,\%

\eta\approx27\,\%

Book answer. 27%

8.3

One mole of a perfect gas in a cylinder with a piston has pressure P, volume V and temperature T. If its temperature is increased by 1\,\mathrm{K}, find the increase in volume, if the initial volume was 100\,\mathrm{m^3}.

Solution

Given

n = 1\ \mathrm{mol}
V = 100\,\mathrm{m^3}
\Delta T = 1\,\mathrm{K}
T = 273\,\mathrm{K} initial temperature taken as 0 ^\circC, the value the book uses

To find

\Delta V

The pressure is constant (the piston is free), so the gas obeys Charles's law, \dfrac{V}{T}=\text{constant}.
Therefore the change in volume is proportional to the change in temperature:
\dfrac{\Delta V}{V}=\dfrac{\Delta T}{T}
Solve for \Delta V.
\Delta V=\dfrac{V\,\Delta T}{T}
Substitute.
\Delta V=\dfrac{100\times1}{273}=0.366\,\mathrm{m^{3}}

\Delta V\approx0.37\,\mathrm{m^{3}}

Book answer. 0.37 m^3

Note. The question does not give the starting temperature. The book's answer 0.37\,\mathrm{m^3} corresponds to T=273\,\mathrm{K}. With T=298\,\mathrm{K} the answer would be 0.34\,\mathrm{m^3}.

8.4

The high temperature of a Carnot engine is 600\,\mathrm{K}. If the engine absorbs 600\,\mathrm{J} of heat at the high temperature, and the low temperature is 400\,\mathrm{K}, find the work done by the engine.

Solution

Given

T_1 = 600\,\mathrm{K} high temperature
T_2 = 400\,\mathrm{K} low temperature
Q_1 = 600\,\mathrm{J} heat absorbed from the hot reservoir

To find

Efficiency of the Carnot engine:
\eta=1-\dfrac{T_2}{T_1}=1-\dfrac{400}{600}=\dfrac{1}{3}
Work done is the efficiency times the heat absorbed:
W=\eta\,Q_1=\dfrac{1}{3}\times600\,\mathrm{J}
Evaluate.
W=200\,\mathrm{J}

W=200\,\mathrm{J}

Book answer. 200 J

Note. The question says the engine absorbs the 600\,\mathrm{J} 'at low temperature of 400\,\mathrm{K}'. An engine absorbs heat from the hot reservoir, so 600\,\mathrm{J} is taken as Q_1 at 600\,\mathrm{K}. This is the only reading that gives the book's 200\,\mathrm{J}.

8.5

A 220\,\mathrm{W} electric immersion heater is used to boil 136\,\mathrm{g} of water. Calculate the time required to bring this water from 23.5\,^\circ\mathrm{C} to its boiling point, ignoring any heat loss.

Solution

Given

P = 220\,\mathrm{W} power of the heater
m = 136\,\mathrm{g}=0.136\,\mathrm{kg}
\theta_1 = 23.5\,^\circ\mathrm{C}
\theta_2 = 100\,^\circ\mathrm{C} boiling point
c = 4186\,\mathrm{J\,kg^{-1}\,K^{-1}} specific heat capacity of water

To find

Heat needed to raise the temperature of the water:
Q=mc\,\Delta\theta
Substitute, with \Delta\theta=100-23.5=76.5\,\mathrm{K}.
Q=0.136\times4186\times76.5=4.36\times10^{4}\,\mathrm{J}
The heater supplies energy at rate P, so Q=Pt. Make t the subject.
t=\dfrac{Q}{P}
Substitute.
t=\dfrac{4.36\times10^{4}}{220}=198\,\mathrm{s}

t\approx198\,\mathrm{s}

Book answer. 198 s

Note. The specific heat capacity of water, 4186\,\mathrm{J\,kg^{-1}\,K^{-1}}, is the standard value used here; the book does not print it in this problem.

8.6

If a heat engine at room temperature has an efficiency of 50\,\%, you change the heat source to increase the efficiency up to 70\,\%. Find the temperature difference between the two heat sources.

Solution

Given

\eta_1 = 50\,\%=0.50
\eta_2 = 70\,\%=0.70
T_c = 298\,\mathrm{K} room temperature, the cold reservoir, unchanged

To find

\Delta T=T_{h2}-T_{h1}

For a Carnot engine \eta=1-\dfrac{T_c}{T_h}, so T_h=\dfrac{T_c}{1-\eta}.
T_h=\dfrac{T_c}{1-\eta}
First source, \eta_1=0.50:
T_{h1}=\dfrac{298}{0.50}=596\,\mathrm{K}
New source, \eta_2=0.70:
T_{h2}=\dfrac{298}{0.30}=993\,\mathrm{K}
Temperature difference between the two sources:
\Delta T=993-596=397\,\mathrm{K}

\Delta T\approx397\,\mathrm{K}

Book answer. 397 K

Note. The answer uses the room temperature 298\,\mathrm{K}, the value that gives the book's 397\,\mathrm{K}. With 300\,\mathrm{K} the answer would be 400\,\mathrm{K}.

8.7

A Carnot engine absorbs an amount Q of heat at temperature T. Find the amount of heat rejected at temperature T/3.

Solution

Given

Q_1 = Q heat absorbed at temperature T
T_1 = T
T_2 = \dfrac{T}{3}

To find

For a Carnot engine the heats are proportional to the absolute temperatures:
\dfrac{Q_2}{Q_1}=\dfrac{T_2}{T_1}
Substitute.
\dfrac{Q_2}{Q}=\dfrac{T/3}{T}=\dfrac{1}{3}
Solve for Q_2.
Q_2=\dfrac{Q}{3}

Q_2=\dfrac{Q}{3}

Book answer. Q/3

8.8

Energy 500\,\mathrm{J} is required to melt 2\,\mathrm{g} of ice at 0\,^\circ\mathrm{C}. Find the change in entropy of 70\,\mathrm{g} of water at 0\,^\circ\mathrm{C}, if it changes into ice in a refrigerator.

Solution

Given

Q = 500\,\mathrm{J} energy to melt 2 g of ice
m = 70\,\mathrm{g} mass of water to freeze
T = 273\,\mathrm{K} 0 ^\circC

To find

\Delta S

Heat per gram of ice melted, from the 2 g that take 500 J:
\dfrac{500}{2}=250\,\mathrm{J\,g^{-1}}
The water freezes, so it gives out heat. The heat removed to freeze 70 g:
Q=70\times250=1.75\times10^{4}\,\mathrm{J}
Heat removed from the water at constant temperature T, so the entropy change is negative:
\Delta S=-\dfrac{Q}{T}
Substitute.
\Delta S=-\dfrac{1.75\times10^{4}}{273}=-64\,\mathrm{J\,K^{-1}}

\Delta S\approx-64\,\mathrm{J\,K^{-1}}

Book answer. -64 J K^-1

Note. The data imply a latent heat of 2.5\times10^{5}\,\mathrm{J\,kg^{-1}}, which differs from the 3.36\times10^{5}\,\mathrm{J\,kg^{-1}} of Example 8.6. The answer follows the data given in the question.

8.9

A gas expands from volume 1\,\mathrm{m^3} to 2\,\mathrm{m^3} at constant atmospheric pressure.

Solution

Given

V_1 = 1\,\mathrm{m^3}
V_2 = 2\,\mathrm{m^3}
P = 1.013\times10^{5}\,\mathrm{Pa} atmospheric pressure, constant

To find

(a) W
(b) the work on a P-V diagram

(a) Calculate the work done by the gas.

The pressure is constant, so the work is W=P\,\Delta V.
W=P\,\Delta V
Change in volume:
\Delta V=V_2-V_1=2-1=1\,\mathrm{m^3}
Substitute.
W=1.013\times10^{5}\times1=1.013\times10^{5}\,\mathrm{J}

W=1.013\times10^{5}\,\mathrm{J}

(b) Represent the work done on a P-V diagram.

The process is isobaric, so the P-V graph is a horizontal line at P=1.013\times10^{5}\,\mathrm{Pa} from V=1\,\mathrm{m^3} to V=2\,\mathrm{m^3}. The work is the shaded area under this line, which is the rectangle 1.013\times10^{5}\times1=1.013\times10^{5}\,\mathrm{J}.

Work as area under the isobar

The shaded rectangle between V=1 and V=2\,\mathrm{m^3} is the work done by the gas.

W=1.013\times10^{5}\,\mathrm{J}

Book answer. 1.013 x 10^5 J

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 81 mark

If the temperature of heat source is doubled than before, efficiency increases by:

A1.5 timesAnswer
B2 times
C2.5 times
D3 times

Why

The efficiency of a Carnot engine is \eta=1-\dfrac{T_2}{T_1}. Take the cold reservoir at T_2=300\,\mathrm{K} and the source at T_1=600\,\mathrm{K}: \eta=1-\dfrac{300}{600}=0.50.

Doubling the source temperature to 1200\,\mathrm{K} gives \eta'=1-\dfrac{300}{1200}=0.75, so \dfrac{\eta'}{\eta}=\dfrac{0.75}{0.50}=1.5 and the answer is A. The efficiency does not double (B): the term T_2/T_1 is halved, not the efficiency, and it can never exceed 1.

Note. The factor depends on the starting temperatures. The value 1.5 follows when the source is initially at twice the temperature of the sink (T_1=2T_2), which is the case taken in the book's MCQ 11, the same question.

2025 Annual I, Set B, MCQ 91 mark

In an adiabatic process which quantity remains constant?

AInternal energy
BVolume
CTemperature
DMassAnswer

Why

In an adiabatic process no heat or mass transfers between the system and its surroundings, so the mass of the gas stays constant (option D). The heat exchanged is zero, Q=0, but heat is not one of the options.

The other quantities all change. The first law gives W=-\Delta U, so the work done by or on the gas changes the internal energy (A) and with it the temperature (C), and the gas expands or is compressed, so the volume (B) changes. The constant-volume process is isochoric and the constant-temperature process is isothermal.

Note. The only quantity in the list that is conserved is the mass. The quantity that is zero in an adiabatic change, the heat Q, is not offered.

2025 Annual I, Set B, Q.2 (ii) OR3 marks

A body at 200\,^{\circ}\mathrm{C} undergoes a reversible isothermal process. The heat energy removed in this process is 7300\,\mathrm{J}. Calculate the entropy change and discuss entropy sign.

Solution

Given

T = 200\,^\circ\mathrm{C}+273=473\,\mathrm{K} constant temperature of the body
Q = 7300\,\mathrm{J} heat removed

To find

\Delta S and its sign

Formula (1 mark). For a reversible process at constant absolute temperature T the entropy change is \Delta S=\dfrac{\Delta Q}{T}. Heat is removed from the body, so \Delta Q=-7300\,\mathrm{J}.
\Delta S=\dfrac{\Delta Q}{T}
Substitution (1 mark).
\Delta S=\dfrac{-7300}{473}
Answer with unit.
\Delta S=-15.4\,\mathrm{J\,K^{-1}}

Sign of the entropy change (1 mark). The sign is negative because heat is removed from the body. Entropy change is positive when heat is added to a system and negative when heat is removed from it. The entropy of the body decreases, while the entropy of the surroundings that receive the heat increases by at least this amount.

\Delta S\approx-15.4\,\mathrm{J\,K^{-1}}

2026 Annual I, Set B, Q.2 (vi) OR1x3 = 3 marks

In thermodynamics briefly describe: i) Reversible process ii) Irreversible process iii) Cyclic process

Solution

Reversible process (1 mark). A process which can be retraced in exactly the reverse order without causing any change in the surroundings. If heat is absorbed in the direct process it is given out in the reverse process, and if work is done by the system in the direct process, it is done on the system in the reverse process. Examples: slow melting, freezing, boiling and condensing (water and ice).

Irreversible process (1 mark). A process which cannot be retraced in exactly the reverse order without causing a change in the surroundings. Examples: any process involving friction or dissipation of energy, and conduction, convection and radiation.

Cyclic process (1 mark). A succession of events which bring the system back to its initial condition. Event A takes the system from condition 1 to condition 2, and event B brings it back from 2 to 1. For a complete cycle the change in internal energy is zero, \Delta U=0. Heat engines work in cycles.

2026 Annual I, Set B, Q.2 (xi)1.5+1.5 = 3 marks

State first law of thermodynamics. Briefly explain the term internal energy.

Solution

First law of thermodynamics (1.5 marks). When heat Q is added to a system, it increases the internal energy \Delta U of the system and also does work W by the system on the environment. It is the law of conservation of energy applied to thermodynamic processes:

Q=\Delta U+W

Here Q is positive when heat is added, W is positive when the gas expands (work done by the system) and \Delta U is positive when the temperature rises.

Internal energy (1.5 marks). The internal energy of a system is the sum of all the forms of molecular energies, kinetic and potential, of its substance. For an ideal gas there are no intermolecular forces, so it is only the translational kinetic energy of the molecules, with average value \tfrac32kT per molecule. It is therefore directly proportional to the absolute temperature. Internal energy is increased by heating and by doing mechanical work, and it is a function of state: its change \Delta U=U_2-U_1 depends only on the initial and final states, not on the path.

2026 Annual I, Set B, Q.2 (xii)2+1 = 3 marks

Briefly explain the relationship between pressure and volume of a gas at constant temperature. Show it graphically as well.

Solution

Relation (2 marks). For a fixed amount of gas at constant temperature, the ideal gas equation PV=nRT has a constant right-hand side, so

PV=\text{constant}\qquad\text{or}\qquad V\propto\dfrac1P

This is Boyle's law: the volume of a given amount of gas at constant temperature is inversely proportional to the pressure applied. If the pressure is increased the volume decreases, and vice versa. For two states of the same gas at the same temperature, P_1V_1=P_2V_2.

Graph (1 mark). The graph of P against V at constant temperature is a curve (a hyperbola, the isotherm). It falls as the volume increases and never touches either axis.

Pressure against volume at constant temperature

Figure 8.7: Boyle's law. At constant temperature the product PV stays constant, so the graph of pressure against volume is a curve (a hyperbola) that falls as the volume increases.

2025 Annual I, Set B, Q.2 (xiii) OR3 marks

Write three statements of second law of thermodynamics. (one in terms of entropy)

Solution

The second law of thermodynamics can be stated in three ways. Two are the classical statements and the third is in terms of entropy.

Kelvin's statement: it is impossible for a heat engine to convert heat taken from a single reservoir entirely into work without leaving any change in the working system.
Clausius's statement: it is impossible to cause heat to flow from a cold body to a hot body without the expenditure of work.
In terms of entropy: if a system undergoes a natural process, it proceeds in the direction that causes the entropy of the system plus the environment to increase. For all reversible processes the entropy stays constant, and for an irreversible process it increases, so the entropy of the universe always increases.

2025 Annual I, Set B, Q.44+2 = 6 marks

Explain the working of refrigerator. Also explain the coefficient of performance of refrigerator.

Solution

Working of a refrigerator (4 marks)

Definition

Working

The working substance, a gas called the refrigerant, takes in heat Q_2 from the cold reservoir at T_2 (the inside of the refrigerator).
A compressor does work W on the refrigerant, raising its pressure and temperature.
The refrigerant gives out heat Q_1 to the hot reservoir at T_1 (the room), where Q_1=Q_2+W.

Figure 8.18: Schematic diagram of a heat pump. A refrigerator works the same way: work W moves heat Q_2 from the cold reservoir to the hot reservoir.

Coefficient of performance (2 marks)

Derivation Coefficient of performance for cooling

Aim. Express the coefficient of performance for cooling, \mathrm{CP}_{\text{cooling}}, in terms of the temperatures.

Definition: heat removed from the cold reservoir divided by the work needed to remove it.
\mathrm{CP}_{\text{cooling}}=\dfrac{Q_2}{W}
From the energy balance W=Q_1-Q_2.
\mathrm{CP}_{\text{cooling}}=\dfrac{Q_2}{Q_1-Q_2}
Heat is proportional to absolute temperature, so \dfrac{Q_2}{Q_1}=\dfrac{T_2}{T_1}. Divide the top and bottom by Q_1, or use Q\propto T directly.
\mathrm{CP}_{\text{cooling}}=\dfrac{T_2}{T_1-T_2}

Result\mathrm{CP}_{\text{cooling}}=\dfrac{T_2}{T_1-T_2}

The closer the inside temperature T_2 is to the room temperature T_1, the larger the coefficient of performance and the less work is needed.

2026 Annual I, Set B, Q.53+3 = 6 marks

a. Discuss working of a refrigerator. b. How increase in Entropy means degradation of energy? Explain with an example.

Solution

(a) Working of a refrigerator (3 marks)

Definition

Working

The working substance, a gas called the refrigerant, takes in heat Q_2 from the cold reservoir at T_2 (the inside of the refrigerator).
A compressor does work W on the refrigerant, raising its pressure and temperature.
The refrigerant gives out heat Q_1 to the hot reservoir at T_1 (the room), where Q_1=Q_2+W.

Figure 8.18: Schematic diagram of a heat pump. A refrigerator works the same way: work W moves heat Q_2 from the cold reservoir to the hot reservoir.

The coefficient of performance for cooling is the ratio of the heat removed from the cold reservoir to the work done:

\text{CP}_{\text{cooling}}=\dfrac{Q_2}{W}=\dfrac{Q_2}{Q_1-Q_2}=\dfrac{T_2}{T_1-T_2}

(b) Increase in entropy means degradation of energy (3 marks)

Entropy S measures the unavailability of a system's thermal energy for conversion into mechanical work, and the degree of disorder. For a reversible process at absolute temperature T, \Delta S=\dfrac{\Delta Q}{T}. In natural (irreversible) processes the entropy of the system plus its surroundings increases. Every time entropy increases, the opportunity to convert some heat into work is lost, so an increase in entropy means the degradation of energy: the energy is still there, but less of it can do useful work.

Example. In a car engine the burning of fuel produces some work, while the remaining heat energy is released into the atmosphere. Before the change all the heat from the fuel was available for work, but afterwards a major part of it is lost to the atmosphere and can no longer be used. The entropy of the universe has increased and the energy has been degraded.

Derivation Energy made unavailable for work by an irreversible flow of heat

Aim. Show that the work lost when heat flows directly from T_1 to T_2 equals T_2\,\Delta S.

Heat Q flows directly from the hot body at T_1 to the cold body at T_2, with no work done. The energy is not available for work now.
Had the same heat been used by a Carnot engine between T_1 and T_2, the maximum work obtainable would have been:
W_{\max}=Q\left(1-\dfrac{T_2}{T_1}\right)
Expand the maximum work.
W_{\max}=Q-\dfrac{QT_2}{T_1}
Write it in terms of entropy, using \Delta S=\dfrac{Q}{T_2}-\dfrac{Q}{T_1} from section 8.10.1.
T_2\,\Delta S=Q-\dfrac{QT_2}{T_1}
Compare the two results: the work lost is equal to T_2\,\Delta S.
W_{\text{lost}}=T_2\,\Delta S

ResultW_{\text{lost}}=T_2\,\Delta S

In a reversible process the entropy is constant and no energy is degraded. Natural processes are irreversible, so energy is always being degraded.

2026 Annual I, Set D, MCQ 91 mark

According to the first law of thermodynamics, if 200 Joules of heat is added to a system and 80 Joules of work is done by the system, the internal energy change is:

A120J decrease
B80J increase
C120 J increaseAnswer
D280J increase

Why

By the first law, Q=\Delta U+W, so \Delta U=Q-W=200-80=120\,\mathrm{J}. The heat added is +200\,\mathrm{J} and the work done by the system is +80\,\mathrm{J}, so the internal energy increases by 120\,\mathrm{J} (option C).

Option B (80\,\mathrm{J}) is just the work. Option D adds the work to the heat instead of subtracting it, and option A has the wrong sign.

2025 Annual I, Set D, MCQ 101 mark

A Carnot engine undergoes ______ processes in one cycle.

ATwo
BThree
CFourAnswer
DSix

Why

A Carnot cycle has four processes: isothermal expansion, adiabatic expansion, isothermal compression and adiabatic compression (option C).

Option A (two) comes from counting only the two kinds of process, isothermal and adiabatic. Each kind occurs twice in the cycle, once as an expansion and once as a compression.

2025 Annual I, Set D, Q.2 (i) OR3 marks

Explain that increase in entropy means degradation of energy.

Solution

Derivation Energy made unavailable for work by an irreversible flow of heat

Aim. Show that the work lost when heat flows directly from T_1 to T_2 equals T_2\,\Delta S.

Heat Q flows directly from the hot body at T_1 to the cold body at T_2, with no work done. The energy is not available for work now.
Had the same heat been used by a Carnot engine between T_1 and T_2, the maximum work obtainable would have been:
W_{\max}=Q\left(1-\dfrac{T_2}{T_1}\right)
Expand the maximum work.
W_{\max}=Q-\dfrac{QT_2}{T_1}
Write it in terms of entropy, using \Delta S=\dfrac{Q}{T_2}-\dfrac{Q}{T_1} from section 8.10.1.
T_2\,\Delta S=Q-\dfrac{QT_2}{T_1}
Compare the two results: the work lost is equal to T_2\,\Delta S.
W_{\text{lost}}=T_2\,\Delta S

ResultW_{\text{lost}}=T_2\,\Delta S

In a reversible process the entropy is constant and no energy is degraded. Natural processes are irreversible, so energy is always being degraded.

Example. In a car engine the burning of fuel gives some work, and the remaining heat is released into the atmosphere. This heat is no longer available for work, so the energy has been degraded and the entropy has increased.

2026 Annual I, Set D, Q.2 (vi) OR1+2 = 3 marks

State first law of thermodynamics. Discuss its significance.

Solution

Statement (1 mark). When heat Q is added to a system, it increases the internal energy \Delta U of the system and also does work W by the system on the environment:

Q=\Delta U+W

Significance (2 marks).

It is the law of conservation of energy for thermodynamic processes: the heat supplied is shared between the increase in internal energy and the work done by the system, and no energy is created or lost.
It shows that the change in internal energy depends only on the initial and final states, not on the path. That is why the change \Delta U is used.
It gives the result for each special process: isothermal (\Delta U=0) Q=W; adiabatic (Q=0) W=-\Delta U; isochoric (W=0) Q=\Delta U; isobaric Q=\Delta U+P\Delta V.
It explains everyday cases: the work of pumping air raises its internal energy, and in human metabolism our internal energy falls when we do work, so we need food.

2025 Annual I, Set D, Q.2 (vii) OR3 marks

Differentiate between reversible and irreversible processes. Also give examples.

Solution

Point	Reversible process	Irreversible process
Meaning	Can be retraced in exactly the reverse order.	Cannot be retraced in exactly the reverse order.
Effect on the surroundings	No change is caused in the surroundings.	A change is caused in the surroundings.
Heat and work	Heat absorbed in the direct process is evolved in the reverse process; work done by the system is done on it in the reverse.	Energy is dissipated, for example as heat by friction; entropy increases.
Examples	Slow melting and freezing of ice, boiling and condensing, slow liquefaction and evaporation.	Friction or dissipation of energy, decay of animals and plants, conduction, convection and radiation.

A really reversible process does not exist in nature, but some changes may be taken as reversible if they are done slowly.

2025 Annual I, Set D, Q.2 (xi)3 marks

State three statements of second law of thermodynamics. (one in terms of entropy)

Solution

The second law of thermodynamics can be stated in three ways. Two are the classical statements and the third is in terms of entropy.

Kelvin's statement: it is impossible for a heat engine to convert heat taken from a single reservoir entirely into work without leaving any change in the working system.
Clausius's statement: it is impossible to cause heat to flow from a cold body to a hot body without the expenditure of work.
In terms of entropy: if a system undergoes a natural process, it proceeds in the direction that causes the entropy of the system plus the environment to increase. For all reversible processes the entropy stays constant, and for an irreversible process it increases, so the entropy of the universe always increases.

2025 Annual I, Set D, Q.2 (xii)3 marks

Calculate entropy change when 800\,\mathrm{g} of ice melts at 0\,^{\circ}\mathrm{C}, latent heat of fusion for ice is 3.36\times10^{5}\,\mathrm{J/kg}. Also discuss sign of entropy.

Solution

Given

m = 800\,\mathrm{g}=0.8\,\mathrm{kg}
L_f = 3.36\times10^{5}\,\mathrm{J\,kg^{-1}} latent heat of fusion of ice
T = 0\,^\circ\mathrm{C}=273\,\mathrm{K} melting temperature

To find

\Delta S and its sign

Heat absorbed (1 mark). The ice melts at constant temperature by absorbing the latent heat.
Q=mL_f=0.8\times3.36\times10^{5}=2.688\times10^{5}\,\mathrm{J}
Entropy change (1 mark). The temperature stays constant at 273\,\mathrm{K}.
\Delta S=\dfrac{Q}{T}=\dfrac{2.688\times10^{5}}{273}
Answer with unit.
\Delta S=9.85\times10^{2}\,\mathrm{J\,K^{-1}}

Sign (1 mark). The entropy change is positive because heat is added to the ice. The molecules are ordered in ice but move randomly in water, so melting increases the disorder and the entropy.

\Delta S\approx+9.85\times10^{2}\,\mathrm{J\,K^{-1}}

2026 Annual I, Set D, Q.51+1+4 = 6 marks

State working principle of Carnot's Engine. With the help of graph discuss its all four processes in detail.

Solution

Working principle (1 mark)

A Carnot engine is a theoretical (ideal) heat engine working on the reversible Carnot cycle between a hot reservoir at T_1 and a cold reservoir at T_2, using an ideal gas as the working substance. Its working principle is the isothermal and adiabatic processes: the cycle consists of isothermal expansion, adiabatic expansion, isothermal compression and adiabatic compression. It is the most efficient engine between the two temperatures.

P-V graph (1 mark)

P-V diagram of the Carnot cycle

The four processes (4 marks)

**Isothermal expansion a\to b:** the gas at T_1 absorbs heat Q_1 from the hot reservoir. Its pressure falls from P_1 to P_2 and its volume rises from V_1 to V_2, so it does work at constant temperature.
**Adiabatic expansion b\to c:** the cylinder is insulated, so Q=0. The gas expands to V_3, does work and its temperature falls from T_1 to T_2 while the pressure falls to P_3.
**Isothermal compression c\to d:** the gas at T_2 is compressed and rejects heat Q_2 to the cold reservoir. The pressure rises to P_4 and the volume falls to V_4. Work is done on the gas.
**Adiabatic compression d\to a:** the gas is compressed in an insulated cylinder, Q=0, and its temperature rises from T_2 to T_1. It returns to P_1, V_1.

Step	Curve	Process	Heat	Temperature
1	ab	Isothermal expansion	Q_1 absorbed	T_1 constant
2	bc	Adiabatic expansion	None	Falls from T_1 to T_2
3	cd	Isothermal compression	Q_2 rejected	T_2 constant
4	da	Adiabatic compression	None	Rises from T_2 to T_1

The working substance returns to its initial state, so \Delta U=0 for the cycle. The net work is the area enclosed by abcda, and by the first law it equals W=Q_1-Q_2. Thermal and mechanical equilibrium are kept at every stage so that the cycle is reversible.

Note. The printed marks (read as 1+1+4) are taken as: working principle, graph, discussion of the four processes.

Exam focus

Explain the working of a refrigerator and its coefficient of performance, \text{CP}=\dfrac{Q_2}{W}=\dfrac{T_2}{T_1-T_2}. (long)
2026 Set B Q.5 (a) (3 marks) and 2025 Set B Q.4 (6 marks, 4+2). Same as comprehensive question 9 (book 8.9).
Go to answer
Entropy and the degradation of energy, with an example. (long)
2026 Set B Q.5 (b) and 2025 Set D Q.2 (i) OR; same as comprehensive question 11 (book 8.11). Use the car engine example.
Go to answer
Carnot's engine: working principle, P-V graph and the four processes. (long)
2026 Set D Q.5 (6 marks), 2025 Set D MCQ 10 (four processes in a cycle). Part of comprehensive question 7 (book 8.7), which also asks the efficiency 1-T_2/T_1.
Go to answer
State the first law of thermodynamics and explain its significance and internal energy. (short)
2026 Set B Q.2 (xi), 2026 Set D Q.2 (vi) OR and the 2026 Set D MCQ 9 numerical use of Q=\Delta U+W. Same as comprehensive question 3 (book 8.3).
Go to answer
Three statements of the second law of thermodynamics, one in terms of entropy. (short)
2025 Set B Q.2 (xiii) OR and 2025 Set D Q.2 (xi), the same question in both sets. Comprehensive question 6 (book 8.6).
Go to answer
Entropy change numericals: \Delta S=\pm Q/T and its sign (melting ice, heat removed at constant temperature). (numerical)
2025 Set B Q.2 (ii) OR (7300\,\mathrm{J} removed at 200\,^\circ\mathrm{C}) and 2025 Set D Q.2 (xii) (800\,\mathrm{g} of ice). Close to numerical 8 (book 8.8) and Example 8.6.
Go to answer
Reversible, irreversible and cyclic processes. (short)
2026 Set B Q.2 (vi) OR (three definitions) and 2025 Set D Q.2 (vii) OR (differences with examples).
Go to answer
Boyle's law with its P-V graph, and the other gas laws. (short)
2026 Set B Q.2 (xii) (2+1 marks, relation and graph). Part of comprehensive question 2 (book 8.2).
Go to answer
Carnot efficiency MCQs: doubling the source temperature. (mcq)
2026 Set B MCQ 8, the same as book exercise MCQ 11. Compute \eta=1-T_2/T_1 before and after.
Go to answer

8.1Thermal equilibrium

Thermal equilibrium

System, surroundings and boundary

Types of systems

8.1.1Internal energy

How the internal energy of a system is increased

Internal energy is a function of state

Sign conventions

8.1.2Ideal gas equation and its applications

Derivation Ideal gas equation in terms of the number of molecules

Value and dimensions of Boltzmann's constant

8.2Kinetic theory of gases

Assumptions of the kinetic theory of gases

Condition for a real gas to behave as an ideal gas

Derivation Density of a gas from the gas equation

8.2.1Work in thermodynamicsOften examined

Pressure-volume work

Derivation Work done by a gas, W=PΔV

Graphical representation

8.2.2Difference between work done by the gas and on the gas

Positive work: work done by the gas

Negative work: work done on the gas

Zero work

8.3The gas laws

8.3.1Boyle's law

8.3.2Charles's law

8.3.3Gay-Lussac's law

8.4First law of thermodynamicsOften examined

Examples of the first law

8.4.1Isothermal processOften examined

Examples of isothermal processes

8.4.2Adiabatic processOften examined

Examples of adiabatic processes

8.4.3Isochoric processOften examined

Examples of isochoric processes

8.4.4Isobaric processOften examined

Derivation First law for an isobaric process

Examples of isobaric processes

The four processes at a glance

8.5Heat engineOften examined

Thermodynamics and the auto-engine

8.6Reversible and irreversible processes

Reversible process

Irreversible process

Cyclic process

8.7Second law of thermodynamicsOften examined

Lord Kelvin's statement

Rudolf Clausius's statement

8.8Carnot's engineOften examined

Step 1: isothermal expansion (ab)

Step 2: adiabatic expansion (bc)

Step 3: isothermal compression (cd)

Step 4: adiabatic compression (da)

8.8.1Efficiency of Carnot engineOften examined

Derivation Efficiency of a Carnot engine

8.8.2Carnot theorem

8.9RefrigeratorOften examined

Coefficient of performance of a refrigerator

Derivation Coefficient of performance in terms of heat and temperature

8.10EntropyOften examined

Examples of entropy changes

8.10.1Sign convention for entropy

Derivation Entropy change when heat flows from a hot body to a cold body

Arrow of time

8.10.2Entropy and energyOften examined

Formula sheet

Key terms

Multiple choice questionsp. 150–151

Short questionsp. 152

Comprehensive questionsp. 152–153

Internal energy

Relation with absolute temperature

Derivation Internal energy of an ideal gas

Boyle's law (constant temperature)

Charles's law (constant pressure)

Gay-Lussac's law (constant volume)

Summary and the ideal gas equation

Statement

Sign conventions

Work done by a gas

Derivation Work done by a gas, $W = P Δ V$