### Entropy of a d-dimensional ideal gas

I’m slowly reading through Kittel and Kroemer to refresh my knowledge of basic statistical mechanics and thermodynamics. They have a nice little problem at the end of Chapter 3 (Problem 11), which is to calculate the entropy of a one-dimensional ideal gas (using the methods outlined in that chapter). It is only a brief step further to calculate the general expression for the entropy of a ${d}$-dimensional ideal gas, which I do here.

The single-particle partition function is calculated using the “particle-in-a-box” solution from quantum mechanics, and the end result is that

$\displaystyle Z_1 = \left( \int_0^\infty \text{d}n\, e^{-\alpha^2 n^2} \right)^d = \left( \frac{ \sqrt{\pi}}{2\alpha}\right)^d, \ \ \ \ \ (1)$

where

$\displaystyle \alpha = \left( \frac{\hbar^2 \pi^2}{2 M L^2 \tau} \right)^{1/2}. \ \ \ \ \ (2)$

Define the ${d}$-dimensional quantum concentration as ${n_{Q_d} = Z_1/L^d}$, then by plugging in we get

$\displaystyle n_{Q_d} = \left( \frac{M\tau}{2\pi \hbar^2} \right)^{d/2}. \ \ \ \ \ (3)$

The free energy is (using the Stirling approximation)

$\displaystyle F = - \tau N \log{(Z_1)} + \tau (N \log{(N)} - N), \ \ \ \ \ (4)$

and thus the entropy is

$\displaystyle \sigma = - \left( \frac{\partial F}{\partial \tau} \right)_V = N \left( \log{\left(\frac{Z_1}{N}\right)} + 1 + \tau \frac{\partial \log{Z_1}}{\partial \tau} \right). \ \ \ \ \ (5)$

Now, we can use the properties of the logarithm to make taking the derivative really easy, since

$\displaystyle \log{Z_1} = \frac{d}{2} \log{\tau} + \cdots \ \ \ \ \ (6)$

where I have dropped all of the terms that don’t depend on the temperature. So

$\displaystyle \frac{\partial \log{Z_1}}{\partial \tau} = \frac{d}{2\tau}, \ \ \ \ \ (7)$

and the final result for the entropy is

$\displaystyle \sigma = N \left( \log{\left( \frac{n_{Q_d}}{n}\right)} + \frac{d}{2} + 1 \right). \ \ \ \ \ (8)$

Here I have defined ${n = N/L^d}$.

For 1,2, and 3 dimensions, we have

$\displaystyle \sigma_1 = N \left( \log{\left( \frac{n_{Q_1}}{n} \right)} + \frac{3}{2} \right), \ \ \ \ \ (9)$

$\displaystyle \sigma_2 = N \left( \log{\left(\frac{n_{Q_2}}{n} \right)} + 2 \right), \ \ \ \ \ (10)$

and

$\displaystyle \sigma_3 = N \left( \log{\left(\frac{n_{Q_3}}{n} \right)} + \frac{5}{2} \right). \ \ \ \ \ (11)$

Amazon Affiliate Link:

Advertisements