# Statistical Mechanics: The Ideal Gas

### The Ideal Gas: Boltzmann’s Approach (The Microcanonical Ensemble)

Consider a monatomic gas of ##N## non-interacting particles with mass ##m## occupying the volume ##V##. Since the particles of the gas do not interact with each other, it is not difficult to explicitly calculate ##\Omega_{E}##. The position of each particle is constrained to be within the volume ##V## and the momentum is constrained to satisfy ##\sum_{i=1}^{N}p_{i}^{2}=2mE##.

In order to count the number of unique locations in the volume, we must somehow quantize the volume. Let’s consider the volume to be quantized as ##v\equiv V/h^{3}##. The number of ways to distribute ##N## particles across ##v## partitions is

$$\frac{(N+v-1)!}{N!(v-1)!}$$

If the number of allowed spatial locations and the number of particles is very large, such that ##v>>1## and ##N>>1##, then Stirling’s approximation can be used to give

$$\frac{(N+v-1)!}{N!(v-1)!}\approx\frac{(N+v)!}{N!v!}\approx\frac{(N+v)^{N+v}}{N^{N}v^{v}}\approx\frac{v^{N}(N+v)^{v}}{N^{N}v^{v}}$$

Now if we consider that the number of spatial configurations is much larger than the number of particles, such that ##v>>N##, then we can use the approximation ##(1+N/v)^{v}\approx e^{N}## along with a final application of Stirling’s approximation to give

$$\frac{v^{N}(N+v)^{v}}{N^{N}v^{v}}=\frac{v^{N}(1+N/v)^{v}}{N^{N}}\approx\frac{v^{N}e^{N}}{N^{N}}\approx\frac{v^{N}}{N!}$$

The size of the allowed phase space is the product of the distinct spacial coordinates ##v^{N}/N!## and the surface of a ##3N## dimensional sphere in momentum space with radius ##\sqrt{2mE}##.

The area of a ##d##-dimensional sphere of radius ##R## is

$$A=\frac{2\pi^{d/2}}{(d/2-1)!}R^{d-1}$$

Thus the available phase space is

$$\Omega_{E}=\frac{v^{N}}{N!}\frac{2\pi^{3N/2}}{(3N/2-1)!}(2mE)^{(3N-1)/2}$$

Since ##N>>1##, this can be approximated as

$$\Omega_{E}=\frac{2}{N!(3N/2)!}(2\pi v^{2/3}mE)^{3N/2}$$

The entropy is obtained from Boltzmann’s formula:

$$S=k\ln\Omega_{E}=k\ln\left[\frac{2}{N!(3N/2)!}(2\pi v^{2/3}mE)^{3N/2}\right]$$

Applying Stirling’s approximation gives

$$S=k\left[\ln 2+\frac{3N}{2}\ln\left(2\pi v^{2/3} mE\right)-N\ln N+N-\frac{3N}{2}\ln\left(\frac{3N}{2}\right)+\frac{3N}{2}\right]$$

In the thermodynamic limit, terms of order 1 can be dropped, giving the Sackur-Tetrode equation:

$$S=k\left[\frac{3N}{2}\ln\left(2\pi v^{2/3} mE\right)-N\ln N-\frac{3N}{2}\ln\left(\frac{3N}{2}\right)+\frac{5N}{2}\right]$$

$$S=Nk\left\{\ln\left[\frac{V}{N}\left(\frac{4\pi mE}{3h^{2}N}\right)^{3/2}\right]+\frac{5}{2}\right\}$$

The properties of the ideal gas can now be found using the combined first and second law: ##TdS=dE+PdV-\mu dN##. The temperature is related to the internal energy as

$$\frac{1}{T}=\frac{\partial S}{\partial E}\bigg|_{N,V}=\frac{3}{2}\frac{Nk}{E}\Longrightarrow E=\frac{3}{2}NkT$$

while the pressure is related to volume and temperature as

$$\frac{P}{T}=\frac{\partial S}{\partial V}\bigg|_{N,E}=\frac{Nk}{V} \Longrightarrow PV=NkT$$

Thus we have arrived at the familiar equation of state for an ideal gas.

### The Ideal Gas: The Canonical Ensemble

We can generalize the definition of the discrete partition function found in part one to a classical continuum system by replacing the sum with an integral over the phase space.

$$Z=\frac{1}{N!h^{3N}}\int\cdots\int\exp\left(-\beta \mathcal{H}\right)d^{3}\mathbf{p}_{1}\cdots d^{3}\mathbf{p}_{N}d^{3}\mathbf{r}_{1}\cdots d^{3}\mathbf{r}_{N}$$

The total energy is represented by the Hamiltonian ##\mathcal{H}=\sum_{i=1}^{N}|\mathbf{p}_{i}|^{2}/2m##. The spatial integrals are easily evaluated to give

$$Z=\frac{V^{N}}{N!h^{3N}}\int\cdots\int\exp\left(-\beta \sum_{i=1}^{N}\frac{|\mathbf{p}_{i}|^{2}}{2m}\right)d^{3}\mathbf{p}_{1}\cdots d^{3}\mathbf{p}_{N}$$

Since the integral for each particle and each component of ##\mathbf{p}## are identical, this can be written as

$$Z=\frac{V^{N}}{N!h^{3N}}\left[\int_{-\infty}^{\infty}\exp\left(-\beta\frac{p^{2}}{2m}\right)dp\right]^{3N}$$

which is easily evaluated to give

$$Z=\frac{V^{N}}{N!}\left(\frac{2\pi m}{h^{2}\beta}\right)^{3N/2}$$

The free energy can be found from the partition function as

$$F=-kT\ln\left[\frac{V^{N}}{N!}\left(\frac{2\pi mkT}{h^{2}}\right)^{3N/2}\right]=-kT\left\{N\ln\left[V\left(\frac{2\pi mkT}{h^{2}}\right)^{3/2}\right]-\ln N!\right\}$$

Again, assuming the thermodynamic limit, Stirling’s approximation can be used to give

$$F=-NkT\left\{\ln\left[V\left(\frac{2\pi mkT}{h^{2}}\right)^{3/2}\right]-\ln N+1\right\}$$

which leads to the following expression for the entropy:

$$S=-\frac{\partial F}{\partial T}=Nk\left\{\frac{3}{2}+\ln\left[V\left(\frac{2\pi mkT}{h^{2}}\right)^{3/2}\right]-\ln N+1\right\}$$

$$S=Nk\left\{\ln\left[\frac{V}{N}\left(\frac{2\pi mkT}{h^{2}}\right)^{3/2}\right]+\frac{5}{2}\right\}$$

The energy can now be related to the temperature of the gas using the definition of the Helmholtz free energy.

$$S=\frac{E-F}{T}$$

$$Nk\left\{\frac{3}{2}+\ln\left[V\left(\frac{2\pi mkT}{h^{2}}\right)^{3/2}\right]-\ln N+1\right\}=\frac{E}{T}+Nk\left\{\ln\left[V\left(\frac{2\pi mkT}{h^{2}}\right)^{3/2}\right]-\ln N+1\right\}$$

$$\frac{3}{2}Nk=\frac{E}{T}\Longrightarrow E=\frac{3}{2}NkT$$

The equation of state is obtained from

$$P=-\frac{\partial F}{\partial V}\bigg|_{T,N}=\frac{kTN}{V}\Longrightarrow PV=NkT$$

We have again arrived at the proper equation of state for an ideal gas. This shows how different thermodynamic ensembles lead to the same equation of state when taken to the thermodynamic limit. For most systems however, there is often only one ensemble that is easy to use and solve for. The ideal gas, because of its simplicity, is a rare example of a system which can be exactly solved in multiple ensemble types.

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