Chapter 3.6, Problem 39P

To determine

The temperature, pressure and chemical potential of an ideal monoatomic gas.

Answer to Problem 39P

  $T=\frac{U}{Nk}$

  $P=\frac{NkT}{A}$

  $\mu =-kT\ln \left(\frac{V}{N}{\left(\frac{2\pi mkT}{h^2}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)$

Explanation of Solution

Given:

The entropy of an ideal gas is,

  $S=Nk\left[\ln \left(\frac{2\pi mAU}{{\left(Nh\right)}^2}\right)+2\right]$

Calculation:

To get the temperature, we partial differentiate with respect to U,

  $\begin{array}{c}\frac{1}{T}={\left(\frac{\partial S}{\partial U}\right)}_{A,N}\\ \frac{1}{T}=\frac{\partial }{\partial U}\left[Nk\left[\ln \left(\frac{2\pi mAU}{{\left(Nh\right)}^2}\right)+2\right]\right]\\ =Nk\left[\frac{{\left(Nh\right)}^2}{2\pi mAU}\frac{2\pi mA}{{\left(Nh\right)}^2}\right]\\ =\frac{Nk}{U}\\ T=\frac{U}{Nk}\end{array}$

Partial differentiating with respect to A, we get the pressure as,

  $\begin{array}{c}P=T{\left(\frac{\partial S}{\partial A}\right)}_{U,N}\\ =T\frac{\partial }{\partial A}\left[Nk\left[\ln \left(\frac{2\pi mAU}{{\left(Nh\right)}^2}\right)+2\right]\right]\\ =NkT\left[\frac{{\left(Nh\right)}^2}{2\pi mAU}\times \frac{2\pi mU}{{\left(Nh\right)}^2}\right]\\ =\frac{NkT}{A}\end{array}$

Partial differentiating the entropy with respect to N, we get the chemical potential as,

  $\begin{array}{c}\mu =-T{\left(\frac{\partial S}{\partial N}\right)}_{U,A}\\ =-T\frac{\partial }{\partial N}\left[Nk\left[\ln \left(\frac{2\pi mAU}{{\left(Nh\right)}^2}\right)+2\right]\right]\\ =-kT\left[\ln \left(\frac{2\pi mAU}{{\left(Nh\right)}^2}\right)+2\right]-kTN\frac{{\left(Nh\right)}^2}{2\pi mAU}\times \frac{2\pi mAU}{h^2}\times \frac{-2}{N^3}\\ =-kT\left[\ln \left(\frac{2\pi mAU}{{\left(Nh\right)}^2}\right)+2\right]+2kT\\ =-kT\left[\ln \left(\frac{2\pi mAU}{{\left(Nh\right)}^2}\right)+2-2\right]\\ =-kT\left[\ln \left(\frac{2\pi mAU}{{\left(Nh\right)}^2}\right)\right]\\ =-kT\left[\ln \left(\frac{2\pi mA\left(NkT\right)}{{\left(Nh\right)}^2}\right)\right]\\ =-kT\ln \left(\frac{V}{N}{\left(\frac{2\pi mkT}{h^2}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)\end{array}$

Conclusion:

The temperature, pressure and entropy are given by:

  $T=\frac{U}{Nk}$

  $P=\frac{NkT}{A}$

  $\mu =-kT\ln \left(\frac{V}{N}{\left(\frac{2\pi mkT}{h^2}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)$