Statistical mechanics and thermodynamics: Physics unit 16 notes (FBISE class 12)

16Statistical mechanics and the kinetic model of a gas

A gas in a container is made of an enormous number of molecules: about $2.7 \times 1 0^{25}$ in every cubic metre of air at $0^{\circ} C$ and normal pressure. Each molecule moves randomly, collides with other molecules and bounces off the walls. Nobody can follow each molecule. Instead we describe the gas with a few quantities that we can measure: pressure $P$ , volume $V$ and temperature $T$ .

Macroscopic properties: Properties of the gas as a whole that we measure with instruments, such as pressure, volume, temperature and entropy.
Microscopic properties: Properties of the individual molecules, such as their mass, speed, momentum and kinetic energy.
Statistical mechanics: The branch of physics that applies statistical methods (averages over a very large number of particles) to explain the behaviour of physical systems. It connects the microscopic details of a system with the macroscopic quantities we measure.

Thermodynamics describes energy, entropy and temperature using measurements only. Statistical mechanics gives thermodynamics a foundation by explaining these quantities from the motion of particles. In this unit it relates the volume, pressure and temperature of a gas to the speed and kinetic energy of its molecules. The same methods are used in many fields, from materials science to biological systems.

Assumptions of the kinetic model of an ideal gas

The derivations in section 16.1 use a simple model of an ideal gas. Its assumptions (postulates) are:

A gas consists of a very large number $N$ of identical molecules, each of mass $m$ .
The size of a molecule is negligible compared with the distance between molecules and the size of the container, so molecules behave like points.
The molecules move randomly in all directions with a range of speeds. No direction is preferred.
There are no forces between the molecules except during collisions (no attraction or repulsion).
All collisions, between molecules and with the walls, are perfectly elastic: kinetic energy and momentum are conserved.
The time of a collision is negligible compared with the time between collisions.
The molecules obey Newton's laws of motion, and gravity on them is negligible.

16.1Pressure exerted by gas moleculesOften examined

The molecules of a gas are in continuous random motion. They collide with one another and with the walls of the container. When a molecule hits a wall and bounces back, its momentum changes. A change of momentum needs a force (Newton's second law): the wall pushes the molecule back. By Newton's third law the molecule pushes the wall outwards with an equal force. An enormous number of these tiny pushes (for air at room conditions, about $3 \times 1 0^{23}$ every second on each square centimetre of wall) add up to a steady outward force on every wall.

Pressure of a gas: The force exerted by the gas molecules per unit area of the walls of the container, as they collide with the walls. SI unit: pascal, $1 Pa = 1 N m^{- 2}$ .

P = \frac{F}{A} = (\frac{Δ p}{Δ t}) \frac{1}{A}

$F$ is the force on a wall of area $A$ , and $Δ p /Δ t$ is the rate at which the molecules give momentum to that wall.

So the pressure depends on two things: how many molecules hit each wall per second, and how much momentum each molecule delivers (how hard it hits).

Top view along the

x

-axis. The molecule hits face-1 with momentum

+ m v_{x 1}

and bounces back with

- m v_{x 1}

. It must travel

2 l

(to face-2 and back) before it hits face-1 again.

Derivation Pressure of an ideal gas: $P = \frac{1}{3} \frac{N m}{V} ⟨ v^{2} ⟩$

Aim. Express the pressure $P$ of an ideal gas in terms of the number of molecules $N$ , the mass $m$ of one molecule, the volume $V$ and the mean square speed $⟨ v^{2} ⟩$ .

Assumptions.

The gas has $N$ identical molecules, each of mass $m$ , in a cubical box of side $l$ . Each face has area $A = l^{2}$ and the volume is $V = l^{3}$ .
Collisions with the walls are perfectly elastic.
Collisions between molecules are ignored on the way across the box. (They only swap momentum between molecules, so they do not change the total.)
The motion is random, and gravity on the molecules is negligible.

A cubical box of side

l

holds

N

molecules. One molecule of mass

m

moves with velocity

v

; its

x

-component

v_{x 1}

carries it towards face-1. Face-2 is the opposite face, a distance

l

away (book Fig. 16.1).

One molecule, one collision. Take molecule 1 moving towards face-1 with $x$ -component of velocity $v_{x 1}$ . Its momentum along $x$ is $m v_{x 1}$ . The collision is perfectly elastic, so it bounces back with velocity $- v_{x 1}$ and momentum $- m v_{x 1}$ . The change in its momentum is final minus initial:
$Δ p = (- m v_{x 1}) - (m v_{x 1}) = - 2 m v_{x 1} (16.1)$
Only the $x$ -component reverses. The $y$ - and $z$ -components do not change, so they exert no force on face-1.
By the law of conservation of momentum, the wall gains the momentum the molecule loses. So each collision gives face-1 a momentum of $+ 2 m v_{x 1}$ .
Time between collisions with face-1. After bouncing, the molecule travels a distance $l$ to face-2, bounces there, and travels $l$ back. Its $x$ -speed stays $v_{x 1}$ , so the time for this round trip of $2 l$ is
$Δ t = \frac{2 l}{v _{x 1}} (16.2)$
Force due to one molecule. By Newton's second law, force is the rate of change of momentum. For the molecule:
$\frac{Δ p}{Δ t} = \frac{- 2 m v _{x 1}}{2 l / v _{x 1}} = - \frac{m v _{x 1}^{2}}{l} (16.3)$
This is the force of the wall on the molecule. It is negative because it points away from face-1.
By Newton's third law, the molecule pushes face-1 with an equal and opposite force:
$F_{x 1} = - (- \frac{m v _{x 1}^{2}}{l}) = \frac{m v _{x 1}^{2}}{l}$
**Force due to all $N$ molecules.** Every molecule does the same with its own $x$ -velocity $v_{x 2}, v_{x 3}, \dots, v_{x N}$ . The total force on face-1 is the sum:
$F_{x} = \frac{m v _{x 1}^{2}}{l} + \frac{m v _{x 2}^{2}}{l} + \dots + \frac{m v _{x N}^{2}}{l} = \frac{m}{l} (v_{x 1}^{2} + v_{x 2}^{2} + \dots + v_{x N}^{2})$
Multiply and divide by $N$ . The bracket divided by $N$ is the mean of the squares of the $x$ -velocities, written $⟨ v_{x}^{2} ⟩$ . Note that $m N$ is the total mass of the gas.
$F_{x} = \frac{m N}{l} (\frac{v _{x 1}^{2} + v _{x 2}^{2} + \dots + v _{x N}^{2}}{N}) (16.4) = \frac{m N}{l} ⟨ v_{x}^{2} ⟩ (16.5)$
From one dimension to three. For every molecule $v^{2} = v_{x}^{2} + v_{y}^{2} + v_{z}^{2}$ (Pythagoras' theorem in three dimensions). Averaging over all molecules gives the mean square speed:
$⟨ v^{2} ⟩ = ⟨ v_{x}^{2} ⟩ + ⟨ v_{y}^{2} ⟩ + ⟨ v_{z}^{2} ⟩ (16.6)$
The motion is random and there is a very large number of molecules, so no direction is special. The three averages are equal, and each is one third of $⟨ v^{2} ⟩$ :
$\Rightarrow \Rightarrow ⟨ v_{x}^{2} ⟩ = ⟨ v_{y}^{2} ⟩ = ⟨ v_{z}^{2} ⟩ ⟨ v^{2} ⟩ = 3 ⟨ v_{x}^{2} ⟩ ⟨ v_{x}^{2} ⟩ = \frac{⟨ v ^{2} ⟩}{3} (16.7)$
Put (16.7) into (16.5). By symmetry the same force acts on every face of the cube, so we drop the subscript $x$ :
$F = \frac{m N}{l} \frac{⟨ v ^{2} ⟩}{3} (16.8)$
Pressure. Pressure is force per unit area, and each face has area $l^{2}$ . By Pascal's law the pressure is the same on every face.
$P = \frac{F}{l ^{2}} = \frac{m N}{l} \frac{⟨ v ^{2} ⟩}{3} \cdot \frac{1}{l ^{2}} = \frac{m N}{3 l ^{3}} ⟨ v^{2} ⟩ (16.9)$
The volume of the cube is $V = l^{3}$ :
$P = \frac{m N}{3 V} ⟨ v^{2} ⟩ (16.10)$

Result

P = \frac{1}{3} \frac{N m}{V} ⟨ v^{2} ⟩ or P V = \frac{1}{3} N m ⟨ v^{2} ⟩

Gas pressure is the momentum carried to the walls by molecules every second, per unit area. It is larger when there are more molecules ( $N$ ), heavier molecules ( $m$ ), faster molecules ( $⟨ v^{2} ⟩$ ), or a smaller volume ( $V$ ). The shape of the box does not matter in the end, only its volume.

Pressure and volume

Decrease the volume of a container at constant temperature and the same molecules are packed into a smaller space. They reach the walls more often, so there are more collisions per second on each square metre and the pressure rises. Increase the volume and the collisions become less frequent, so the pressure falls. In the formula, $P = \frac{N m ⟨ v ^{2} ⟩}{3 V}$ and $⟨ v^{2} ⟩$ is fixed by the temperature, so $P \propto \frac{1}{V}$ . This is Boyle's law explained by molecular motion.

16.1Pressure in terms of average translational kinetic energyOften examined

A molecule moving from one place to another has translational kinetic energy \frac12mv^{2}. (Molecules with two or more atoms can also rotate and vibrate, but only motion from place to place carries momentum to the walls.) The average over all the molecules is the average translational kinetic energy \left\langle\frac12mv^{2}\right\rangle, also written \langle K.E\rangle.

Derivation Pressure and average translational kinetic energy

Aim. Show that P=\dfrac{2}{3}N_0\left\langle\tfrac12mv^{2}\right\rangle, so that P\propto\langle K.E\rangle.

Start from the pressure of an ideal gas.
P=\dfrac{mN}{3V}\langle v^{2}\rangle\qquad(16.10)
Multiply and divide the right-hand side by 2 so that the kinetic energy \frac12mv^{2} appears. The mass m is the same for every molecule, so it can go inside the average.
P=\dfrac{2}{3}\,\dfrac{N}{V}\left\langle\tfrac12mv^{2}\right\rangle\qquad(16.11)
N/V is the number of molecules per unit volume (the number density), written N_0. Its unit is \mathrm{m^{-3}}.
P=\dfrac{2}{3}N_0\left\langle\tfrac12mv^{2}\right\rangle
For a gas with a fixed number of molecules per unit volume, \frac23N_0 is a constant.
P=\text{constant}\times\left\langle\tfrac12mv^{2}\right\rangle\;\Rightarrow\;P\propto\langle K.E\rangle

ResultP=\dfrac{2}{3}N_0\left\langle\tfrac12mv^{2}\right\rangle\qquad\Rightarrow\qquad P\propto\langle K.E\rangle

The pressure exerted by the gas molecules on the walls of the container is directly proportional to the average translational kinetic energy of the molecules. In words, pressure is two thirds of the translational kinetic energy stored in each cubic metre of the gas.

16.1Pressure in terms of densityOften examined

The density \rho of a gas is its mass per unit volume. The total mass of N molecules is Nm, so \rho=\dfrac{Nm}{V}. SI unit: \mathrm{kg\,m^{-3}}.

Derivation Pressure in terms of density

Aim. Write the pressure of a gas in terms of its density \rho and the mean square speed \langle v^{2}\rangle.

Start from equation (16.10) and separate the factor \frac{mN}{V}.
P=\dfrac{mN}{3V}\langle v^{2}\rangle=\dfrac{1}{3}\left(\dfrac{mN}{V}\right)\langle v^{2}\rangle
mN is the total mass of the gas, so mN/V is its density.
\dfrac{mN}{V}=\rho
Substitute.
P=\dfrac{1}{3}\rho\langle v^{2}\rangle\qquad(16.12)

ResultP=\dfrac{1}{3}\rho\langle v^{2}\rangle

At constant temperature \langle v^{2}\rangle does not change, so the pressure of a gas is directly proportional to its density, P\propto\rho. Squeeze a gas into half its volume: its density doubles and its pressure doubles (Boyle's law). The formula also lets us find molecular speeds from two easily measured quantities, P and \rho.

Example 16.1: pressure of air

The density of air is \rho=1.296\,\mathrm{kg\,m^{-3}} at a temperature of 0\,^\circ\mathrm{C}. The root mean square velocity of air molecules is 484\,\mathrm{m\,s^{-1}}. Determine the pressure of air.

Given

\rho = 1.296\,\mathrm{kg\,m^{-3}} density of air
v_{\mathrm{rms}} = 484\,\mathrm{m\,s^{-1}} rms speed
T = 0\,^\circ\mathrm{C}=273\,\mathrm{K} not needed in the formula

To find

Pressure P

The rms speed is the square root of the mean square speed, so \langle v^{2}\rangle=v_{\mathrm{rms}}^{2} (section 16.2). Use the pressure-density formula.
P=\dfrac{1}{3}\rho\langle v^{2}\rangle=\dfrac{1}{3}\rho\,v_{\mathrm{rms}}^{2}
Substitute the values in SI units.
P=\dfrac{1}{3}\times1.296\times(484)^{2}=\dfrac{1}{3}\times1.296\times234\,256
Evaluate.
P=1.012\times10^{5}\,\mathrm{Pa}

P\approx1.01\times10^{5}\,\mathrm{Pa}

16.1Temperature in terms of average translational kinetic energyOften examined

The ideal gas law connects the macroscopic quantities P, V and T. Kinetic theory connects P and V with the motion of molecules. Putting the two together tells us what temperature means at the level of molecules.

PV=nRT\qquad(16.13)

n: number of moles; V: volume; T: absolute temperature in kelvin; R=8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}: universal gas constant.

Derivation Ideal gas law in terms of molecules: PV=NkT

Aim. Write the ideal gas law using the number of molecules N instead of the number of moles n.

One mole contains Avogadro's number of molecules, N_A=6.022\times10^{23}\,\mathrm{mol^{-1}}. So N molecules make
n=\dfrac{N}{N_A}
Substitute in PV=nRT and group the constants.
PV=N\left(\dfrac{R}{N_A}\right)T
The gas constant per molecule is the Boltzmann constant k.
k=\dfrac{R}{N_A}=\dfrac{8.314}{6.022\times10^{23}}=1.38\times10^{-23}\,\mathrm{J\,K^{-1}}
Hence
PV=NkT\qquad(16.14)

Result\begin{gathered}PV=NkT\\k=\dfrac{R}{N_A}=1.38\times10^{-23}\,\mathrm{J\,K^{-1}}\end{gathered}

R is the gas constant for one mole; k is the gas constant for one molecule. Use PV=nRT when you know the number of moles and PV=NkT when you know the number of molecules.

Derivation Temperature and average translational kinetic energy

Aim. Show that T=\dfrac{2}{3k}\left\langle\tfrac12mv^{2}\right\rangle, that is, the average translational kinetic energy of a molecule is \tfrac32kT.

Kinetic theory. Multiply equation (16.11) by V.
PV=\dfrac{2N}{3}\left\langle\tfrac12mv^{2}\right\rangle\qquad(16.15)
Ideal gas law. For the same gas,
PV=NkT\qquad(16.14)
Both right-hand sides equal PV, so they are equal to each other.
NkT=\dfrac{2N}{3}\left\langle\tfrac12mv^{2}\right\rangle
Cancel N and make T the subject.
T=\dfrac{2}{3k}\left\langle\tfrac12mv^{2}\right\rangle\qquad(16.16)
\frac{2}{3k} is a constant, so
T=\text{constant}\times\left\langle\tfrac12mv^{2}\right\rangle\;\Rightarrow\;T\propto\langle K.E\rangle
Rearranged, the same result gives the average translational kinetic energy of one molecule:
\left\langle\tfrac12mv^{2}\right\rangle=\dfrac{3}{2}kT

Result\begin{aligned}T&=\dfrac{2}{3k}\left\langle\tfrac12mv^{2}\right\rangle\\\Longleftrightarrow\quad\left\langle\tfrac12mv^{2}\right\rangle&=\dfrac{3}{2}kT\end{aligned}

The absolute temperature of an ideal gas is directly proportional to the average translational kinetic energy of its molecules. Temperature is a measure of how fast the molecules move on average. The average kinetic energy depends only on T, not on the type of gas: at the same temperature, hydrogen and oxygen molecules have the same average kinetic energy.

Average translational kinetic energy against absolute temperature

A straight line through the origin: \langle K.E\rangle\propto T. The slope is \tfrac32k=2.07\times10^{-23}\,\mathrm{J\,K^{-1}} for every gas. At 0\,\mathrm{K} the average kinetic energy is zero, and it cannot go lower.

Assignment 16.1: kinetic energy at 320 K

Calculate the average translational kinetic energy of a gas molecule at a temperature 320\,\mathrm{K}.

Given

T = 320\,\mathrm{K}
k = 1.38\times10^{-23}\,\mathrm{J\,K^{-1}} Boltzmann constant

To find

\langle K.E\rangle

The average translational kinetic energy depends only on the absolute temperature.
\langle K.E\rangle=\dfrac{3}{2}kT
Substitute and evaluate.
\begin{aligned}\langle K.E\rangle&=\dfrac{3}{2}\times1.38\times10^{-23}\times320\\&=6.62\times10^{-21}\,\mathrm{J}\end{aligned}

\langle K.E\rangle=6.62\times10^{-21}\,\mathrm{J}

16.2Root mean square speed of an ideal gasOften examined

According to the kinetic theory of gases, the molecules of a gas do not all move at the same speed. Collisions keep changing each molecule's speed, so at any instant some molecules are slow and some are very fast. To describe them with one number we use an average. Two averages are used.

Mean square speed: \langle v^{2}\rangle: the average of the squares of the speeds of all the molecules. Unit \mathrm{m^{2}\,s^{-2}}.
Root mean square (rms) speed: v_{\mathrm{rms}}: the square root of the mean square speed of the gas molecules. Unit \mathrm{m\,s^{-1}}.
Average speed: v_{\mathrm{av}}: the arithmetic mean of the speeds of the molecules of a gas at a given temperature. Unit \mathrm{m\,s^{-1}}.

\langle v^{2}\rangle=\dfrac{v_1^{2}+v_2^{2}+v_3^{2}+\dots+v_N^{2}}{N}

v_{\mathrm{rms}}=\sqrt{\langle v^{2}\rangle}=\sqrt{\dfrac{v_1^{2}+v_2^{2}+v_3^{2}+\dots+v_N^{2}}{N}}\qquad(16.17)

v_{\mathrm{av}}=\dfrac{v_1+v_2+v_3+\dots+v_N}{N}\qquad(16.18)

v_1,v_2,\dots,v_N are the speeds of the N molecules.

Example 16.3: average speed and rms speed

Four molecules of a gas have speeds of 2\,\mathrm{km\,s^{-1}}, 4\,\mathrm{km\,s^{-1}}, 6\,\mathrm{km\,s^{-1}} and 8\,\mathrm{km\,s^{-1}}. Calculate their average speed and root mean square speed.

Given

v_1 = 2\,\mathrm{km\,s^{-1}}
v_2 = 4\,\mathrm{km\,s^{-1}}
v_3 = 6\,\mathrm{km\,s^{-1}}
v_4 = 8\,\mathrm{km\,s^{-1}}
N = 4

To find

v_{\mathrm{av}} and v_{\mathrm{rms}}

Average speed: add the speeds and divide by the number of molecules.
\begin{aligned}v_{\mathrm{av}}&=\dfrac{v_1+v_2+v_3+v_4}{N}\\&=\dfrac{2+4+6+8}{4}=5\,\mathrm{km\,s^{-1}}\end{aligned}
Root mean square speed: square each speed, take the mean, then the square root.
\begin{aligned}v_{\mathrm{rms}}&=\sqrt{\dfrac{v_1^{2}+v_2^{2}+v_3^{2}+v_4^{2}}{N}}\\&=\sqrt{\dfrac{4+16+36+64}{4}}=\sqrt{30}\end{aligned}
Evaluate.
v_{\mathrm{rms}}=5.48\,\mathrm{km\,s^{-1}}

The rms speed (5.48\,\mathrm{km\,s^{-1}}) is larger than the average speed (5\,\mathrm{km\,s^{-1}}). Squaring gives extra weight to the faster molecules.

v_{\mathrm{av}}=5\,\mathrm{km\,s^{-1}},\qquad v_{\mathrm{rms}}=5.48\,\mathrm{km\,s^{-1}}

Derivation rms speed in terms of temperature: v_{\mathrm{rms}}=\sqrt{3kT/m}=\sqrt{3RT/M}

Aim. Express v_{\mathrm{rms}} in terms of the absolute temperature T and the mass of a molecule m (or the molar mass M).

Start from the pressure of an ideal gas (16.10) and multiply both sides by V.
PV=\dfrac{mN}{3}\langle v^{2}\rangle\qquad(16.19)
The ideal gas law for N molecules is
PV=NkT\qquad(16.20)
Equate the two right-hand sides.
\dfrac{mN}{3}\langle v^{2}\rangle=NkT
Cancel N and make \langle v^{2}\rangle the subject.
\langle v^{2}\rangle=\dfrac{3kT}{m}\qquad(16.21)
Take the square root of both sides. By definition \sqrt{\langle v^{2}\rangle}=v_{\mathrm{rms}}.
v_{\mathrm{rms}}=\sqrt{\dfrac{3kT}{m}}\qquad(16.22)
To use the molar mass, put k=R/N_A. The product mN_A is the mass of one mole, the molar mass M (in \mathrm{kg\,mol^{-1}}).
v_{\mathrm{rms}}=\sqrt{\dfrac{3RT}{mN_A}}=\sqrt{\dfrac{3RT}{M}}\qquad(16.23)

Resultv_{\mathrm{rms}}=\sqrt{\dfrac{3kT}{m}}=\sqrt{\dfrac{3RT}{M}}

For a given gas v_{\mathrm{rms}}\propto\sqrt{T}: to double the rms speed, the absolute temperature must be made four times larger. At a given temperature v_{\mathrm{rms}}\propto1/\sqrt{M}: a gas of smaller molar mass moves faster. Hydrogen (M=2) is \sqrt{32/2}=4 times faster than oxygen (M=32) at the same temperature.

Derivation rms speed in terms of pressure and density: v_{\mathrm{rms}}=\sqrt{3P/\rho}

Aim. Express v_{\mathrm{rms}} using the pressure P and density \rho of the gas.

Start from the pressure-density relation (16.12).
P=\dfrac{1}{3}\rho\langle v^{2}\rangle
Make \langle v^{2}\rangle the subject.
\langle v^{2}\rangle=\dfrac{3P}{\rho}
Take the square root; the left side is v_{\mathrm{rms}}.
v_{\mathrm{rms}}=\sqrt{\dfrac{3P}{\rho}}\qquad(16.24)
Combine with (16.22) and (16.23).
v_{\mathrm{rms}}=\sqrt{\dfrac{3kT}{m}}=\sqrt{\dfrac{3RT}{M}}=\sqrt{\dfrac{3P}{\rho}}\qquad(16.25)

Resultv_{\mathrm{rms}}=\sqrt{\dfrac{3P}{\rho}}

At constant temperature the ratio P/\rho stays constant (doubling the pressure doubles the density), so changing the pressure alone does not change v_{\mathrm{rms}}. Only the temperature and the kind of gas decide the rms speed.

Root mean square speed (km/s) against temperature (K)

v_{\mathrm{rms}}=\sqrt{3RT/M} rises as \sqrt T (the curves bend over). At any temperature the lightest gas is fastest: hydrogen is \sqrt{32/2}=4 times faster than oxygen.

Derivation Ratio of rms speeds of two gases

Aim. Show that at the same temperature \dfrac{(v_{\mathrm{rms}})_1}{(v_{\mathrm{rms}})_2}=\sqrt{\dfrac{m_2}{m_1}}.

Write the rms speed of each gas at the same temperature T.
(v_{\mathrm{rms}})_1=\sqrt{\dfrac{3kT}{m_1}},\qquad(v_{\mathrm{rms}})_2=\sqrt{\dfrac{3kT}{m_2}}
Divide the first by the second. 3kT is the same for both and cancels.
\dfrac{(v_{\mathrm{rms}})_1}{(v_{\mathrm{rms}})_2}=\sqrt{\dfrac{3kT/m_1}{3kT/m_2}}=\sqrt{\dfrac{m_2}{m_1}}
Since m=M/N_A, the same ratio holds for the molar masses.
\dfrac{(v_{\mathrm{rms}})_1}{(v_{\mathrm{rms}})_2}=\sqrt{\dfrac{M_2}{M_1}}

Result\dfrac{(v_{\mathrm{rms}})_1}{(v_{\mathrm{rms}})_2}=\sqrt{\dfrac{m_2}{m_1}}=\sqrt{\dfrac{M_2}{M_1}}

The ratio of rms speeds equals the square root of the inverse ratio of the masses. For hydrogen and oxygen: \sqrt{32/2}=4, so v_{\mathrm{H_2}}:v_{\mathrm{O_2}}=4:1. This is why light gases diffuse faster (Graham's law).

Example 16.2: rms speed of oxygen at STP

Calculate the root mean square speed of oxygen molecules in the air at standard temperature and pressure.

Given

T = 0\,^\circ\mathrm{C}=273\,\mathrm{K}
P = 1.01\times10^{5}\,\mathrm{Pa}
M = 32\,\mathrm{g\,mol^{-1}}=32\times10^{-3}\,\mathrm{kg\,mol^{-1}} molar mass of \mathrm{O_2}
k = 1.38\times10^{-23}\,\mathrm{J\,K^{-1}}

To find

v_{\mathrm{rms}} of oxygen molecules

Find the mass of one oxygen molecule: the molar mass shared among Avogadro's number of molecules.
m=\dfrac{M}{N_A}=\dfrac{32\times10^{-3}}{6.022\times10^{23}}=5.31\times10^{-26}\,\mathrm{kg}
Use the rms speed formula in terms of k and m.
v_{\mathrm{rms}}=\sqrt{\dfrac{3kT}{m}}=\sqrt{\dfrac{3\times1.38\times10^{-23}\times273}{5.31\times10^{-26}}}
Evaluate.
v_{\mathrm{rms}}=\sqrt{2.13\times10^{5}}=461\,\mathrm{m\,s^{-1}}
Check with the molar form, which avoids finding m.
v_{\mathrm{rms}}=\sqrt{\dfrac{3RT}{M}}=\sqrt{\dfrac{3\times8.314\times273}{0.032}}=461\,\mathrm{m\,s^{-1}}

v_{\mathrm{rms}}\approx461\,\mathrm{m\,s^{-1}}

Assignment 16.2: rms speed of nitrogen at STP

Calculate the root mean square speed of nitrogen gas molecules in air under standard conditions of pressure and temperature.

Given

T = 273\,\mathrm{K}
M = 28\,\mathrm{g\,mol^{-1}}=0.028\,\mathrm{kg\,mol^{-1}} molar mass of \mathrm{N_2}
R = 8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}

To find

v_{\mathrm{rms}}

Use the molar form of the rms speed.
v_{\mathrm{rms}}=\sqrt{\dfrac{3RT}{M}}
Substitute with M in \mathrm{kg\,mol^{-1}}.
v_{\mathrm{rms}}=\sqrt{\dfrac{3\times8.314\times273}{0.028}}=\sqrt{2.43\times10^{5}}
Evaluate.
v_{\mathrm{rms}}=493\,\mathrm{m\,s^{-1}}

Nitrogen (M=28) is a little lighter than oxygen (M=32), so it is a little faster: 493 against 461\,\mathrm{m\,s^{-1}}. Check: 461\times\sqrt{32/28}=493.

v_{\mathrm{rms}}\approx493\,\mathrm{m\,s^{-1}}

16.3Behaviour of ideal gases and the birth of statistical mechanics

In the ideal gas model the size of the gas molecules is negligible, there are no interactions between them, and their collisions with each other and with the walls are perfectly elastic. A gas that obeys the ideal gas law PV=nRT is called an ideal gas.

The ideal gas law governs the macroscopic properties of a gas: temperature, pressure, volume, quantity of gas and entropy. No real gas is truly ideal, but real gases come close to ideal behaviour at low pressure and high temperature, when the molecules are far apart and move fast.

Kinetic theory gave the temperature of an ideal gas as T=\dfrac{m}{3k}\langle v^{2}\rangle. This one equation says that temperature, a macroscopic property of the whole gas, is fixed by the motion of individual particles, a microscopic quantity. Results like this, which connect microscopic and macroscopic quantities, gave rise to the branch of physics called statistical mechanics.

How statistical mechanics links the two levels of description
Microscopic cause (molecules)	Macroscopic observable (whole gas)	Link
Momentum delivered to the walls by colliding molecules	Pressure	P=\frac13\rho\langle v^{2}\rangle
Average translational kinetic energy of the molecules	Temperature	\left\langle\frac12mv^{2}\right\rangle=\frac32kT
Number of molecules N	Amount of gas	n=N/N_A
Space in which the molecules move	Volume	PV=NkT
Number of ways the molecules can be arranged	Entropy (disorder)	more arrangements, more entropy

16.3Behaviour of non-ideal gases: the Van der Waals equationOften examined

The ideal gas law works well for large volumes, high temperatures and low pressures. It fails to explain the behaviour of gases at high pressure, high density and low temperature, where the molecules move slowly and interact with each other. It also fails where strong gravitational forces act, as inside stars. To account for these deviations from ideal behaviour, the ideal gas law had to be modified. A gas that does not obey PV=nRT is a real or non-ideal gas.

Van der Waals equation

The equation PV=nRT can be used for real gases at high temperatures and low pressures, where intermolecular forces are negligibly small. Van der Waals modified it so that it can be used for real gases over wide ranges of temperature and pressure. He assumed that gas molecules are hard spheres with a definite volume, and that two molecules attract each other, especially at low temperatures when they move slowly. He made two corrections.

(i) Volume correction

The size of the gas molecules is not negligible. Part of the volume of the container is occupied by the molecules themselves, so the space available for their motion is slightly less than the volume V of the container. For one mole, the effective (free) volume is V-b, where b is a constant for each gas, roughly the volume of one mole of the gas molecules.

Volume correction. In an ideal gas the molecules are points, so all of V is free. Real molecules have a size; packed together they would fill a volume b per mole, so only V-b is free for motion (book Fig. 16.2).

(ii) Pressure correction

At low temperature the molecules attract each other. Inside the container, the pulls on a molecule from its neighbours cancel. But a molecule near a wall has neighbours on one side only, so it feels a net inward force (figure below). It strikes the wall with less momentum, so the measured pressure P is less than it would be without attraction. To get the effective pressure of the gas, a correction is added: the effective pressure is \left(P+\dfrac{a}{V^{2}}\right) for one mole.

Pressure correction. A molecule inside the gas is pulled equally in all directions, so the pulls cancel. A molecule about to strike the wall has neighbours on one side only, so it feels a net inward pull and hits the wall with less momentum (book Fig. 16.3).

Derivation The Van der Waals equation

Aim. Correct the ideal gas equation for the finite size of the molecules and for the attraction between them.

Start from the ideal gas equation for one mole of gas.
PV=RT
Volume correction. The molecules can move only in the free volume, so replace V by V-b.
V\;\to\;V-b
Pressure correction. The measured pressure is reduced by attraction, so the effective pressure is larger than P by a/V^{2}.
P\;\to\;P+\dfrac{a}{V^{2}}
Put both corrections into PV=RT to get the Van der Waals equation for 1 mole.
\left(P+\dfrac{a}{V^{2}}\right)(V-b)=RT\qquad(16.26)
For n moles the excluded volume is nb and the pressure correction becomes an^{2}/V^{2}; the right side becomes nRT.
\left(P+\dfrac{an^{2}}{V^{2}}\right)(V-nb)=nRT\qquad(16.27)

Result\left(P+\dfrac{an^{2}}{V^{2}}\right)(V-nb)=nRT

The constant a measures how strongly the molecules attract each other and b measures their size. Both are empirical constants (found by experiment) whose values are different for different gases. In SI units a is in \mathrm{Pa\,m^{6}\,mol^{-2}} and b in \mathrm{m^{3}\,mol^{-1}}; in practice \mathrm{L^{2}\,atm\,mol^{-2}} and \mathrm{L\,mol^{-1}} are common.

Reduction to the ideal gas equation

The Van der Waals equation describes real gases, but it still works for ideal gases. Real gases approach ideal behaviour at high temperature and low pressure. At high temperature or at very low pressure the volume V is very large, so an^{2}/V^{2} is negligible compared with P, and nb is negligible compared with V. Then the equation reduces to PV=nRT.

Ideal and real gases compared
Feature	Ideal gas	Real (non-ideal) gas
Size of molecules	Negligible (points)	Finite; excluded volume nb
Forces between molecules	None	Attractive forces, important at low temperature
Equation	PV=nRT	\left(P+\frac{an^{2}}{V^{2}}\right)(V-nb)=nRT
When it applies	A model; real gases approach it	Always; deviates most from ideal at high P and low T

Example 16.4: pressure from the Van der Waals equation

Calculate the pressure exerted by one mole of a gas at room temperature 300\,\mathrm{K} with fixed volume 0.022\,\mathrm{m^{3}}, using the Van der Waals equation. a=3.59\,\mathrm{L^{2}\,atm\,mol^{-2}} and b=0.0427\,\mathrm{L\,mol^{-1}}.

Given

n = 1\,\mathrm{mol}
T = 300\,\mathrm{K}
V = 0.022\,\mathrm{m^{3}}=0.022\times1000\,\mathrm{L}=22\,\mathrm{L} using 1\,\mathrm{m^{3}}=1000\,\mathrm{L}
a = 3.59\,\mathrm{L^{2}\,atm\,mol^{-2}}
b = 0.0427\,\mathrm{L\,mol^{-1}}
R = 0.0821\,\mathrm{L\,atm\,mol^{-1}\,K^{-1}} gas constant in litre-atmosphere units

To find

Pressure P (in atm)

Make P the subject of the Van der Waals equation. Because a, b and R are in litres and atmospheres, keep V in litres; the answer comes out in atmospheres.
\begin{aligned}&\left(P+\dfrac{an^{2}}{V^{2}}\right)(V-nb)=nRT\\\Rightarrow\;&P=\dfrac{nRT}{V-nb}-\dfrac{an^{2}}{V^{2}}\end{aligned}
Substitute.
\begin{aligned}P&=\dfrac{1\times0.0821\times300}{22-(1)(0.0427)}-\dfrac{3.59\times(1)^{2}}{(22)^{2}}\\&=\dfrac{24.63}{21.9573}-\dfrac{3.59}{484}\end{aligned}
Evaluate each term and subtract.
P=1.1217-0.0074=1.114\,\mathrm{atm}

P\approx1.114\,\mathrm{atm}

Assignment 16.3: temperature of helium

(a) Using the Van der Waals equation, calculate the temperature of 20.0 moles of helium in a 10.0 litre cylinder at 120 atmosphere pressure. Van der Waals constants for helium: a=0.0341\,\mathrm{L^{2}\,atm\,mol^{-2}}; b=0.0237\,\mathrm{L\,mol^{-1}}. (b) Compare this value with the temperature calculated from the ideal gas equation.

Given

n = 20.0\,\mathrm{mol}
V = 10.0\,\mathrm{L}
P = 120\,\mathrm{atm}
a = 0.0341\,\mathrm{L^{2}\,atm\,mol^{-2}}
b = 0.0237\,\mathrm{L\,mol^{-1}}
R = 0.0821\,\mathrm{L\,atm\,mol^{-1}\,K^{-1}}

To find

(a) T from Van der Waals; (b) T from PV=nRT

(a) Make T the subject of the Van der Waals equation.
T=\dfrac{\left(P+\dfrac{an^{2}}{V^{2}}\right)(V-nb)}{nR}
Pressure correction term.
\begin{aligned}\dfrac{an^{2}}{V^{2}}&=\dfrac{0.0341\times(20.0)^{2}}{(10.0)^{2}}=0.136\,\mathrm{atm}\\P+\dfrac{an^{2}}{V^{2}}&=120+0.136=120.136\,\mathrm{atm}\end{aligned}
Volume correction term.
\begin{aligned}nb&=20.0\times0.0237=0.474\,\mathrm{L}\\V-nb&=10.0-0.474=9.526\,\mathrm{L}\end{aligned}
Substitute. Here nR=20.0\times0.0821=1.642\,\mathrm{L\,atm\,K^{-1}}.
T=\dfrac{120.136\times9.526}{1.642}=\dfrac{1144.4}{1.642}=697\,\mathrm{K}
(b) Ideal gas equation.
T=\dfrac{PV}{nR}=\dfrac{120\times10.0}{1.642}=731\,\mathrm{K}

Comparison. The ideal gas equation gives a temperature about 34\,\mathrm{K} (about 5\%) higher than the Van der Waals value. For helium the attraction between atoms is very weak (a is small), so the pressure correction is tiny (0.136\,\mathrm{atm} out of 120\,\mathrm{atm}). But at this high pressure the atoms' own volume (nb=0.474\,\mathrm{L}, almost 5\% of the cylinder) is not negligible, and that is what makes the ideal gas answer wrong.

T_{\text{VdW}}\approx697\,\mathrm{K},\qquad T_{\text{ideal}}\approx731\,\mathrm{K}

16.4Gravitational effects on the ideal gas model

In the ideal gas model at the laboratory scale, gravity is much weaker than the other factors that act on a gas, such as the kinetic energy of the molecules and the pressure of the gas. That is why it makes sense to ignore gravity for an ideal gas.

On a large scale, for example in a star, gravity plays the main role in stability. A star is a huge ball of hot gas held together only by its own gravity. Accounting for gravity requires much more complex models than the ideal gas model to understand the behaviour of the gas molecules.

Hydrostatic equilibrium

In the core of a star, heat produced by nuclear fusion (which gives a high thermal gas pressure), radiation pressure, and degeneracy pressure (a quantum effect, section 16.5.1) together produce a net outward pressure. Gravity pulls all the matter of the star inwards. When the outward pressure balances the inward pull of gravity, the star is in hydrostatic equilibrium. This ensures the stability of the star: it neither collapses nor flies apart.

Hydrostatic equilibrium in a star. The outward pressure from the hot core balances the inward pull of gravity at every depth, so the star neither collapses nor expands (book Fig. 16.4).

Hydrostatic equilibrium: The state of a star in which the net outward pressure (thermal gas pressure, radiation pressure and degeneracy pressure) is balanced by the inward pull of its own gravity, so the star is stable.

16.5Behaviour of matter under extreme physical conditions

In the universe there are many places where matter has evolved into extreme physical conditions. Extreme conditions change the behaviour of matter and lead to unique and unusual states of matter, such as degenerate matter, and to phenomena like Bose-Einstein condensation, superconductivity and superfluidity. These appear at extremely high pressure or at extremely low temperature, in materials of very high density, or in materials under very strong gravitational force.

Under such conditions, atoms can break down into sub-atomic particles, and these can form unusual states of matter. Such matter is often made mostly of one kind of sub-atomic particle. For example, in a neutron star, strong gravity crushes the atoms and electrons combine with protons, so the star is made almost entirely of neutrons (neutron degenerate matter).

Extreme conditions and the states of matter they produce
Extreme condition	Unusual state or effect	Where it is found
Extremely high pressure and density (strong gravity)	Electron degenerate matter	White dwarfs
Even higher pressure and density	Neutron degenerate matter	Neutron stars
Temperature very close to absolute zero	Bose-Einstein condensate	Laboratory experiments with cooled atoms
Extremely low temperature	Superfluidity (zero viscosity)	Liquid helium below about 2.2\,\mathrm{K}
Temperature below a critical value T_c	Superconductivity (zero resistance)	Mercury below 4.2\,\mathrm{K}, lead below 7.2\,\mathrm{K}

16.5.1Degenerate matterOften examined

Extremely high pressure can force particles so close together that quantum mechanical effects, particularly the Pauli exclusion principle, become significant.

Degenerate matter forms when gravity squeezes atoms so tightly, under the extremely high pressure inside a dying star, that they change from their normal state. Normally, an atom is made of a nucleus with electrons around it, and there is a lot of empty space. When the pressure is extremely high, the atoms are crushed together and the electrons are pressed tightly against the nuclei. All the low-energy states fill up, so, according to the Pauli exclusion principle, further electrons are forced into higher and higher energy states. The electrons then resist being compressed any further, against gravity and other forces. This resistance is called degeneracy pressure. It counteracts gravity and other compressive forces.

Each energy level can hold at most 2 electrons (spin up and spin down). In ordinary matter only the low levels are needed. When matter is squeezed to extreme density, the low levels are full, so extra electrons must go into high-energy states. These fast electrons resist compression: this is degeneracy pressure.

Degenerate matter: Matter that is packed so tightly that its electrons (or neutrons) are forced into a tightly packed configuration, filling all the low-energy states, so that quantum mechanical effects dominate over classical mechanics.
Degeneracy pressure: The resistance to further compression shown by particles of degenerate matter. It arises from the Pauli exclusion principle, not from heat, and it balances gravity in white dwarfs and neutron stars.

Degenerate matter is typically formed in the cores of white dwarfs and neutron stars, where gravity compresses the material to incredibly high densities.

What distinguishes degenerate matter from regular matter
Property	Regular matter	Degenerate matter
Packing	Atoms with much empty space around the nucleus	Particles packed as tightly as the Pauli principle allows
Origin of pressure	Thermal motion; for a gas P\propto T	Pauli exclusion principle (degeneracy pressure)
Effect of cooling	Pressure falls when it is cooled	Pressure stays even close to absolute zero
Typical density	Up to about 2\times10^{4}\,\mathrm{kg\,m^{-3}} (the densest metals)	About 10^{9}\,\mathrm{kg\,m^{-3}} (white dwarf) to 10^{17}\,\mathrm{kg\,m^{-3}} (neutron star)
Physics that describes it	Classical mechanics works	Quantum mechanics dominates

16.5.1White dwarfsOften examined

White dwarfs are formed from the remnants of low- to intermediate-mass stars, such as our Sun. When such a star has used up all of its nuclear fusion fuel, it throws off its outer layers, and its core contracts under gravity to form a white dwarf.

In a white dwarf, the high gravitational force is balanced by electron degeneracy pressure, which arises from the Pauli exclusion principle. Gravity pulls the material together while the electrons are forced into high-energy states, and the star reaches an equilibrium state.

How white dwarfs and neutron stars form. In both, gravity is finally balanced by degeneracy pressure (Pauli exclusion principle): of electrons in a white dwarf, of neutrons in a neutron star.

16.5.2Neutron starsOften examined

Neutron stars are formed from the remnants of more massive stars. In a neutron star, matter is compressed even further by gravity, so much that even electron degeneracy pressure cannot stop the collapse. Electrons and protons start to combine to form neutrons:

\mathrm{p}+\mathrm{e^{-}}\;\longrightarrow\;\mathrm{n}+\nu_e

A proton captures an electron and becomes a neutron; a neutrino \nu_e is also emitted.

The core becomes almost pure neutrons. Neutrons also obey the Pauli exclusion principle, so a neutron degeneracy pressure arises. It balances gravity and prevents further compression. This stable state of matter is known as a neutron star.

Size: only about 20\,\mathrm{km} in diameter.
Mass: about 1.5 times the mass of the Sun.
Density: of the order of 10^{17}\,\mathrm{kg\,m^{-3}}, about the density of an atomic nucleus.
Temperature: newly formed "hot neutron stars" can have a surface temperature of several million degrees Celsius. Inside, a newly formed neutron star is at about 10^{11} to 10^{12}\,\mathrm{K}, but the huge number of neutrinos it emits carries away so much energy that within a few years an isolated neutron star cools to about 10^{6}\,\mathrm{K}.

White dwarfs and neutron stars compared
Feature	White dwarf	Neutron star
Formed from	Remnant of a low- to intermediate-mass star	Remnant of a more massive star
How it forms	Fuel used up, outer layers thrown off, core contracts	Core collapses; electrons and protons combine into neutrons
What holds it up	Electron degeneracy pressure	Neutron degeneracy pressure
Made mostly of	Nuclei (such as carbon and oxygen) and degenerate electrons	Neutrons
Size	About the size of the Earth	About 20\,\mathrm{km} across
Density	About 10^{9}\,\mathrm{kg\,m^{-3}}	About 10^{17}\,\mathrm{kg\,m^{-3}}

16.5.3Bose-Einstein condensationOften examined

A Bose-Einstein condensate (BEC) is a state of matter that forms at extremely low temperatures, close to absolute zero. At such temperatures the kinetic energies of the particles decrease greatly. Particles called bosons can then all condense into the same quantum state, the lowest energy state. The whole group behaves like one single particle. This state of matter was predicted by S. N. Bose and Albert Einstein in the 1920s.

Fermions: Particles that obey the Pauli exclusion principle: no two identical fermions can be in the same quantum state. Examples: electrons, protons, neutrons.
Bosons: Particles that do not obey the Pauli exclusion principle: any number of them can occupy the same quantum state. Examples: photons, W bosons, Z bosons, and whole atoms such as helium-4.
Bose-Einstein condensate: A state of matter at temperatures close to absolute zero in which a group of bosons all occupy the same (lowest) quantum state.

Ten fermions must stack up the energy levels two at a time. Ten bosons cooled close to absolute zero can all drop into the lowest state together, forming a Bose-Einstein condensate.

This is different from degenerate matter. In degenerate matter the particles are fermions (for example electrons), and two fermions cannot exist in the same state, so they are forced into higher and higher states. In a Bose-Einstein condensate the particles are bosons, and they all crowd into the lowest state together.

The Bose-Einstein condensation state of matter shows unique properties such as superfluidity and superconductivity.

16.5.4Superfluidity

A notable property of a Bose-Einstein condensate is superfluidity: the condensate shows zero viscosity, which allows it to flow without resistance. Superfluidity is observed when certain fluids are cooled to extremely low temperatures, close to absolute zero.

Superfluidity: The property of a fluid, cooled close to absolute zero, of having zero viscosity so that it flows without any resistance.
Viscosity: The internal friction of a fluid, which resists its flow.
Vortex: A region in a fluid in which the flow revolves around an axis line, which may be straight or curved.

Zero viscosity gives a superfluid special qualities:

It can pass through incredibly small spaces (very fine pores and cracks) at a steady speed, without the aid of any outside force.
Once it is set moving, it keeps flowing, because there is no friction to stop it.
It can form persistent vortices: whirlpools that keep rotating without slowing down.

16.5.5SuperconductivityOften examined

Under certain conditions, Bose-Einstein condensation leads to superconductivity, in which electrical resistance drops to zero and current flows without resistance. This happens when two electrons form a boson-like pair called a Cooper pair. Cooper pairs can occupy the same quantum state and move through the material without scattering.

Below the superconducting transition temperature, the paired electrons form a condensate: a single quantum state occupied on a macroscopic scale, which flows without resistance.

Superconductivity: The property of certain materials to conduct current without energy loss when they are cooled below a critical temperature (transition temperature).
Superconductors: Substances whose resistivity becomes zero at very low specific temperatures.
Critical temperature T_c: The low temperature at and below which the resistivity \rho of a substance becomes zero. It is also called the superconducting transition temperature.
Cooper pair: Two electrons bound together in a superconductor that together behave like a boson.

Resistivity against temperature (K)

An ordinary metal keeps some resistivity even close to 0\,\mathrm{K}. A superconductor's resistivity falls suddenly to exactly zero at the critical temperature T_c (4.2\,\mathrm{K} for mercury) and stays zero below it (book Fig. 16.8).

Discovery and progress

1911: the Dutch physicist Heike Kamerlingh Onnes discovered superconductivity when he observed the sudden disappearance of electrical resistance in mercury at extremely low temperature (4.2\,\mathrm{K}).
1988: new ceramics made of thallium, calcium, barium, copper and oxygen showed superconductivity at critical temperatures up to 125\,\mathrm{K}.
Lanthanum superhydride (\mathrm{LaH_{10}}), at a pressure of about 170 gigapascals, shows superconductivity with a critical temperature of 250\,\mathrm{K}, the highest confirmed critical temperature for a superconducting material so far.

Table 16.1: critical temperatures of some superconductors
Substance	T_c
Mercury (Hg)	4.2\,\mathrm{K}
Aluminum (Al)	1.18\,\mathrm{K}
Tin (Sn)	3.72\,\mathrm{K}
Lead (Pb)	7.2\,\mathrm{K}

Applications of superconductors

Powerful electromagnets (up to about 32\,\mathrm{T}), for example in MRI scanners in hospitals.
Particle accelerators, which use superconducting magnets to steer fast particles.
Magnetic levitation (maglev) trains, which float above the track with no wheel friction.
Small but powerful electric motors.
Fast computer chips.
Potentially more efficient power transmission lines, since a wire with zero resistance wastes no energy as heat (I^{2}R=0).

Researchers continue to search for superconductors that work at higher temperatures, which would make them practical for everyday use.

Formula sheet

Pressure (definition)

P=\dfrac{F}{A}=\left(\dfrac{\Delta p}{\Delta t}\right)\dfrac{1}{A}

F: force on a wall; A: area of the wall; \Delta p/\Delta t: rate of transfer of momentum to the wall

Unit: \mathrm{Pa}=\mathrm{N\,m^{-2}}

Momentum change in one collision with a wall

\Delta p=-2mv_{x}

m: mass of a molecule; v_x: its velocity component towards the wall. The wall receives +2mv_x

Unit: \mathrm{kg\,m\,s^{-1}}

Time between collisions with the same face

\Delta t=\dfrac{2l}{v_{x}}

l: side of the cubical box

Unit: \mathrm{s}

Force of all molecules on one face

F=\dfrac{mN}{l}\langle v_x^{2}\rangle=\dfrac{mN}{l}\,\dfrac{\langle v^{2}\rangle}{3}

N: number of molecules; \langle v_x^{2}\rangle=\frac13\langle v^{2}\rangle for random motion

Unit: \mathrm{N}

Mean square speed

\langle v^{2}\rangle=\dfrac{v_1^{2}+v_2^{2}+\dots+v_N^{2}}{N}=\langle v_x^{2}\rangle+\langle v_y^{2}\rangle+\langle v_z^{2}\rangle

v_1,\dots,v_N: speeds of the N molecules

Unit: \mathrm{m^{2}\,s^{-2}}

Pressure of an ideal gas

P=\dfrac{mN}{3V}\langle v^{2}\rangle\quad\text{or}\quad PV=\dfrac13Nm\langle v^{2}\rangle

V: volume of the gas

Unit: \mathrm{Pa}

Pressure and average translational kinetic energy

P=\dfrac{2}{3}N_0\left\langle\tfrac12mv^{2}\right\rangle

N_0=N/V: number of molecules per unit volume

Unit: \mathrm{Pa}

Pressure and density

P=\dfrac13\rho\langle v^{2}\rangle

\rho=Nm/V: density of the gas

Unit: \mathrm{Pa}

Ideal gas law (moles)

PV=nRT,\qquad n=\dfrac{N}{N_A}

n: number of moles; R=8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}; T: absolute temperature; N_A=6.022\times10^{23}\,\mathrm{mol^{-1}}

Units: P in \mathrm{Pa}, V in \mathrm{m^{3}}, T in \mathrm{K}

Ideal gas law (molecules)

PV=NkT,\qquad k=\dfrac{R}{N_A}

k=1.38\times10^{-23}\,\mathrm{J\,K^{-1}}: Boltzmann constant

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

Temperature and average translational kinetic energy

T=\dfrac{2}{3k}\left\langle\tfrac12mv^{2}\right\rangle\quad\Longleftrightarrow\quad\left\langle\tfrac12mv^{2}\right\rangle=\dfrac32kT

\left\langle\frac12mv^{2}\right\rangle: average translational kinetic energy of one molecule

Units: \mathrm{J} (kinetic energy), \mathrm{K} (temperature)

Translational kinetic energy of one mole

E=\dfrac32RT

R: universal gas constant

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

Root mean square speed (definition)

v_{\mathrm{rms}}=\sqrt{\langle v^{2}\rangle}=\sqrt{\dfrac{v_1^{2}+v_2^{2}+\dots+v_N^{2}}{N}}

square root of the mean square speed

Unit: \mathrm{m\,s^{-1}}

Average speed

v_{\mathrm{av}}=\dfrac{v_1+v_2+\dots+v_N}{N}

arithmetic mean of the speeds; v_{\mathrm{rms}}\ge v_{\mathrm{av}}

Unit: \mathrm{m\,s^{-1}}

Root mean square speed

v_{\mathrm{rms}}=\sqrt{\dfrac{3kT}{m}}=\sqrt{\dfrac{3RT}{M}}=\sqrt{\dfrac{3P}{\rho}}

m: mass of one molecule (kg); M: molar mass (\mathrm{kg\,mol^{-1}}); \rho: density

Unit: \mathrm{m\,s^{-1}}

Mass of one molecule

m=\dfrac{M}{N_A}

M: molar mass in \mathrm{kg\,mol^{-1}} (for oxygen 0.032)

Unit: \mathrm{kg}

Ratios of rms speeds

\dfrac{(v_{\mathrm{rms}})_1}{(v_{\mathrm{rms}})_2}=\sqrt{\dfrac{M_2}{M_1}}\ \text{(same }T\text{)},\qquad\dfrac{v_2}{v_1}=\sqrt{\dfrac{T_2}{T_1}}\ \text{(same gas)}

M_1,M_2: molar masses; T_1,T_2: absolute temperatures

no unit (ratio)

Van der Waals equation (1 mole)

\left(P+\dfrac{a}{V^{2}}\right)(V-b)=RT

a: attraction constant; b: excluded volume per mole

Units: a in \mathrm{Pa\,m^{6}\,mol^{-2}} or \mathrm{L^{2}\,atm\,mol^{-2}}; b in \mathrm{m^{3}\,mol^{-1}} or \mathrm{L\,mol^{-1}}

Van der Waals equation (n moles)

\left(P+\dfrac{an^{2}}{V^{2}}\right)(V-nb)=nRT

n: number of moles; reduces to PV=nRT when V is very large

Units: with R=0.0821\,\mathrm{L\,atm\,mol^{-1}\,K^{-1}}, use P in atm and V in L

Neutron formation in a neutron star

\mathrm{p}+\mathrm{e^{-}}\longrightarrow\mathrm{n}+\nu_e

a proton captures an electron and becomes a neutron; \nu_e: neutrino

Unit: none

Key terms

Statistical mechanics: The branch of physics that applies statistical methods to systems of a very large number of particles, connecting microscopic details (motion, energy, interactions of particles) with macroscopic observables (temperature, pressure, volume, entropy).
Thermodynamics: The study of heat, energy, entropy and temperature from measurements. Without statistical mechanics it is a phenomenological theory (built only on observation).
Macroscopic properties: Properties of a system as a whole that can be measured, such as pressure, volume and temperature.
Microscopic properties: Properties of individual particles, such as mass, speed, momentum and kinetic energy.
Ideal gas: A gas that obeys PV=nRT exactly: its molecules have negligible size, do not interact, and collide perfectly elastically. Real gases approach it at low pressure and high temperature.
Pressure of a gas: The force exerted by gas molecules per unit area of the walls as they collide with them, P=F/A. Unit: pascal (\mathrm{N\,m^{-2}}).
Elastic collision: A collision in which both momentum and kinetic energy are conserved.
Mean square speed: \langle v^{2}\rangle: the average of the squares of the speeds of all the molecules of a gas.
Number density: N_0=N/V: the number of molecules per unit volume.
Average translational kinetic energy: \left\langle\frac12mv^{2}\right\rangle: the average kinetic energy of the molecules due to their motion from place to place; equal to \frac32kT.
Absolute zero: 0\,\mathrm{K} (-273.15\,^\circ\mathrm{C}), the temperature at which the average translational kinetic energy of the molecules is zero; no lower temperature is possible.
Boltzmann constant: k=R/N_A=1.38\times10^{-23}\,\mathrm{J\,K^{-1}}: the gas constant per molecule.
Avogadro's number: N_A=6.022\times10^{23}\,\mathrm{mol^{-1}}: the number of molecules (or atoms) in one mole.
Universal gas constant: R=8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}: the gas constant per mole.
Root mean square (rms) speed: v_{\mathrm{rms}}: the square root of the mean square speed of the molecules of a gas, \sqrt{\langle v^{2}\rangle}=\sqrt{3kT/m}.
Average speed: v_{\mathrm{av}}: the arithmetic mean of the speeds of the molecules of a gas at a given temperature.
Molar mass: M: the mass of one mole of a substance, M=mN_A; in \mathrm{kg\,mol^{-1}} in formulas.
Real (non-ideal) gas: A gas that does not obey PV=nRT, especially at high pressure and low temperature, because its molecules have a finite size and attract each other.
Van der Waals equation: \left(P+\frac{an^{2}}{V^{2}}\right)(V-nb)=nRT: the ideal gas equation corrected for the volume of the molecules (b) and the attraction between them (a).
Volume correction: Replacing V by V-nb, because the molecules themselves occupy part of the volume; b is the excluded volume per mole.
Pressure correction: Replacing P by P+an^{2}/V^{2}, because attraction pulls molecules near a wall inwards and reduces the measured pressure.
Hydrostatic equilibrium: The balance in a star between the net outward pressure (thermal, radiation and degeneracy pressure) and the inward pull of gravity, which keeps the star stable.
Pauli exclusion principle: Two electrons (in general, two identical fermions) cannot occupy the same quantum state.
Degenerate matter: Matter packed so tightly, under extremely high pressure, that its electrons (or neutrons) are forced into a tightly packed configuration and quantum effects dominate over classical mechanics.
Degeneracy pressure: The resistance of degenerate matter to further compression, arising from the Pauli exclusion principle; it counteracts gravity in white dwarfs and neutron stars.
White dwarf: The remnant of a low- to intermediate-mass star that has used up its fusion fuel and thrown off its outer layers; gravity is balanced by electron degeneracy pressure.
Neutron star: The remnant of a more massive star in which electrons and protons have combined into neutrons; gravity is balanced by neutron degeneracy pressure. About 20\,\mathrm{km} across, density about 10^{17}\,\mathrm{kg\,m^{-3}}.
Fermions: Particles such as electrons, protons and neutrons that obey the Pauli exclusion principle.
Bosons: Particles such as photons, W and Z bosons (and atoms such as helium-4) that do not obey the Pauli exclusion principle, so any number can share one quantum state.
Bose-Einstein condensate (BEC): A state of matter formed at temperatures close to absolute zero in which a group of bosons occupy the same quantum state; predicted by Bose and Einstein in the 1920s.
Superfluidity: The property of a Bose-Einstein condensate (or a fluid cooled close to absolute zero) of having zero viscosity, so that it flows without resistance.
Viscosity: The internal friction of a fluid, which resists its flow.
Vortex: A region in a fluid in which the flow revolves around an axis line, which may be straight or curved.
Superconductivity: The property of certain materials to conduct current without energy loss (zero resistance) when cooled below a critical temperature.
Superconductors: Substances whose resistivity becomes zero at very low specific temperatures.
Critical temperature: T_c: the temperature at and below which the resistivity of a substance becomes zero; also called the superconducting transition temperature.
Cooper pair: A pair of electrons in a superconductor that behaves like a boson; Cooper pairs share one quantum state and move without scattering.
High-temperature superconductor: A superconductor with a critical temperature above about 77\,\mathrm{K} (liquid nitrogen), for example the thallium-barium-calcium-copper oxide ceramic (125\,\mathrm{K}).

Multiple choice questionsp. 40–41

Encircle the correct option.

1

Which of the following is the correct relation between pressure P and density \rho of the gas, at constant temperature?

AP\propto\dfrac{1}{\rho}
BP\propto\dfrac{1}{\rho^{2}}
CP\propto\rhoAnswer
DP\propto\rho^{2}

Why

P=\frac13\rho\langle v^{2}\rangle, and at constant temperature \langle v^{2}\rangle is constant, so P\propto\rho: doubling the density doubles the pressure. Options A and B would mean a denser gas exerts less pressure, the opposite of Boyle's law.

2

Which of the following is the correct formula for the root mean square speed of a molecule in a gas at the absolute temperature T?

Av_{\mathrm{rms}}=1.5\sqrt{\dfrac{kT}{m}}
Bv_{\mathrm{rms}}=1.5\sqrt{\dfrac{mk}{T}}
Cv_{\mathrm{rms}}=1.73\sqrt{\dfrac{m}{kT}}
Dv_{\mathrm{rms}}=1.73\sqrt{\dfrac{kT}{m}}Answer

Why

v_{\mathrm{rms}}=\sqrt{\dfrac{3kT}{m}}=\sqrt3\,\sqrt{\dfrac{kT}{m}} and \sqrt3=1.73. Option A uses 1.5=\frac32, which belongs to the kinetic energy formula \frac32kT, not to the speed; options B and C have the fraction upside down.

3

Which of the following motions of ideal gas molecules determines the temperature?

ATranslational motionAnswer
BRotational motion
CVibrational motion
DAll types of motion

Why

Kinetic theory gives T=\dfrac{2}{3k}\left\langle\tfrac12mv^{2}\right\rangle: the absolute temperature is fixed by the average translational kinetic energy (motion from place to place). Only translational motion carries molecules to the walls and appears in this relation.

4

Which of the following gases possesses the maximum root mean square velocity at the same temperature?

AOxygen
BCarbon dioxide
CHydrogenAnswer
DHelium

Why

At the same temperature v_{\mathrm{rms}}=\sqrt{3RT/M}\propto1/\sqrt{M}. Hydrogen has the smallest molar mass (2\,\mathrm{g\,mol^{-1}}), compared with helium (4), oxygen (32) and carbon dioxide (44), so its molecules are the fastest.

5

Real gases deviate from ideal behavior because gas molecules:

Aare colorless
Bare spherical
Chave intermolecular forces of attractionAnswer
Dhave high speeds

Why

The ideal gas model assumes no forces between molecules (and negligible size). Real molecules attract each other, which lowers the pressure, especially at low temperature. Colour and shape are irrelevant, and high speeds (high temperature) actually make a gas behave more ideally.

6

The root mean square speed of gas at 27\,^\circ\mathrm{C} is v_{\mathrm{rms}}. If the temperature of the gas is raised to 327\,^\circ\mathrm{C}, then which of the following will be the root mean square speed of gas?

A\dfrac{1}{\sqrt2}\,v_{\mathrm{rms}}
Bv_{\mathrm{rms}}
C\sqrt2\,v_{\mathrm{rms}}Answer
D2\,v_{\mathrm{rms}}

Why

v_{\mathrm{rms}}\propto\sqrt{T} with T in kelvin. 27\,^\circ\mathrm{C}=300\,\mathrm{K} and 327\,^\circ\mathrm{C}=600\,\mathrm{K}, so T doubles and the speed becomes \sqrt{600/300}=\sqrt2 times larger. Option D is the trap for students who forget the square root.

7

Which of the following is the ratio of root mean square speed of oxygen \mathrm{O_2} and hydrogen \mathrm{H_2} at same temperature?

A1 : 2
B2 : 1
C1 : 4Answer
D4 : 1

Why

\dfrac{v_{\mathrm{O_2}}}{v_{\mathrm{H_2}}}=\sqrt{\dfrac{M_{\mathrm{H_2}}}{M_{\mathrm{O_2}}}}=\sqrt{\dfrac{2}{32}}=\dfrac14, so the ratio is 1:4. Option D, 4:1, is hydrogen to oxygen, the reverse of what is asked.

8

Density of a gas is 5\,\mathrm{kg\,m^{-3}} at a pressure of 6\times10^{5}\,\mathrm{Pa}. Which of the following is the root mean square velocity of the gas molecules?

A600\,\mathrm{m\,s^{-1}}Answer
B400\,\mathrm{m\,s^{-1}}
C300\,\mathrm{m\,s^{-1}}
D180\,\mathrm{m\,s^{-1}}

Why

v_{\mathrm{rms}}=\sqrt{\dfrac{3P}{\rho}}=\sqrt{\dfrac{3\times6\times10^{5}}{5}}=\sqrt{3.6\times10^{5}}=600\,\mathrm{m\,s^{-1}}. Forgetting the factor 3 gives about 346\,\mathrm{m\,s^{-1}}, which is not an option.

9

The average kinetic energy of gas molecules is:

Adirectly proportional to pressure of gas
Binversely proportional to volume of gas
Cinversely proportional to the absolute temperature of gas
Ddirectly proportional to the absolute temperature of gasAnswer

Why

\left\langle\tfrac12mv^{2}\right\rangle=\tfrac32kT, so the average kinetic energy is directly proportional to the absolute temperature. Changing P or V alone at constant T does not change it.

10

The pressure is exerted by the gas on the walls of the container because its molecules:

Alose kinetic energy
Bstick to the walls
Con collision with the walls, there is a change in momentum of gas moleculesAnswer
Don collision with the walls, gas molecules exert no force on wall

Why

When a molecule bounces off a wall its momentum changes by 2mv_x. The rate of change of momentum is a force on the wall, and force per unit area is pressure. The collisions are elastic, so the molecules do not lose kinetic energy (A) or stick (B), and they certainly exert a force (D).

11

The rms speed of the molecules of enclosed gas is v_{\mathrm{rms}}. Which of the following will be the rms speed of gas molecule, if pressure is doubled keeping the temperature same?

A\tfrac12\,v_{\mathrm{rms}}
Bv_{\mathrm{rms}}Answer
C2\,v_{\mathrm{rms}}
D4\,v_{\mathrm{rms}}

Why

v_{\mathrm{rms}}=\sqrt{3kT/m} depends only on the temperature and the mass of a molecule. Doubling the pressure at constant temperature also doubles the density, so v_{\mathrm{rms}}=\sqrt{3P/\rho} stays the same.

12

A neutron star has a super high ______.

AdensityAnswer
Boutput of light
Cinput of neutrons
Doutput of x-rays

Why

A neutron star packs about 1.5 times the mass of the Sun into a sphere about 20\,\mathrm{km} across, so its density is about 10^{17}\,\mathrm{kg\,m^{-3}}, the density of an atomic nucleus.

13

Which of the following is the electrical resistance of the superconductor below critical temperature?

AFinite
BLarge
CZeroAnswer
DInfinity

Why

Below its critical temperature T_c the resistivity of a superconductor becomes exactly zero, so a current flows with no energy loss.

14

What is the process by which a Bose-Einstein Condensate is formed?

AFusion
BFission
CDecomposition
DCooling to extremely low temperatureAnswer

Why

A Bose-Einstein condensate forms when bosons are cooled close to absolute zero, so that they all settle into the same lowest quantum state. Fusion and fission are nuclear reactions and decomposition is a chemical change.

Short questionsp. 42

13 marks

Mention the different ways of increasing the number of molecular collisions per unit time in a gas.

Solution

The number of collisions per unit time (between molecules and with the walls) depends on how many molecules there are in each unit volume and how fast they move. It can be increased in three ways:

Raise the temperature. The molecules move faster (v_{\mathrm{rms}}\propto\sqrt{T}), so they reach the walls and meet other molecules more often.
Reduce the volume (compress the gas). The number density N/V increases and each molecule has a shorter distance to travel between collisions.
Add more gas to the same container. More molecules per unit volume means more collisions every second.

In short, the number of collisions on each unit area of wall per second is proportional to (number density) \times (molecular speed), N_0v_{\mathrm{av}}. Anything that increases either factor increases the collision rate. Increasing the pressure at constant temperature also works, because it increases the density.

23 marks

By reducing the volume of a gas at a constant temperature, the pressure of the gas increases. Explain it on the basis of kinetic theory.

Solution

At constant temperature the average kinetic energy of the molecules, \frac32kT, does not change. So each molecule still hits the wall with the same speed and gives it the same momentum, 2mv_x, per collision.

When the volume is reduced, the same number of molecules is packed into a smaller space. The number of molecules per unit volume increases, and each molecule has a shorter distance to travel between the walls, so the time between its collisions, \Delta t=2l/v_x, becomes shorter. The molecules therefore strike each unit area of the walls more often. More momentum is delivered per second to each unit area, so the force per unit area, which is the pressure, increases.

\begin{gathered}P=\dfrac{1}{3}\,\dfrac{N}{V}\,m\langle v^{2}\rangle\\\text{with }N,\ m,\ \langle v^{2}\rangle\text{ constant:}\quad P\propto\dfrac{1}{V}\end{gathered}

Halving the volume doubles the pressure. This is Boyle's law explained by the kinetic theory.

33 marks

What do you mean by the root mean square speed of the molecules of a gas? Is the root mean square speed the same as the average speed?

Solution

The root mean square (rms) speed is the square root of the mean of the squares of the speeds of all the molecules of a gas:

v_{\mathrm{rms}}=\sqrt{\dfrac{v_1^{2}+v_2^{2}+\dots+v_N^{2}}{N}}=\sqrt{\dfrac{3kT}{m}}

No, it is not the same as the average speed. The average speed is the simple arithmetic mean of the speeds, v_{\mathrm{av}}=\dfrac{v_1+v_2+\dots+v_N}{N}. Squaring gives extra weight to the faster molecules, so v_{\mathrm{rms}} is always greater than v_{\mathrm{av}}, unless every molecule has the same speed.

Example: for four molecules with speeds 2, 4, 6 and 8\,\mathrm{km\,s^{-1}}, v_{\mathrm{av}}=5\,\mathrm{km\,s^{-1}} but v_{\mathrm{rms}}=\sqrt{30}=5.48\,\mathrm{km\,s^{-1}}.

43 marks

Why is the temperature below absolute zero not possible?

Solution

According to kinetic theory, the absolute temperature of a gas is directly proportional to the average translational kinetic energy of its molecules:

\left\langle\tfrac12mv^{2}\right\rangle=\dfrac32kT

Kinetic energy \frac12mv^{2} can never be negative, because m is positive and v^{2} cannot be negative. Its smallest possible value is zero, when the molecular motion stops (in this model). That happens at T=0\,\mathrm{K} (-273.15\,^\circ\mathrm{C}), called absolute zero. A temperature below 0\,\mathrm{K} would need a negative kinetic energy, which is impossible.

The gas laws say the same: at constant pressure V\propto T, so the volume of an ideal gas would shrink to zero at 0\,\mathrm{K}, and a lower temperature would need a negative volume. Absolute zero can be approached but never actually reached.

53 marks

Estimate the average kinetic energy of a helium atom at the temperature on the surface of the sun (6000\,\mathrm{K}).

Solution

The average translational kinetic energy of a gas particle depends only on the absolute temperature, not on its mass, so the mass of the helium atom is not needed.

Given

T = 6000\,\mathrm{K}
k = 1.38\times10^{-23}\,\mathrm{J\,K^{-1}}

To find

\langle K.E\rangle

Use the kinetic energy-temperature relation.
\langle K.E\rangle=\dfrac32kT
Substitute and evaluate.
\langle K.E\rangle=\dfrac32\times1.38\times10^{-23}\times6000=1.24\times10^{-19}\,\mathrm{J}

This is about 0.78\,\mathrm{eV} (using 1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J}). Any gas atom at 6000\,\mathrm{K}, helium or not, has this same average kinetic energy.

\langle K.E\rangle\approx1.24\times10^{-19}\,\mathrm{J}

63 marks

Show that the ratio of the root mean square speeds of molecules of two different gases at a certain temperature is equal to the square root of the inverse ratio of their masses.

Solution

The rms speed of the molecules of a gas at absolute temperature T is v_{\mathrm{rms}}=\sqrt{3kT/m}, where m is the mass of one molecule.

Write it for gas 1 (molecular mass m_1) and gas 2 (molecular mass m_2) at the same temperature T.
(v_{\mathrm{rms}})_1=\sqrt{\dfrac{3kT}{m_1}},\qquad(v_{\mathrm{rms}})_2=\sqrt{\dfrac{3kT}{m_2}}
Divide the first equation by the second. 3kT is common and cancels.
\dfrac{(v_{\mathrm{rms}})_1}{(v_{\mathrm{rms}})_2}=\sqrt{\dfrac{3kT/m_1}{3kT/m_2}}=\sqrt{\dfrac{m_2}{m_1}}

So the ratio of the rms speeds equals the square root of the inverse ratio of the masses (and, since m=M/N_A, of the molar masses). Example: for hydrogen and oxygen, \sqrt{32/2}=4, so hydrogen molecules move 4 times faster than oxygen molecules at the same temperature.

73 marks

Differentiate between the formations of white dwarfs and neutron stars.

Solution

Both form when a star has used up its nuclear fuel and gravity makes its core collapse. They differ in the mass of the original star and in the pressure that finally stops the collapse.

Feature	White dwarf	Neutron star
Parent star	Low- to intermediate-mass star (like the Sun)	More massive star
Process	Fusion fuel used up; outer layers thrown off; core contracts	Core collapses so strongly that electrons and protons combine to form neutrons
Collapse stopped by	Electron degeneracy pressure	Neutron degeneracy pressure
Result	About Earth-sized, density about 10^{9}\,\mathrm{kg\,m^{-3}}	About 20\,\mathrm{km} across, density about 10^{17}\,\mathrm{kg\,m^{-3}}

83 marks

Why do the gases at low temperatures and high pressure show large deviations from ideal behaviour?

Solution

The ideal gas equation assumes that (i) the molecules have negligible volume and (ii) there are no forces between them. Both assumptions fail at low temperature and high pressure.

High pressure: the molecules are crowded close together. Their own volume becomes a noticeable fraction of the container, so the free space for their motion is V-nb, not V.
Low temperature: the molecules move slowly and stay near each other long enough for intermolecular attraction to act. A molecule about to hit a wall is pulled back by its neighbours, so it strikes the wall with less momentum and the measured pressure is less than nRT/V.

Because of these two effects the gas no longer obeys PV=nRT. Van der Waals corrected the equation for both: \left(P+\dfrac{an^{2}}{V^{2}}\right)(V-nb)=nRT.

93 marks

What distinguishes degenerate matter from regular matter?

Solution

Degenerate matter is matter packed so tightly, by extremely high pressure, that quantum mechanical effects dominate over classical mechanics. It differs from regular matter in three ways:

Packing: in regular matter each atom has a nucleus with electrons around it and much empty space. In degenerate matter the atoms are crushed and the particles fill all the low-energy states, so further electrons are forced into high-energy states (Pauli exclusion principle).
Source of pressure: the pressure of regular matter comes from thermal motion and falls when it is cooled. The pressure of degenerate matter (degeneracy pressure) comes from the Pauli exclusion principle and remains even near absolute zero.
Density: about 10^{9}\,\mathrm{kg\,m^{-3}} in white dwarfs and 10^{17}\,\mathrm{kg\,m^{-3}} in neutron stars, far more than any regular matter (at most about 2\times10^{4}\,\mathrm{kg\,m^{-3}}).

102026 Annual I, Set B, Q.2 (v)3 marks

Show that the temperature of ideal gas is directly proportional to the average translational kinetic energy of gas molecules.

Solution

From kinetic theory, the pressure of an ideal gas of N molecules in volume V is P=\dfrac23\,\dfrac{N}{V}\left\langle\tfrac12mv^{2}\right\rangle.

Multiply both sides by V.
PV=\dfrac{2N}{3}\left\langle\tfrac12mv^{2}\right\rangle
The ideal gas law for N molecules (k is the Boltzmann constant):
PV=NkT
Equate the two expressions for PV and cancel N.
NkT=\dfrac{2N}{3}\left\langle\tfrac12mv^{2}\right\rangle\;\Rightarrow\;T=\dfrac{2}{3k}\left\langle\tfrac12mv^{2}\right\rangle

Since \dfrac{2}{3k} is a constant, T\propto\left\langle\tfrac12mv^{2}\right\rangle: the absolute temperature of an ideal gas is directly proportional to the average translational kinetic energy of its molecules. Equivalently, \left\langle\tfrac12mv^{2}\right\rangle=\tfrac32kT.

113 marks

What happens to the electrical resistance of a superconductor when it is cooled below its critical temperature?

Solution

As a superconductor is cooled, its resistance first falls gradually, like that of any metal. At the critical temperature T_c the resistance drops suddenly to exactly zero, and it stays zero at all lower temperatures.

Below T_c the electrons join into Cooper pairs, which behave like bosons and form a single condensate. The pairs move through the material without scattering, so no energy is lost as heat (I^{2}R=0). A current started in a superconducting loop can keep flowing for years. For example, mercury becomes a superconductor below 4.2\,\mathrm{K}.

Resistivity against temperature (K)

An ordinary metal keeps some resistivity even close to 0\,\mathrm{K}. A superconductor's resistivity falls suddenly to exactly zero at the critical temperature T_c (4.2\,\mathrm{K} for mercury) and stays zero below it (book Fig. 16.8).

123 marks

What are some potential applications of superconductors in transportation?

Solution

Superconductors carry large currents with no energy loss and make very strong electromagnets. In transportation they can be used in:

Magnetic levitation (maglev) trains: superconducting electromagnets lift the train above the track and push it forward. With no wheel friction, maglev trains can run faster than 500\,\mathrm{km\,h^{-1}}, quietly and smoothly.
Electric motors for ships, aircraft and vehicles: superconducting motors are small and light but very powerful.
Power supply for electric railways and vehicle charging: superconducting cables would carry current with zero resistance, so no energy is wasted as heat.

133 marks

Provide an example of a high-temperature superconductor.

Solution

A high-temperature superconductor is a material whose critical temperature is above about 77\,\mathrm{K}, the boiling point of liquid nitrogen. It can be cooled with cheap liquid nitrogen instead of expensive liquid helium.

Example: the ceramic made of thallium, barium, calcium, copper and oxygen (discovered in 1988), with T_c=125\,\mathrm{K}. Another well-known example is yttrium barium copper oxide, \mathrm{YBa_2Cu_3O_7} (YBCO), with T_c\approx92\,\mathrm{K}.

Under very high pressure (about 170\,\mathrm{GPa}), lanthanum superhydride \mathrm{LaH_{10}} is superconducting up to 250\,\mathrm{K}, the highest confirmed critical temperature so far.

Comprehensive questionsp. 42

18 marks

What is the pressure of gas? How do gas molecules exert pressure on the walls of a container?

Solution

Pressure of a gas

The pressure of a gas is the force exerted by the gas molecules per unit area of the walls of the container as they collide with the walls. It equals the rate at which momentum is transferred to the wall, per unit area:

P=\dfrac{F}{A}=\left(\dfrac{\Delta p}{\Delta t}\right)\dfrac{1}{A}

SI unit: pascal, 1\,\mathrm{Pa}=1\,\mathrm{N\,m^{-2}}.

How gas molecules exert pressure

The molecules of a gas are in continuous random motion, moving in all directions with different speeds. They collide with one another and with the walls of the container, and all the collisions are perfectly elastic. Each time a molecule bounces off a wall its momentum changes. A change of momentum needs a force: the wall pushes the molecule back (Newton's second law), and the molecule pushes the wall outwards with an equal force (Newton's third law).

Top view along the x-axis. The molecule hits face-1 with momentum +mv_{x1} and bounces back with -mv_{x1}. It must travel 2l (to face-2 and back) before it hits face-1 again.

Force due to one molecule

A molecule of mass m moves towards face-1 of a cubical box of side l with velocity component v_x. Its momentum is mv_x; after the elastic collision it is -mv_x.
\Delta p=(-mv_x)-(mv_x)=-2mv_x
By conservation of momentum, the wall receives momentum +2mv_x in each collision.
The molecule travels 2l (to the opposite face and back) before it hits face-1 again.
\Delta t=\dfrac{2l}{v_x}
The force of this one molecule on the wall is the momentum it gives per unit time.
F_1=\dfrac{2mv_x}{2l/v_x}=\dfrac{mv_x^{2}}{l}

Pressure due to all the molecules

Every molecule does the same. For air at room conditions there are about 3\times10^{23} collisions every second on each square centimetre of wall, and they add up to a steady force. Adding the forces of all N molecules, and using \langle v_x^{2}\rangle=\frac13\langle v^{2}\rangle for random motion, gives the force on one face, F=\dfrac{mN}{3l}\langle v^{2}\rangle. Dividing by the area of the face, l^{2}, with V=l^{3}:

P=\dfrac{1}{3}\,\dfrac{Nm}{V}\,\langle v^{2}\rangle=\dfrac13\rho\langle v^{2}\rangle

The full step-by-step derivation is given in comprehensive question 2.

Factors on which gas pressure depends

Number of molecules per unit volume (density): more molecules strike the walls every second, so the pressure is larger (P\propto\rho).
Speed of the molecules (temperature): faster molecules hit harder and more often, so P\propto\langle v^{2}\rangle\propto T at constant volume.
Volume: in a smaller volume the molecules reach the walls more often, so P\propto1/V at constant temperature (Boyle's law).

28 marks

Derive an expression for the pressure of an ideal gas and show that pressure is directly proportional to the average translational kinetic energy of gas molecules.

Solution

Assumptions

The gas has N identical molecules, each of mass m, in a cubical box of side l.
The molecules are points moving randomly; there are no forces between them except in collisions.
All collisions are perfectly elastic, and gravity is negligible.

Derivation of the pressure

Consider one molecule moving towards face-1, find the force it exerts, add the forces of all the molecules, and divide by the area of the face.

Derivation Pressure of an ideal gas: P=\dfrac{1}{3}\dfrac{Nm}{V}\langle v^{2}\rangle

Aim. Express the pressure P of an ideal gas in terms of the number of molecules N, the mass m of one molecule, the volume V and the mean square speed \langle v^{2}\rangle.

Assumptions.

The gas has N identical molecules, each of mass m, in a cubical box of side l. Each face has area A=l^{2} and the volume is V=l^{3}.
Collisions with the walls are perfectly elastic.
Collisions between molecules are ignored on the way across the box. (They only swap momentum between molecules, so they do not change the total.)
The motion is random, and gravity on the molecules is negligible.

A cubical box of side l holds N molecules. One molecule of mass m moves with velocity v; its x-component v_{x1} carries it towards face-1. Face-2 is the opposite face, a distance l away (book Fig. 16.1).

One molecule, one collision. Take molecule 1 moving towards face-1 with x-component of velocity v_{x1}. Its momentum along x is mv_{x1}. The collision is perfectly elastic, so it bounces back with velocity -v_{x1} and momentum -mv_{x1}. The change in its momentum is final minus initial:
\Delta p=(-mv_{x1})-(mv_{x1})=-2mv_{x1}\qquad(16.1)
Only the x-component reverses. The y- and z-components do not change, so they exert no force on face-1.
By the law of conservation of momentum, the wall gains the momentum the molecule loses. So each collision gives face-1 a momentum of +2mv_{x1}.
Time between collisions with face-1. After bouncing, the molecule travels a distance l to face-2, bounces there, and travels l back. Its x-speed stays v_{x1}, so the time for this round trip of 2l is
\Delta t=\dfrac{2l}{v_{x1}}\qquad(16.2)
Force due to one molecule. By Newton's second law, force is the rate of change of momentum. For the molecule:
\dfrac{\Delta p}{\Delta t}=\dfrac{-2mv_{x1}}{2l/v_{x1}}=-\dfrac{mv_{x1}^{2}}{l}\qquad(16.3)
This is the force of the wall on the molecule. It is negative because it points away from face-1.
By Newton's third law, the molecule pushes face-1 with an equal and opposite force:
F_{x1}=-\left(-\dfrac{mv_{x1}^{2}}{l}\right)=\dfrac{mv_{x1}^{2}}{l}
**Force due to all N molecules.** Every molecule does the same with its own x-velocity v_{x2},v_{x3},\dots,v_{xN}. The total force on face-1 is the sum:
\begin{aligned}F_x&=\dfrac{mv_{x1}^{2}}{l}+\dfrac{mv_{x2}^{2}}{l}+\dots+\dfrac{mv_{xN}^{2}}{l}\\&=\dfrac{m}{l}\left(v_{x1}^{2}+v_{x2}^{2}+\dots+v_{xN}^{2}\right)\end{aligned}
Multiply and divide by N. The bracket divided by N is the mean of the squares of the x-velocities, written \langle v_x^{2}\rangle. Note that mN is the total mass of the gas.
\begin{aligned}F_x&=\dfrac{mN}{l}\left(\dfrac{v_{x1}^{2}+v_{x2}^{2}+\dots+v_{xN}^{2}}{N}\right)\quad(16.4)\\&=\dfrac{mN}{l}\langle v_x^{2}\rangle\quad(16.5)\end{aligned}
From one dimension to three. For every molecule v^{2}=v_x^{2}+v_y^{2}+v_z^{2} (Pythagoras' theorem in three dimensions). Averaging over all molecules gives the mean square speed:
\langle v^{2}\rangle=\langle v_x^{2}\rangle+\langle v_y^{2}\rangle+\langle v_z^{2}\rangle\qquad(16.6)
The motion is random and there is a very large number of molecules, so no direction is special. The three averages are equal, and each is one third of \langle v^{2}\rangle:
\begin{aligned}&\langle v_x^{2}\rangle=\langle v_y^{2}\rangle=\langle v_z^{2}\rangle\\\Rightarrow\;&\langle v^{2}\rangle=3\langle v_x^{2}\rangle\\\Rightarrow\;&\langle v_x^{2}\rangle=\dfrac{\langle v^{2}\rangle}{3}\qquad(16.7)\end{aligned}
Put (16.7) into (16.5). By symmetry the same force acts on every face of the cube, so we drop the subscript x:
F=\dfrac{mN}{l}\,\dfrac{\langle v^{2}\rangle}{3}\qquad(16.8)
Pressure. Pressure is force per unit area, and each face has area l^{2}. By Pascal's law the pressure is the same on every face.
P=\dfrac{F}{l^{2}}=\dfrac{mN}{l}\,\dfrac{\langle v^{2}\rangle}{3}\cdot\dfrac{1}{l^{2}}=\dfrac{mN}{3\,l^{3}}\langle v^{2}\rangle\qquad(16.9)
The volume of the cube is V=l^{3}:
P=\dfrac{mN}{3V}\langle v^{2}\rangle\qquad(16.10)

ResultP=\dfrac{1}{3}\,\dfrac{Nm}{V}\,\langle v^{2}\rangle\quad\text{or}\quad PV=\dfrac{1}{3}Nm\langle v^{2}\rangle

Gas pressure is the momentum carried to the walls by molecules every second, per unit area. It is larger when there are more molecules (N), heavier molecules (m), faster molecules (\langle v^{2}\rangle), or a smaller volume (V). The shape of the box does not matter in the end, only its volume.

Pressure and average translational kinetic energy

Derivation Pressure and average translational kinetic energy

Aim. Show that P=\dfrac{2}{3}N_0\left\langle\tfrac12mv^{2}\right\rangle, so that P\propto\langle K.E\rangle.

Start from the pressure of an ideal gas.
P=\dfrac{mN}{3V}\langle v^{2}\rangle\qquad(16.10)
Multiply and divide the right-hand side by 2 so that the kinetic energy \frac12mv^{2} appears. The mass m is the same for every molecule, so it can go inside the average.
P=\dfrac{2}{3}\,\dfrac{N}{V}\left\langle\tfrac12mv^{2}\right\rangle\qquad(16.11)
N/V is the number of molecules per unit volume (the number density), written N_0. Its unit is \mathrm{m^{-3}}.
P=\dfrac{2}{3}N_0\left\langle\tfrac12mv^{2}\right\rangle
For a gas with a fixed number of molecules per unit volume, \frac23N_0 is a constant.
P=\text{constant}\times\left\langle\tfrac12mv^{2}\right\rangle\;\Rightarrow\;P\propto\langle K.E\rangle

ResultP=\dfrac{2}{3}N_0\left\langle\tfrac12mv^{2}\right\rangle\qquad\Rightarrow\qquad P\propto\langle K.E\rangle

The pressure exerted by the gas molecules on the walls of the container is directly proportional to the average translational kinetic energy of the molecules. In words, pressure is two thirds of the translational kinetic energy stored in each cubic metre of the gas.

Conclusion

P=\dfrac23N_0\left\langle\tfrac12mv^{2}\right\rangle, where N_0=N/V is the number of molecules per unit volume. For a given number density, \frac23N_0 is constant, so P\propto\langle K.E\rangle: the pressure exerted by the gas molecules on the walls of the container is directly proportional to the average translational kinetic energy of the molecules. For example, heating a gas in a closed container increases the kinetic energy of its molecules, and its pressure rises in the same proportion.

38 marks

Describe the root mean square speed of the ideal gas molecule and derive its expressions.

Solution

Root mean square speed

The molecules of a gas move with many different speeds, and collisions keep changing them. The root mean square (rms) speed is the square root of the mean of the squares of the speeds of all the molecules. If v_1,v_2,\dots,v_N are the speeds of N molecules:

\begin{gathered}\langle v^{2}\rangle=\dfrac{v_1^{2}+v_2^{2}+\dots+v_N^{2}}{N}\\v_{\mathrm{rms}}=\sqrt{\langle v^{2}\rangle}=\sqrt{\dfrac{v_1^{2}+v_2^{2}+\dots+v_N^{2}}{N}}\end{gathered}

It is different from the average speed v_{\mathrm{av}}=\dfrac{v_1+v_2+\dots+v_N}{N}. Squaring gives more weight to fast molecules, so v_{\mathrm{rms}}>v_{\mathrm{av}} (for speeds 2,4,6,8\,\mathrm{km\,s^{-1}}: v_{\mathrm{av}}=5 and v_{\mathrm{rms}}=5.48\,\mathrm{km\,s^{-1}}). The rms speed is the useful one because the kinetic energy depends on v^{2}: \frac12mv_{\mathrm{rms}}^{2} is the average kinetic energy of a molecule.

Expressions in terms of temperature

Derivation rms speed in terms of temperature: v_{\mathrm{rms}}=\sqrt{3kT/m}=\sqrt{3RT/M}

Aim. Express v_{\mathrm{rms}} in terms of the absolute temperature T and the mass of a molecule m (or the molar mass M).

Start from the pressure of an ideal gas (16.10) and multiply both sides by V.
PV=\dfrac{mN}{3}\langle v^{2}\rangle\qquad(16.19)
The ideal gas law for N molecules is
PV=NkT\qquad(16.20)
Equate the two right-hand sides.
\dfrac{mN}{3}\langle v^{2}\rangle=NkT
Cancel N and make \langle v^{2}\rangle the subject.
\langle v^{2}\rangle=\dfrac{3kT}{m}\qquad(16.21)
Take the square root of both sides. By definition \sqrt{\langle v^{2}\rangle}=v_{\mathrm{rms}}.
v_{\mathrm{rms}}=\sqrt{\dfrac{3kT}{m}}\qquad(16.22)
To use the molar mass, put k=R/N_A. The product mN_A is the mass of one mole, the molar mass M (in \mathrm{kg\,mol^{-1}}).
v_{\mathrm{rms}}=\sqrt{\dfrac{3RT}{mN_A}}=\sqrt{\dfrac{3RT}{M}}\qquad(16.23)

Resultv_{\mathrm{rms}}=\sqrt{\dfrac{3kT}{m}}=\sqrt{\dfrac{3RT}{M}}

For a given gas v_{\mathrm{rms}}\propto\sqrt{T}: to double the rms speed, the absolute temperature must be made four times larger. At a given temperature v_{\mathrm{rms}}\propto1/\sqrt{M}: a gas of smaller molar mass moves faster. Hydrogen (M=2) is \sqrt{32/2}=4 times faster than oxygen (M=32) at the same temperature.

Expression in terms of pressure and density

Derivation rms speed in terms of pressure and density: v_{\mathrm{rms}}=\sqrt{3P/\rho}

Aim. Express v_{\mathrm{rms}} using the pressure P and density \rho of the gas.

Start from the pressure-density relation (16.12).
P=\dfrac{1}{3}\rho\langle v^{2}\rangle
Make \langle v^{2}\rangle the subject.
\langle v^{2}\rangle=\dfrac{3P}{\rho}
Take the square root; the left side is v_{\mathrm{rms}}.
v_{\mathrm{rms}}=\sqrt{\dfrac{3P}{\rho}}\qquad(16.24)
Combine with (16.22) and (16.23).
v_{\mathrm{rms}}=\sqrt{\dfrac{3kT}{m}}=\sqrt{\dfrac{3RT}{M}}=\sqrt{\dfrac{3P}{\rho}}\qquad(16.25)

Resultv_{\mathrm{rms}}=\sqrt{\dfrac{3P}{\rho}}

At constant temperature the ratio P/\rho stays constant (doubling the pressure doubles the density), so changing the pressure alone does not change v_{\mathrm{rms}}. Only the temperature and the kind of gas decide the rms speed.

Conclusions

v_{\mathrm{rms}}\propto\sqrt{T} for a given gas: four times the absolute temperature doubles the rms speed.
v_{\mathrm{rms}}\propto1/\sqrt{M} at a given temperature: hydrogen molecules are 4 times faster than oxygen molecules.
At constant temperature v_{\mathrm{rms}} does not depend on the pressure, because P/\rho stays constant.
Typical value: oxygen at STP has v_{\mathrm{rms}}=\sqrt{3\times8.314\times273/0.032}=461\,\mathrm{m\,s^{-1}}.

Root mean square speed (km/s) against temperature (K)

v_{\mathrm{rms}}=\sqrt{3RT/M} rises as \sqrt T (the curves bend over). At any temperature the lightest gas is fastest: hydrogen is \sqrt{32/2}=4 times faster than oxygen.

42026 Annual I, Set B, Q.2 (v)8 marks

Derive an expression for the average translational kinetic energy of the ideal gas molecule.

Solution

Idea

Kinetic theory gives the pressure of a gas in terms of the motion of its molecules. The ideal gas law gives the pressure in terms of temperature. Comparing the two gives the average translational kinetic energy in terms of temperature. Assumptions: N identical point molecules of mass m in random motion, no intermolecular forces, and perfectly elastic collisions.

Step 1: pressure from kinetic theory

Derivation Step 1: PV from kinetic theory

Aim. Find PV for an ideal gas of N molecules, each of mass m, in a cubical box of side l (volume V=l^{3}).

A molecule with x-velocity v_x bounces elastically off a face: its momentum changes by -2mv_x, so it gives the face +2mv_x. It returns to the same face after travelling 2l.
\Delta p=-2mv_x,\qquad\Delta t=\dfrac{2l}{v_x}
Force of one molecule on the face = momentum given per unit time (Newton's second and third laws).
F_1=\dfrac{2mv_x}{2l/v_x}=\dfrac{mv_x^{2}}{l}
Add the forces of all N molecules and write the sum as N times the mean of the squares.
F=\dfrac{m}{l}\left(v_{x1}^{2}+v_{x2}^{2}+\dots+v_{xN}^{2}\right)=\dfrac{mN}{l}\langle v_x^{2}\rangle
The motion is random, so \langle v_x^{2}\rangle=\langle v_y^{2}\rangle=\langle v_z^{2}\rangle=\frac13\langle v^{2}\rangle.
F=\dfrac{mN}{3l}\langle v^{2}\rangle
Pressure is force per unit area; the area of a face is l^{2} and l^{3}=V.
P=\dfrac{F}{l^{2}}=\dfrac{mN}{3V}\langle v^{2}\rangle
Multiply by V, and multiply and divide by 2 to bring in the kinetic energy.
PV=\dfrac{2N}{3}\left\langle\tfrac12mv^{2}\right\rangle

ResultPV=\dfrac{2}{3}N\left\langle\tfrac12mv^{2}\right\rangle

Kinetic theory says that PV equals two thirds of the total translational kinetic energy of all the molecules.

Step 2: comparison with the ideal gas law

The ideal gas law for N molecules is PV=nRT with n=N/N_A, that is PV=N\left(\dfrac{R}{N_A}\right)T=NkT, where k=R/N_A=1.38\times10^{-23}\,\mathrm{J\,K^{-1}} is the Boltzmann constant.

Derivation Temperature and average translational kinetic energy

Aim. Show that T=\dfrac{2}{3k}\left\langle\tfrac12mv^{2}\right\rangle, that is, the average translational kinetic energy of a molecule is \tfrac32kT.

Kinetic theory. Multiply equation (16.11) by V.
PV=\dfrac{2N}{3}\left\langle\tfrac12mv^{2}\right\rangle\qquad(16.15)
Ideal gas law. For the same gas,
PV=NkT\qquad(16.14)
Both right-hand sides equal PV, so they are equal to each other.
NkT=\dfrac{2N}{3}\left\langle\tfrac12mv^{2}\right\rangle
Cancel N and make T the subject.
T=\dfrac{2}{3k}\left\langle\tfrac12mv^{2}\right\rangle\qquad(16.16)
\frac{2}{3k} is a constant, so
T=\text{constant}\times\left\langle\tfrac12mv^{2}\right\rangle\;\Rightarrow\;T\propto\langle K.E\rangle
Rearranged, the same result gives the average translational kinetic energy of one molecule:
\left\langle\tfrac12mv^{2}\right\rangle=\dfrac{3}{2}kT

Result\begin{aligned}T&=\dfrac{2}{3k}\left\langle\tfrac12mv^{2}\right\rangle\\\Longleftrightarrow\quad\left\langle\tfrac12mv^{2}\right\rangle&=\dfrac{3}{2}kT\end{aligned}

The absolute temperature of an ideal gas is directly proportional to the average translational kinetic energy of its molecules. Temperature is a measure of how fast the molecules move on average. The average kinetic energy depends only on T, not on the type of gas: at the same temperature, hydrogen and oxygen molecules have the same average kinetic energy.

Final expression and conclusions

\left\langle\tfrac12mv^{2}\right\rangle=\dfrac32kT

The average translational kinetic energy of a gas molecule is directly proportional to the absolute temperature.
It does not depend on the mass or the type of the molecule: at the same temperature all gases have the same average translational kinetic energy per molecule.
For one mole the total translational kinetic energy is N_A\cdot\frac32kT=\frac32RT.
Example: at 300\,\mathrm{K}, \left\langle\frac12mv^{2}\right\rangle=\frac32\times1.38\times10^{-23}\times300=6.21\times10^{-21}\,\mathrm{J}.

58 marks

Describe the modification of the ideal gas model to discuss the behaviour of non-ideal gases using the Van der Waals equation.

Solution

The ideal gas model

In the ideal gas model the molecules have negligible size, there are no forces between them, and their collisions are perfectly elastic. Such a gas obeys PV=nRT at all temperatures and pressures. Real gases obey it well only at high temperature and low pressure.

Why real gases deviate

At high pressure, high density and low temperature the molecules are close together and move slowly. Then (i) the volume of the molecules themselves is no longer negligible, and (ii) the attraction between molecules becomes important. The ideal gas law also fails where strong gravity acts, as in stars. Van der Waals modified the ideal gas equation, treating the molecules as hard spheres of definite volume that attract each other, and made two corrections.

(i) Volume correction

Part of the volume of the container is occupied by the molecules, so the space available for their motion is less than V. For one mole the free volume is V-b, where b is roughly the volume of one mole of the molecules; for n moles it is V-nb.

Volume correction. In an ideal gas the molecules are points, so all of V is free. Real molecules have a size; packed together they would fill a volume b per mole, so only V-b is free for motion (book Fig. 16.2).

(ii) Pressure correction

A molecule inside the gas is pulled equally in all directions. A molecule about to hit the wall is pulled back by its neighbours, which are on one side only. It strikes the wall with less momentum, so the measured pressure P is less than the ideal value. The correction is proportional to the square of the number density, so the effective pressure is P+\dfrac{a}{V^{2}} for one mole and P+\dfrac{an^{2}}{V^{2}} for n moles.

Pressure correction. A molecule inside the gas is pulled equally in all directions, so the pulls cancel. A molecule about to strike the wall has neighbours on one side only, so it feels a net inward pull and hits the wall with less momentum (book Fig. 16.3).

Van der Waals equation

Derivation The Van der Waals equation

Aim. Correct the ideal gas equation for the finite size of the molecules and for the attraction between them.

Start from the ideal gas equation for one mole of gas.
PV=RT
Volume correction. The molecules can move only in the free volume, so replace V by V-b.
V\;\to\;V-b
Pressure correction. The measured pressure is reduced by attraction, so the effective pressure is larger than P by a/V^{2}.
P\;\to\;P+\dfrac{a}{V^{2}}
Put both corrections into PV=RT to get the Van der Waals equation for 1 mole.
\left(P+\dfrac{a}{V^{2}}\right)(V-b)=RT\qquad(16.26)
For n moles the excluded volume is nb and the pressure correction becomes an^{2}/V^{2}; the right side becomes nRT.
\left(P+\dfrac{an^{2}}{V^{2}}\right)(V-nb)=nRT\qquad(16.27)

Result\left(P+\dfrac{an^{2}}{V^{2}}\right)(V-nb)=nRT

The constant a measures how strongly the molecules attract each other and b measures their size. Both are empirical constants (found by experiment) whose values are different for different gases. In SI units a is in \mathrm{Pa\,m^{6}\,mol^{-2}} and b in \mathrm{m^{3}\,mol^{-1}}; in practice \mathrm{L^{2}\,atm\,mol^{-2}} and \mathrm{L\,mol^{-1}} are common.

Reduction to the ideal gas equation

At high temperature or very low pressure the volume V is very large. Then an^{2}/V^{2} is negligible compared with P, and nb is negligible compared with V, so the Van der Waals equation reduces to PV=nRT. The equation therefore describes real gases and still works for ideal gases.

This is an example of the SLO idea that the ideal gas model is a base that can be modified: adding the size of molecules and their attraction explains real gases, and adding gravity helps explain the behaviour of gases in stars.

Note. The book prints "Wan der Waals"; the correct name is Van der Waals.

68 marks

Describe the behaviour of matter under extreme physical conditions. What is degenerate matter, and how are white dwarf and neutron stars formed?

Solution

Matter under extreme conditions

Extreme physical conditions, such as extremely high pressure and density (under strong gravity) or extremely low temperature, change the behaviour of matter and produce unusual states of matter. Atoms can break down into sub-atomic particles, and matter may end up made mostly of one kind of particle. Examples:

Degenerate matter in white dwarfs and neutron stars (extremely high pressure).
Bose-Einstein condensates close to absolute zero.
Superfluidity (zero viscosity) and superconductivity (zero resistance) at very low temperatures.

Degenerate matter

At extremely high pressure particles are pushed so close together that the Pauli exclusion principle becomes important: two electrons cannot occupy the same quantum state. When gravity crushes atoms inside a dying star, all the low-energy states fill up and further electrons are forced into high-energy states. The electrons resist further compression. This resistance, called degeneracy pressure, counteracts gravity.

Each energy level can hold at most 2 electrons (spin up and spin down). In ordinary matter only the low levels are needed. When matter is squeezed to extreme density, the low levels are full, so extra electrons must go into high-energy states. These fast electrons resist compression: this is degeneracy pressure.

Degenerate matter is matter packed so tightly that its electrons (or neutrons) are forced into a tightly packed configuration, so quantum effects dominate over classical mechanics. It is found in the cores of white dwarfs and neutron stars.

Formation of white dwarfs

White dwarfs form from the remnants of low- to intermediate-mass stars. When such a star has used up all its fusion fuel, it throws off its outer layers and its core contracts. The contraction stops when electron degeneracy pressure balances gravity, and the star settles into equilibrium as a white dwarf, about the size of the Earth, with density about 10^{9}\,\mathrm{kg\,m^{-3}}.

Formation of neutron stars

Neutron stars form from the remnants of more massive stars. Gravity is so strong that even electron degeneracy pressure cannot stop the collapse. Matter is compressed further until electrons and protons combine to form neutrons:

\mathrm{p}+\mathrm{e^{-}}\;\longrightarrow\;\mathrm{n}+\nu_e

The core becomes almost pure neutrons. Neutron degeneracy pressure, also due to the Pauli exclusion principle, balances gravity and prevents further compression. A neutron star is only about 20\,\mathrm{km} in diameter but about 1.5 times as massive as the Sun, with a density of about 10^{17}\,\mathrm{kg\,m^{-3}}.

How white dwarfs and neutron stars form. In both, gravity is finally balanced by degeneracy pressure (Pauli exclusion principle): of electrons in a white dwarf, of neutrons in a neutron star.

78 marks

Discuss superfluidity and superconductivity on the basis of Bose-Einstein condensation.

Solution

Bose-Einstein condensation

A Bose-Einstein condensate (BEC) is a state of matter formed at temperatures close to absolute zero. The particles' kinetic energies become very small, and bosons (particles that do not obey the Pauli exclusion principle, such as photons or helium-4 atoms) all settle into the same lowest quantum state. The whole group then behaves like a single particle. Fermions such as electrons cannot do this, because no two of them can share a state.

Ten fermions must stack up the energy levels two at a time. Ten bosons cooled close to absolute zero can all drop into the lowest state together, forming a Bose-Einstein condensate.

Superfluidity

A notable property of a Bose-Einstein condensate is superfluidity: the condensate has zero viscosity, so it flows without any resistance. Because the atoms share one quantum state and move together, there is no internal friction. Superfluidity is observed when certain fluids are cooled close to absolute zero; liquid helium-4 becomes superfluid below about 2.17\,\mathrm{K}.

It passes through incredibly small spaces at a steady speed without any outside force.
Once moving, it keeps flowing, since nothing slows it down.
It forms persistent vortices (regions where the flow revolves around an axis line).

Superconductivity

In certain materials cooled below a critical temperature, electrons join in pairs called Cooper pairs. A Cooper pair behaves like a boson, so the pairs can all occupy the same quantum state and form a condensate. This condensate moves through the material without scattering, so the electrical resistance drops to zero. This is superconductivity: conduction of current without energy loss below a critical temperature.

The critical temperature T_c is the temperature at and below which the resistivity of the substance becomes zero. Examples: mercury 4.2\,\mathrm{K} (discovered by Kamerlingh Onnes in 1911), lead 7.2\,\mathrm{K}, thallium-based ceramics 125\,\mathrm{K}, and \mathrm{LaH_{10}} at about 170\,\mathrm{GPa}, 250\,\mathrm{K}.

Resistivity against temperature (K)

An ordinary metal keeps some resistivity even close to 0\,\mathrm{K}. A superconductor's resistivity falls suddenly to exactly zero at the critical temperature T_c (4.2\,\mathrm{K} for mercury) and stays zero below it (book Fig. 16.8).

Applications of superconductors

Powerful electromagnets (up to about 32\,\mathrm{T}), for example in MRI machines and particle accelerators.
Magnetic levitation (maglev) trains and small but powerful electric motors.
Fast computer chips and more efficient power transmission lines with no I^{2}R loss.

Two effects of Bose-Einstein condensation
Feature	Superfluidity	Superconductivity
Bosons that condense	Atoms (for example helium-4)	Cooper pairs of electrons
What becomes zero	Viscosity	Electrical resistance
Result	Flow without friction	Current without energy loss

Numerical problemsp. 42–43

Convert every temperature to kelvin and every molar mass to kg per mol before substituting.

13 marks

The mass of a helium atom is 6.64\times10^{-27}\,\mathrm{kg}. Calculate the root mean square speed of helium atom in a gas at a temperature of 15\,^\circ\mathrm{C}.

Solution

Given

m = 6.64\times10^{-27}\,\mathrm{kg} mass of one helium atom
T = 15\,^\circ\mathrm{C}=15+273=288\,\mathrm{K}
k = 1.38\times10^{-23}\,\mathrm{J\,K^{-1}}

To find

v_{\mathrm{rms}}

The mass of one atom is given, so use the form with k and m.
v_{\mathrm{rms}}=\sqrt{\dfrac{3kT}{m}}
Substitute, with T in kelvin.
v_{\mathrm{rms}}=\sqrt{\dfrac{3\times1.38\times10^{-23}\times288}{6.64\times10^{-27}}}=\sqrt{\dfrac{1.192\times10^{-20}}{6.64\times10^{-27}}}
Evaluate.
v_{\mathrm{rms}}=\sqrt{1.80\times10^{6}}=1.34\times10^{3}\,\mathrm{m\,s^{-1}}

v_{\mathrm{rms}}\approx1.34\times10^{3}\,\mathrm{m\,s^{-1}}

Book answer. 1.34\times10^{3}\,\mathrm{m\,s^{-1}}

23 marks

At which temperature will the root mean square velocity of the oxygen molecules become equal to the escape velocity of the earth (11.2\,\mathrm{km\,s^{-1}})? (Mass of one molecule of oxygen is 5.3\times10^{-26}\,\mathrm{kg} and Boltzmann constant k=1.38\times10^{-23}\,\mathrm{J\,K^{-1}}.)

Solution

Given

v_{\mathrm{rms}} = 11.2\,\mathrm{km\,s^{-1}}=1.12\times10^{4}\,\mathrm{m\,s^{-1}} set equal to the escape velocity
m = 5.3\times10^{-26}\,\mathrm{kg}
k = 1.38\times10^{-23}\,\mathrm{J\,K^{-1}}

To find

Temperature T

Start from the rms speed and square both sides to remove the root.
v_{\mathrm{rms}}=\sqrt{\dfrac{3kT}{m}}\;\Rightarrow\;v_{\mathrm{rms}}^{2}=\dfrac{3kT}{m}
Make T the subject.
T=\dfrac{m\,v_{\mathrm{rms}}^{2}}{3k}
Substitute in SI units.
\begin{aligned}T&=\dfrac{5.3\times10^{-26}\times(1.12\times10^{4})^{2}}{3\times1.38\times10^{-23}}\\&=\dfrac{5.3\times10^{-26}\times1.254\times10^{8}}{4.14\times10^{-23}}\end{aligned}
Evaluate.
T=\dfrac{6.65\times10^{-18}}{4.14\times10^{-23}}=1.6\times10^{5}\,\mathrm{K}

T\approx1.6\times10^{5}\,\mathrm{K}

Book answer. 1.6\times10^{5}\,\mathrm{K}

33 marks

The mass of a molecule of a gas is 6.4\times10^{-27}\,\mathrm{kg}. Calculate the root mean square speed of gas molecule and kinetic energy per molecule at temperature 400\,\mathrm{K}.

Solution

Given

m = 6.4\times10^{-27}\,\mathrm{kg}
T = 400\,\mathrm{K}
k = 1.38\times10^{-23}\,\mathrm{J\,K^{-1}}

To find

v_{\mathrm{rms}} and average kinetic energy per molecule

rms speed.
\begin{aligned}v_{\mathrm{rms}}&=\sqrt{\dfrac{3kT}{m}}=\sqrt{\dfrac{3\times1.38\times10^{-23}\times400}{6.4\times10^{-27}}}\\&=\sqrt{2.59\times10^{6}}\end{aligned}
Evaluate.
v_{\mathrm{rms}}=1.61\times10^{3}\,\mathrm{m\,s^{-1}}
Kinetic energy per molecule. The average translational kinetic energy depends only on T.
\begin{aligned}\langle K.E\rangle&=\dfrac32kT=\dfrac32\times1.38\times10^{-23}\times400\\&=8.28\times10^{-21}\,\mathrm{J}\end{aligned}
Check with \frac12mv_{\mathrm{rms}}^{2}: it gives the same value.
\tfrac12\times6.4\times10^{-27}\times2.59\times10^{6}=8.28\times10^{-21}\,\mathrm{J}

v_{\mathrm{rms}}\approx1.6\times10^{3}\,\mathrm{m\,s^{-1}},\quad\langle K.E\rangle\approx8.3\times10^{-21}\,\mathrm{J}

Book answer. 1.6\times10^{3}\,\mathrm{m\,s^{-1}}, 8.3\times10^{-21}\,\mathrm{J}

43 marks

At which temperature the root mean square speed of gas molecules becomes double than its speed at 27\,^\circ\mathrm{C} (pressure of gas is kept constant)?

Solution

Given

T_1 = 27\,^\circ\mathrm{C}=300\,\mathrm{K}
v_2 = 2v_1

To find

T_2

For a given gas v_{\mathrm{rms}}=\sqrt{3kT/m}, so v_{\mathrm{rms}}\propto\sqrt{T}. Write the ratio for the two temperatures.
\dfrac{v_2}{v_1}=\sqrt{\dfrac{T_2}{T_1}}
Square both sides and put v_2/v_1=2.
\dfrac{T_2}{T_1}=\left(\dfrac{v_2}{v_1}\right)^{2}=2^{2}=4
Find T_2 in kelvin, then convert to Celsius.
T_2=4\times300=1200\,\mathrm{K}=1200-273=927\,^\circ\mathrm{C}

T_2=1200\,\mathrm{K}=927\,^\circ\mathrm{C}

Book answer. 927\,^\circ\mathrm{C}

53 marks

Determine the root mean square speed of argon atoms at temperature 40\,^\circ\mathrm{C}. The molar mass of argon is 39.95\,\mathrm{g\,mol^{-1}}.

Solution

Given

T = 40\,^\circ\mathrm{C}=313\,\mathrm{K}
M = 39.95\,\mathrm{g\,mol^{-1}}=39.95\times10^{-3}\,\mathrm{kg\,mol^{-1}}
R = 8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}

To find

v_{\mathrm{rms}}

The molar mass is given, so use the molar form.
v_{\mathrm{rms}}=\sqrt{\dfrac{3RT}{M}}
Substitute with M in \mathrm{kg\,mol^{-1}}.
\begin{aligned}v_{\mathrm{rms}}&=\sqrt{\dfrac{3\times8.314\times313}{39.95\times10^{-3}}}=\sqrt{\dfrac{7807}{0.03995}}\\&=\sqrt{1.954\times10^{5}}\end{aligned}
Evaluate.
v_{\mathrm{rms}}=442\,\mathrm{m\,s^{-1}}

v_{\mathrm{rms}}\approx442\,\mathrm{m\,s^{-1}}

Book answer. 442\,\mathrm{m\,s^{-1}}

63 marks

Calculate the number of gas molecules in a cubic meter of gas at standard temperature and pressure (STP).

Solution

Given

V = 1\,\mathrm{m^{3}}
T = 273\,\mathrm{K}
P = 1.01\times10^{5}\,\mathrm{Pa}
k = 1.38\times10^{-23}\,\mathrm{J\,K^{-1}}

To find

Number of molecules N

Use the ideal gas law in terms of molecules and make N the subject.
PV=NkT\;\Rightarrow\;N=\dfrac{PV}{kT}
Substitute.
N=\dfrac{1.01\times10^{5}\times1}{1.38\times10^{-23}\times273}=\dfrac{1.01\times10^{5}}{3.767\times10^{-21}}
Evaluate.
N=2.68\times10^{25}\ \text{molecules}

N\approx2.68\times10^{25}\ \text{molecules}

Book answer. 2.68\times10^{25} molecules

73 marks

A vessel A contains hydrogen and another vessel B whose volume is twice of vessel A contains same mass of oxygen at the same temperature. Find (i) the ratio of the root mean square speeds of hydrogen and oxygen gases, (ii) the ratio of pressures of gases in vessels A and B. Molecular mass of hydrogen and oxygen are 2 and 32 respectively.

Solution

Given

V_A = V
V_B = 2V
m_A = m_B=m same mass of gas
M_{\mathrm{H}} = 2\,\mathrm{g\,mol^{-1}}
M_{\mathrm{O}} = 32\,\mathrm{g\,mol^{-1}}
T_A = T_B=T

To find

(i) v_{\mathrm{H}}:v_{\mathrm{O}}; (ii) P_A:P_B

(i) At the same temperature v_{\mathrm{rms}}=\sqrt{3RT/M}\propto1/\sqrt{M}.
\dfrac{v_{\mathrm{H}}}{v_{\mathrm{O}}}=\sqrt{\dfrac{M_{\mathrm{O}}}{M_{\mathrm{H}}}}=\sqrt{\dfrac{32}{2}}=\sqrt{16}=4
(ii) Number of moles of each gas: same mass m, different molar masses.
n_A=\dfrac{m}{2},\qquad n_B=\dfrac{m}{32}
Use P=\dfrac{nRT}{V} for each vessel.
\begin{gathered}P_A=\dfrac{(m/2)RT}{V}\\P_B=\dfrac{(m/32)RT}{2V}=\dfrac{(m/64)RT}{V}\end{gathered}
Divide. m, R, T and V cancel.
\dfrac{P_A}{P_B}=\dfrac{1/2}{1/64}=32

\text{(i) } v_{\mathrm{H}}:v_{\mathrm{O}}=4:1,\qquad\text{(ii) } P_A:P_B=32:1

Book answer. 4:1 and 32:1

FBISE 2026 questions

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

2026 Annual I, Set B, Q.2 (ii)3 marks

What happens to pressure when temperature and molecular speed of gas molecules increase?

Solution

The pressure increases (for a gas kept in a fixed volume).

When the temperature rises, the average translational kinetic energy of the molecules, \frac32kT, increases, so the molecules move faster (v_{\mathrm{rms}}\propto\sqrt{T}). Faster molecules raise the pressure in two ways:

each collision with a wall gives a larger change of momentum, 2mv_x (the molecules hit harder);
the molecules reach the walls more often, because the time between collisions with the same wall, 2l/v_x, is shorter.

So more momentum is delivered to each unit area of the wall every second, and the pressure rises. Mathematically:

P=\dfrac{1}{3}\,\dfrac{Nm}{V}\,\langle v^{2}\rangle=\dfrac{2}{3}N_0\left\langle\tfrac12mv^{2}\right\rangle\;\propto\;T\quad(\text{constant }V)

At constant volume P\propto\langle v^{2}\rangle\propto T: doubling the absolute temperature doubles the pressure, and doubling the rms speed makes the pressure four times larger. (If the gas is free to expand at constant pressure, its volume increases instead.)

2026 Annual I, Set B, Q.2 (v)3 marks

Derive the relation between absolute temperature and average kinetic energy of gas molecules.

Solution

From kinetic theory, the pressure of an ideal gas of N molecules, each of mass m, in volume V is P=\dfrac13\,\dfrac{Nm}{V}\langle v^{2}\rangle.

Multiply by V, and multiply and divide by 2 to bring in the kinetic energy.
PV=\dfrac{2N}{3}\left\langle\tfrac12mv^{2}\right\rangle\qquad(1)
The ideal gas law with n=N/N_A and k=R/N_A:
PV=nRT=NkT\qquad(2)
Equate (1) and (2) and cancel N.
NkT=\dfrac{2N}{3}\left\langle\tfrac12mv^{2}\right\rangle\;\Rightarrow\;T=\dfrac{2}{3k}\left\langle\tfrac12mv^{2}\right\rangle
\frac{2}{3k} is constant, so
T\propto\langle K.E\rangle\qquad\text{or}\qquad\left\langle\tfrac12mv^{2}\right\rangle=\dfrac32kT

The absolute temperature of an ideal gas is directly proportional to the average translational kinetic energy of its molecules. Temperature is therefore a measure of the average kinetic energy of molecular motion; at 0\,\mathrm{K} this energy would be zero.

T=\dfrac{2}{3k}\left\langle\tfrac12mv^{2}\right\rangle

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

Calculate root mean square speed of oxygen molecule at standard temperature and pressure.

Solution

Given

T = 0\,^\circ\mathrm{C}=273\,\mathrm{K} standard temperature
P = 1.01\times10^{5}\,\mathrm{Pa} standard pressure
M = 32\,\mathrm{g\,mol^{-1}}=0.032\,\mathrm{kg\,mol^{-1}} molar mass of \mathrm{O_2}
R = 8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}

To find

v_{\mathrm{rms}}

Formula (from the paper's formula sheet):
v_{\mathrm{rms}}=\sqrt{\dfrac{3RT}{M}}
Substitute, with M in \mathrm{kg\,mol^{-1}}:
\begin{aligned}v_{\mathrm{rms}}&=\sqrt{\dfrac{3\times8.314\times273}{0.032}}=\sqrt{\dfrac{6809}{0.032}}\\&=\sqrt{2.128\times10^{5}}\end{aligned}
Answer with unit:
v_{\mathrm{rms}}=461\,\mathrm{m\,s^{-1}}

Alternative method: find the mass of one molecule, then use v_{\mathrm{rms}}=\sqrt{3kT/m}.

\begin{gathered}m=\dfrac{M}{N_A}=\dfrac{0.032}{6.022\times10^{23}}=5.31\times10^{-26}\,\mathrm{kg}\\v_{\mathrm{rms}}=\sqrt{\dfrac{3\times1.38\times10^{-23}\times273}{5.31\times10^{-26}}}=461\,\mathrm{m\,s^{-1}}\end{gathered}

This is Example 16.2 of the book. Marks: formula 1, substitution 1, answer with unit 1.

v_{\mathrm{rms}}\approx461\,\mathrm{m\,s^{-1}}

2026 Annual I, Set D, Q.2 (ii)2+1 = 3 marks

How is the pressure related to the density of the gas? Also give its mathematical expression.

Solution

Relation (2 marks)

At constant temperature, the pressure of a gas is directly proportional to its density, P\propto\rho.

Reason: density \rho=\dfrac{Nm}{V} tells how many molecules (how much mass) there are in each unit volume. In a denser gas more molecules strike each unit area of the wall every second. At constant temperature each molecule delivers the same momentum per collision, so more collisions per second give a larger force per unit area, that is, a larger pressure. Doubling the density doubles the pressure.

Kinetic theory gives the pressure of an ideal gas as
P=\dfrac13\left(\dfrac{Nm}{V}\right)\langle v^{2}\rangle
Nm is the total mass of the gas, so Nm/V=\rho:
P=\dfrac13\rho\langle v^{2}\rangle

Mathematical expression (1 mark)

P=\dfrac{1}{3}\rho\langle v^{2}\rangle

\rho: density of the gas; \langle v^{2}\rangle: mean square speed of the molecules, which is constant at constant temperature.

2026 Annual I, Set D, Q.2 (xii)3 marks

Calculate the average translational kinetic energy of a gas molecule at a temperature 47\,^\circ\mathrm{C}.

Solution

Given

T = 47\,^\circ\mathrm{C}=47+273=320\,\mathrm{K}
k = 1.38\times10^{-23}\,\mathrm{J\,K^{-1}}

To find

\langle K.E\rangle

Formula: the average translational kinetic energy depends only on the absolute temperature.
\langle K.E\rangle=\left\langle\tfrac12mv^{2}\right\rangle=\dfrac32kT
Substitute, with T in kelvin.
\langle K.E\rangle=\dfrac32\times1.38\times10^{-23}\times320
Answer with unit.
\langle K.E\rangle=6.62\times10^{-21}\,\mathrm{J}

This is Assignment 16.1 of the book (320\,\mathrm{K}). The paper's formula sheet gives it as T=\frac{2}{3k}\left(\frac12mv^{2}\right); rearranging gives the same \frac32kT. Marks: formula 1, substitution 1, answer with unit 1.

\langle K.E\rangle\approx6.62\times10^{-21}\,\mathrm{J}

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 16, old syllabus1 mark

Which of the following is a boson?

AProton
BNeutrino
CPhotonAnswer
DPion

Why

Bosons are particles that do not obey the Pauli exclusion principle (they have whole-number spin), so any number of them can share one quantum state. The photon (spin 1) is the standard example, and the book names photons as bosons. The proton and the neutrino have spin \frac12, so they are fermions.

Note. Strictly, the pion (a meson with spin 0) is also a boson, so option D is also correct. The photon is the intended textbook answer.

Exam focus

Derive P=\dfrac13\dfrac{Nm}{V}\langle v^{2}\rangle for an ideal gas and show that P\propto\langle K.E\rangle. (long)
The central derivation of the unit (comprehensive Q2 and an SLO); every step carries marks.
Go to answer
Derive T=\dfrac{2}{3k}\left\langle\tfrac12mv^{2}\right\rangle, the relation between absolute temperature and average kinetic energy. (short)
Asked in 2026 (Set B Q.2 (v)); also short Q10 and comprehensive Q4.
Go to answer
Relate pressure and density and write P=\frac13\rho\langle v^{2}\rangle. (short)
Asked in 2026 (Set D Q.2 (ii)); also MCQ 1, and it leads to v_{\mathrm{rms}}=\sqrt{3P/\rho}.
Go to answer
Define rms speed and derive v_{\mathrm{rms}}=\sqrt{3kT/m}=\sqrt{3RT/M}=\sqrt{3P/\rho}. (long)
Comprehensive Q3; these formulas drive most of the numericals.
Go to answer
Calculate the rms speed of a gas at a given temperature (for example oxygen at STP). (numerical)
Asked in 2026 (Set B Q.2 (xi)); numericals 1, 3 and 5 use the same method.
Go to answer
Calculate the average translational kinetic energy of a molecule at a given temperature. (numerical)
Asked in 2026 (Set D Q.2 (xii), 47\,^\circ\mathrm{C}); convert to kelvin and use \frac32kT.
Go to answer
Explain the effect of temperature and molecular speed on gas pressure. (short)
Asked in 2026 (Set B Q.2 (ii)); short Q1 and Q2 test the same kinetic picture.
Go to answer
Ratio problems with v_{\mathrm{rms}}\propto\sqrt{T} and v_{\mathrm{rms}}\propto1/\sqrt{M}. (numerical)
MCQs 6 and 7 and numericals 4 and 7 are quick ratio questions.
Go to answer
Van der Waals equation: the volume and pressure corrections, and when it reduces to PV=nRT. (long)
Comprehensive Q5 and short Q8.
Go to answer
Degenerate matter and the formation of white dwarfs and neutron stars. (long)
Comprehensive Q6, short Q7 and Q9; learn the comparison table.
Go to answer
Bose-Einstein condensate, superfluidity and superconductivity (critical temperature, graph, applications). (long)
Comprehensive Q7, short Q11 to Q13 and MCQs 13 and 14; a boson MCQ also appeared in the 2026 old-syllabus paper.
Go to answer

16Statistical mechanics and the kinetic model of a gas

Assumptions of the kinetic model of an ideal gas

16.1Pressure exerted by gas moleculesOften examined

Derivation Pressure of an ideal gas: P=31​VNm​⟨v2⟩

Pressure and volume

16.1Pressure in terms of average translational kinetic energyOften examined

Derivation Pressure and average translational kinetic energy

16.1Pressure in terms of densityOften examined

Derivation Pressure in terms of density

16.1Temperature in terms of average translational kinetic energyOften examined

Derivation Ideal gas law in terms of molecules: PV=NkT

Derivation Temperature and average translational kinetic energy

16.2Root mean square speed of an ideal gasOften examined

Derivation rms speed in terms of temperature: v_{\mathrm{rms}}=\sqrt{3kT/m}=\sqrt{3RT/M}

Derivation rms speed in terms of pressure and density: v_{\mathrm{rms}}=\sqrt{3P/\rho}

Derivation Ratio of rms speeds of two gases

16.3Behaviour of ideal gases and the birth of statistical mechanics

16.3Behaviour of non-ideal gases: the Van der Waals equationOften examined

Van der Waals equation

(i) Volume correction

(ii) Pressure correction

Derivation The Van der Waals equation

Reduction to the ideal gas equation

16.4Gravitational effects on the ideal gas model

Hydrostatic equilibrium

16.5Behaviour of matter under extreme physical conditions

16.5.1Degenerate matterOften examined

16.5.1White dwarfsOften examined

16.5.2Neutron starsOften examined

16.5.3Bose-Einstein condensationOften examined

16.5.4Superfluidity

16.5.5SuperconductivityOften examined

Discovery and progress

Applications of superconductors

Formula sheet

Key terms

Multiple choice questionsp. 40–41

Short questionsp. 42

Comprehensive questionsp. 42

Pressure of a gas

How gas molecules exert pressure

Pressure due to all the molecules

Factors on which gas pressure depends

Assumptions

Derivation of the pressure

Derivation Pressure of an ideal gas: P=\dfrac{1}{3}\dfrac{Nm}{V}\langle v^{2}\rangle

Pressure and average translational kinetic energy

Derivation Pressure and average translational kinetic energy

Conclusion

Root mean square speed

Expressions in terms of temperature

Derivation rms speed in terms of temperature: v_{\mathrm{rms}}=\sqrt{3kT/m}=\sqrt{3RT/M}

Expression in terms of pressure and density

Derivation rms speed in terms of pressure and density: v_{\mathrm{rms}}=\sqrt{3P/\rho}

Conclusions

Idea

Step 1: pressure from kinetic theory

Derivation Step 1: PV from kinetic theory

Step 2: comparison with the ideal gas law

Derivation Temperature and average translational kinetic energy

Final expression and conclusions

The ideal gas model

Why real gases deviate

(i) Volume correction

(ii) Pressure correction

Van der Waals equation

Derivation The Van der Waals equation

Reduction to the ideal gas equation

Matter under extreme conditions

Degenerate matter

Formation of white dwarfs

Formation of neutron stars

Bose-Einstein condensation

Superfluidity

Superconductivity

Applications of superconductors

Numerical problemsp. 42–43

FBISE 2026 questions

Relation (2 marks)

Mathematical expression (1 mark)

Derivation Pressure of an ideal gas: $P = \frac{1}{3} \frac{N m}{V} ⟨ v^{2} ⟩$