# Semi-Classical/Classical derivation of ideal gas partition function

1. May 14, 2014

### CAF123

In the semi-classical treatment of the ideal gas, we write the partition function for the system as $$Z = \frac{Z(1)^N}{N!}$$ where $Z(1)$ is the single particle partition function and $N$ is the number of particles. It is semi-classical in the sense that we consider the indistinguishability of the particles, so we divide by $N!$.

The resulting expression for the entropy of the system is $$S = Nk \left(\ln \left[\left(\frac{V}{N}\right) \left(\frac{2\pi mkT}{h^2}\right)^{3/2} \right] + \frac{5}{2}\right)$$

Now consider a fully classical analysis. There $Z(1) = \sum_{E} \exp (-\frac{E}{kT})$, where $E = p^2/2m$ (assuming no interaction potentials). The problem can be mapped to an integral over the phase space of the Hamiltonian to give $$Z(1) \rightarrow \int \exp \left(-\frac{1}{2mkT} (p_x^2 + p_y^2 + p_z^2) \right)\text{d}^3 \underline{p} \,\text{d}^3 \underline{x}$$ This can then be rewritten like $$\int \exp \left(-\frac{p_x^2}{2mkT}\right) \text{d}p_x \int \exp \left(-\frac{p_y^2}{2mkT}\right) \text{d}p_y \int \exp \left(-\frac{p_z^2}{2mkT} \right) \text{d}p_z \cdot V$$ where $V$ is the volume of the container. Those are Gaussian integrals and so evaluation is immediate. The result is that $Z(1) = (2\pi mkT)^{3/2} V$. The corresponding entropy can be calculated and the result is that $$S = Nk \left(\frac{3}{2} + \ln\left(\frac{(2\pi mkT)^{3/2}}{V}\right)\right).$$

What is the significance of the factors 5/2 in the semi-classical treatment and the factor 3/2 in the classical treatment and why are they different? They look like the number of degrees of freedom a monatomic and diatomic molecule would have at room temperature, but I think this is a coincidence.

Many thanks.

2. May 15, 2014

### WannabeNewton

It's just a coincidence. The factor of $N!$ in $Z$ ends up giving a factor of $-k_B T N$ in the free energy $F = -k_B T \ln Z$ which then gives a factor of $k_B N$ in $S = -\frac{\partial F}{\partial T}$ which gets added on to the factor of $\frac{3}{2} k_B N$ coming from $\frac{\partial }{\partial T} \ln Z_1$ present also in the classical gas, this giving an overall factor of $\frac{5}{2}$. It has nothing to do with the degrees of freedom of the classical and semi-classical monoatomic gases as both have a factor of $\frac{3}{2}$ in $\bar{E} = -\frac{\partial }{\partial \beta} \ln Z$ in accordance with equipartition.