Chapter 1

Kinetic Theory of Gases

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Theory

1.1 Basic assumptions of kinetic theory

The kinetic theory explains the macroscopic behaviour of a gas (pressure, temperature, heat capacity) from the motion of its molecules. It rests on the following assumptions.

Even an infinitesimally small volume of gas contains a very large number of molecules, so averages over molecules are meaningful.
A gas is made up of identical molecules that behave as rigid, perfectly elastic, hard spheres.
The molecules move continuously in random directions. All directions of motion are equally probable (isotropy).
The size of a molecule is much smaller than the average distance between molecules.
Molecules exert no force on each other except during a collision, so between collisions they move in straight lines with constant speed.
Collisions between molecules, and with the walls, are perfectly elastic.
Molecular speeds range from \(0\) to \(\infty\).
The duration of a collision is negligible compared with the time between collisions.

1.2 Pressure exerted by a gas

Set-up. Consider a cubical container of side \(L\) (volume \(V=L^3\)) holding \(N\) molecules, each of mass \(m\). Group the molecules so that \(n_i\) molecules per unit volume have velocity \(\vec v_i=(v_{ix},v_{iy},v_{iz})\), with \(\sum_in_i=n=N/V\) and \(v_i^2=v_{ix}^2+v_{iy}^2+v_{iz}^2\).

Step 1 — Momentum transferred in one collision. A molecule moving towards the wall perpendicular to \(x\) with velocity component \(v_{ix}\) rebounds elastically with \(-v_{ix}\). The change in its momentum is

\[\Delta p_{ix}=mv_{ix}-(-mv_{ix})=2mv_{ix},\]

and by Newton’s third law the same momentum is delivered to the wall.

Step 2 — Rate of collisions. After hitting the wall the molecule travels to the opposite face and back, a distance \(2L\), before it strikes the same wall again. The time between successive hits is \(\Delta t=2L/v_{ix}\).

Step 3 — Force from one molecule.

\[f_{ix}=\frac{\Delta p_{ix}}{\Delta t}=\frac{2mv_{ix}}{2L/v_{ix}}=\frac{mv_{ix}^2}{L}.\]

Step 4 — Pressure from all molecules. The number of molecules of group \(i\) in the box is \(n_iL^3\). Summing their forces and dividing by the wall area \(L^2\):

\[P_x=\frac{1}{L^2}\sum_i n_iL^3\frac{mv_{ix}^2}{L}=m\sum_in_iv_{ix}^2=mn\langle v_x^2\rangle,\qquad \langle v_x^2\rangle=\frac{\sum_in_iv_{ix}^2}{\sum_in_i}.\]

Step 5 — Isotropy. Since \(\langle v_x^2\rangle+\langle v_y^2\rangle+\langle v_z^2\rangle=\langle v^2\rangle\) and no direction is preferred, \(\langle v_x^2\rangle=\langle v_y^2\rangle=\langle v_z^2\rangle=\frac13\langle v^2\rangle\). The pressure is the same on every wall:

\[\boxed{P=\frac13nm\langle v^2\rangle=\frac13\frac{Nm\langle v^2\rangle}{V}},\qquad PV=\frac13Nm\langle v^2\rangle .\]

In two dimensions. For a gas confined to an area \(A\), the “wall” is a line of length \(L\) and isotropy gives \(\langle v_x^2\rangle=\langle v_y^2\rangle=\frac12\langle v^2\rangle\). Repeating the steps gives the 2D pressure (force per unit length)

\[P=\frac12\frac{Nm\langle v^2\rangle}{A},\qquad PA=\frac12Nm\langle v^2\rangle .\]

In \(d\) dimensions, \(P=\frac1d\,nm\langle v^2\rangle\).

Photon gas. For particles with \(E=pc\) moving at speed \(c\) with momentum parallel to velocity, the momentum-flux argument above gives \(P=\frac13n\langle\vec p\cdot\vec v\rangle\). With \(\vec p\cdot\vec v=pc=E\):

\[P=\frac13n\langle E\rangle\quad\Rightarrow\quad PV=\frac13E .\]

More generally, for \(E\propto p^s\) we have \(\vec p\cdot\vec v=p\,\frac{dE}{dp}=sE\), so \(PV=\frac{s}{d}E\). The non-relativistic gas (\(s=2\), \(d=3\)) gives \(PV=\frac23E\).

1.3 Gas laws for ideal gases

Law	Statement
Boyle	At constant \(T\), \(P\propto 1/V\) for a given mass of gas
Charles	At constant \(P\), \(V\propto T\)
Avogadro	Equal volumes of all gases at the same \(T\) and \(P\) contain equal numbers of molecules
Graham	At the same \(P\) and \(T\), rates of diffusion satisfy \(r_1/r_2=\sqrt{\rho_2/\rho_1}\)
Dalton	The total pressure of a mixture is the sum of the partial pressures: \(P=P_1+P_2+\cdots\)

Combining these gives the ideal-gas equation

\[PV=nRT=Nk_BT,\]

where \(n\) is the number of moles, \(R=8.314\) J mol⁻¹ K⁻¹, and \(k_B=R/N_A=8.314/6.022\times10^{23}=1.38\times10^{-23}\) J K⁻¹.

1.4 Kinetic interpretation of temperature

The molecules have only translational kinetic energy and no mutual potential energy, so the mean energy per molecule is \(\langle E\rangle=\frac12m\langle v^2\rangle\). Write the pressure formula in terms of \(\langle E\rangle\):

\[P=\frac13nm\langle v^2\rangle=\frac23n\Big(\frac12m\langle v^2\rangle\Big)=\frac23n\langle E\rangle\quad\Rightarrow\quad PV=\frac23N\langle E\rangle .\]

Compare with \(PV=Nk_BT\):

\[\boxed{\langle E\rangle=\frac12m\langle v^2\rangle=\frac32k_BT} .\]

Temperature is therefore a direct measure of the mean translational kinetic energy of the molecules. For one mole, \(\langle E\rangle=\frac32\frac{RT}{N_A}\).

Equipartition of energy

Statement. In classical thermal equilibrium, each independent quadratic term in the energy (such as \(\frac12mv_x^2\), \(\frac12I\omega^2\) or \(\frac12kx^2\)) has an average value of \(\frac12k_BT\).

Proof for one term. For an energy \(\epsilon=aq^2\), the Boltzmann average is

\[\langle aq^2\rangle=\frac{\int aq^2e^{-\beta aq^2}dq}{\int e^{-\beta aq^2}dq}=-\frac{\partial}{\partial\beta}\ln\int e^{-\beta aq^2}dq=-\frac{\partial}{\partial\beta}\ln\sqrt{\frac{\pi}{a\beta}}=\frac{1}{2\beta}=\frac12k_BT .\]

Generalisation. The same method with \(\epsilon=a|q|^s\) gives \(\int e^{-\beta a|q|^s}dq\propto\beta^{-1/s}\), so \(\langle a|q|^s\rangle=k_BT/s\). For example, a potential \(V=\gamma r^3\) in 3D gives \(\langle V\rangle=k_BT\).

Degrees of freedom of molecules.

Molecule	Translational \(f\)	Rotational \(f\)	Vibration (when excited)	\(C_V\) (rigid)	\(\gamma=C_P/C_V\) (rigid)
Monatomic	3	0	none	\(\frac32R\)	\(\frac53\)
Diatomic / linear	3	2	\(3N_a-5\) modes	\(\frac52R\)	\(\frac75\)
Non-linear polyatomic	3	3	\(3N_a-6\) modes	\(3R\)	\(\frac43\)

Each vibrational mode contributes \(k_BT\) (kinetic \(\frac12k_BT\) + potential \(\frac12k_BT\)), i.e. \(R\) per mole to \(C_V\). In general \(C_V=\frac f2R\), \(C_P=C_V+R\), and \(\gamma=1+\frac2f\), where a vibrational mode counts as \(f=2\).

Mixture of ideal gases. Since energy is additive, \(C_V^{\rm mix}=\dfrac{\sum n_iC_{V,i}}{\sum n_i}\) and \(C_P^{\rm mix}=C_V^{\rm mix}+R\).

Exam tip. “Low temperature” means vibrations are frozen out. “High temperature” (but below dissociation) means every vibrational mode is active and adds \(k_B\) per molecule to \(C_V\).

Example 1.1 Electromagnetic radiation in a container with mirror walls can be treated as a gas of photons moving with constant speed \(c\), whose energy is related to momentum by \(E=pc\), with \(\vec p\) parallel to the velocity. Show that if the container is full of radiation, \(PV=\frac13E\).

Solution.
From §1.2, the pressure is the momentum flux: \(P=\frac13n\langle\vec p\cdot\vec v\rangle\). For photons \(v=c\) and \(\vec p\parallel\vec v\), so \(\vec p\cdot\vec v=pc=E\). Hence \(P=\frac13n\langle pc\rangle=\frac13\frac NV\langle E_{\rm photon}\rangle\), and with \(E=N\langle E_{\rm photon}\rangle\) the total energy, \(PV=\frac13E\).

1.5 Maxwell–Boltzmann distribution of molecular velocities

Derivation

Assumptions.

The probability that a molecule has velocity component in \([v_x,v_x+dv_x]\) is \(f(v_x)dv_x\), and the three components are statistically independent.
Space is isotropic, so the joint distribution can depend only on the speed: \(f(v_x)f(v_y)f(v_z)=F(v^2)\).

Step 1 — Functional equation. For fixed speed, \(v_x^2+v_y^2+v_z^2=\) const, so \(v_xdv_x+v_ydv_y+v_zdv_z=0\), and \(d[f(v_x)f(v_y)f(v_z)]=0\). Dividing by \(f(v_x)f(v_y)f(v_z)\):

\[\frac{f'(v_x)}{f(v_x)}dv_x+\frac{f'(v_y)}{f(v_y)}dv_y+\frac{f'(v_z)}{f(v_z)}dv_z=0 .\]

Step 2 — Lagrange multiplier. Add \(2\beta\) times the constraint. Because \(dv_x\), \(dv_y\), \(dv_z\) can now be treated as independent, each bracket vanishes:

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Chapter 1 Kinetic Theory of Gases