The ideal gas from statistical mechanics

Once you have q, you can forget the microscopic detail

Intuition

Statistical mechanics is a recipe. List the energy levels a molecule can have. Add up the scores to get q. After that, every bulk number such as energy, pressure and heat capacity is a slope or a derivative of q. A gas is the simplest test of the recipe.

The lecture summarises progress so far: equilibrium is the most likely configuration, and the partition function is a sum over microstates that holds what we usually want to know. All of classical thermodynamics (heat, work, entropy, free energy) can be derived from it.

Before the gas, one rule about combining parts. The partition function of two isolated systems is the product $q_{\text{total}} = q_1 q_2$ (the lecture sets this as homework to show). For $N$ components that can be told apart, like numbered balls, $Q = q^N$. For $N$ components that cannot be told apart, like molecules of one gas, swapping two gives the same state, so you divide out the repeats:

$Q = \frac{q^N}{N!}$

This is the same double-counting fix as in the weight formula $W = N!/\prod n_i!$.

A gas molecule is a particle in a box

For an atomic gas the only motion is translation. The lecture uses the particle in a box. In 1D the levels are

$\varepsilon_i = \frac{i^2h^2}{8mL^2}$

A smaller box (smaller $L$) means higher energies: squeeze a guitar string and the pitch rises. In a cube of side $L$ you need three quantum numbers:

$\varepsilon_{i_x,i_y,i_z} = \frac{h^2}{8mL^2}\left(i_x^2 + i_y^2 + i_z^2\right)$

The 1D partition function is $q = \sum_{i=1}^{\infty} e^{-\beta\varepsilon_i}$, which cannot be evaluated exactly. But if many states are occupied, the steps are tiny and you can swap the sum for an integral. Check with helium in a 1 m³ box: the characteristic temperature is

$\Theta_{\text{trans}} \sim \frac{h^2}{8mL^2k} \sim 10^{-18}\ \text{K}$

so $T \gg \Theta$ for any sensible temperature. Then

$q \approx \int_0^\infty e^{-\beta i^2h^2/8mL^2}\,di = \left(\frac{2\pi m}{h^2\beta}\right)^{1/2}L$

In 3D, $q_{\text{trans}} = q_xq_yq_z$, giving

$q_{\text{trans}} = \left(\frac{2\pi m}{h^2\beta}\right)^{3/2}V$

Define the thermal wavelength $\Lambda = \left(\dfrac{h^2\beta}{2\pi m}\right)^{1/2}$. Then

$q_{\text{trans}} = \frac{V}{\Lambda^3}$

For 1 m³ of gas, $q_{\text{trans}} = 10^{30}$. In plain words, a molecule in that box has about $10^{30}$ reachable translational states.

The energy comes out as 3/2 NkT

The earlier result from the partition-function topic: $E = NkT^2\dfrac{d\ln q}{dT}$. Put in $q_{\text{trans}}$. Since $\beta = 1/kT$, $q_{\text{trans}} \propto T^{3/2}V$, so $\ln q_{\text{trans}} = \tfrac{3}{2}\ln T + \text{constants}$ and $\dfrac{d\ln q}{dT} = \dfrac{3}{2T}$. Therefore

$E_{\text{trans}} = NkT^2\cdot\frac{3}{2T} = \frac{3}{2}NkT = \frac{3}{2}nRT$

Because $U = E$, the internal energy of an ideal gas depends only on its temperature. The lecture notes that $h$ dropped out.

Try it: worked example

Show that C_V for a monatomic ideal gas is 3/2 Nk.

Show answer and steps

AnswerC_V = (dU/dT) at constant V = 3/2 Nk = 3/2 nR.

1. Heat capacity is $C = dQ/dT$.
2. At constant volume no expansion work is done, so $dQ = dU$ and $C_V = (dU/dT)_V$.
3. For the ideal monatomic gas $U = \tfrac{3}{2}NkT$.
4. So $C_V = \dfrac{d\left(\tfrac{3}{2}NkT\right)}{dT} = \dfrac{3}{2}Nk = \dfrac{3}{2}nR$.

The lecture sets “show that $C_P = \tfrac{5}{2}Nk$” as homework, so no working is given here.

Pressure and the ideal gas law drop out of the same recipe

What is pressure, microscopically? It is the generalised force linked to volume: the work on an expanding gas is $dW = -p\,dV$. From the first-law topic, $dU = \sum_i n_i\,d\varepsilon_i + \sum_i\varepsilon_i\,dn_i$, where the first sum is the work. So $-p\,dV = \sum_i n_i\,d\varepsilon_i$, and

$p = -\sum_i n_i\frac{d\varepsilon_i}{dV}$

Squeeze the box and every level rises, which is the push you feel. Because $L^2 = V^{2/3}$, the levels are

$\varepsilon = \frac{h^2}{8m}\left(i_x^2+i_y^2+i_z^2\right)V^{-2/3} = aV^{-2/3}$

where $a$ does not depend on $V$. Differentiate:

$\frac{d\varepsilon_i}{dV} = -\frac{2}{3}aV^{-5/3} = -\frac{2}{3}\varepsilon_iV^{-1}$

Substitute and use $E = \sum n_i \varepsilon_i = \tfrac{3}{2}NkT$:

$p = \sum_i n_i\frac{2}{3}\varepsilon_iV^{-1} = \frac{2}{3}V^{-1}E = \frac{NkT}{V} = \frac{nRT}{V}$

This derives the ideal gas law from purely microscopic principles.

A second route uses the Helmholtz energy. From $dA = -S\,dT - p\,dV$, $p = -(\partial A/\partial V)_T$. With $A = -kT\ln Q = -kT\ln(q^N/N!)$:

$p = NkT\left(\frac{\partial\ln q}{\partial V}\right)_T = NkT\left(\frac{\partial}{\partial V}\ln\frac{V}{\Lambda^3}\right)_T = \frac{NkT}{V}$

Both routes agree. Polyatomic gases can also vibrate, rotate or have excited electronic states, but if they are ideal they still obey $pV = NkT$. Real gases deviate because of interactions. Interactions matter at high $p$, when molecules cannot avoid each other, and at low $T$, when energy-lowering interactions beat the entropy of expansion. The lecture promises more on this next week.

Pitfall

Do not carry $\tfrac{3}{2}NkT$ over to a polyatomic gas. That value is for translation only, which is all a monatomic gas has. The ideal gas law $pV = NkT$ is the one that stays true for polyatomic ideal gases.

Exam detail

Lecture 17 summary: statistical mechanics of translation and vibration can be derived from quantum-mechanical models; statistical mechanics predicts the ideal gas law; heat capacity can be calculated, including its temperature dependence for simple solids. Slide numbers: $\Theta_{\text{trans}} \sim 10^{-18}$ K, $q_{\text{trans}} = 10^{30}$ for 1 m³.

Recall

What two results fall out of putting q_trans = V/Lambda^3 into the statistical-mechanics recipe?

The energy $E = \tfrac{3}{2}NkT$ (so $C_V = \tfrac{3}{2}Nk$) and the pressure $p = NkT/V$, the ideal gas law.