Question

Starting from the formula for $C_V$ derived in Problem 7.70(b), calculate the entropy, Helmholtz free energy, and pressure of a Bose gas for $T<T_c$. Notice that the pressure is independent of volume; how can this be the case?

Short answer

Entropy of Bose gas , $S=3.3537{\left(\frac{2\pi m}{h^2}\right)}^{\frac{3}{2}}V·{\left(K_{B}T\right)}^{\frac{3}{2}}{K}_B$

Helmholtz free energy, $F=-1.3415{\left(\frac{2\pi m}{h^2}\right)}^{\frac{1}{2}}V·{\left(K_{B}T\right)}^{\frac{5}{2}}$

Pressure of a Bose gas,$P=1.3415{\left(\frac{2\pi m}{h^2}\right)}^{\frac{3}{2}}{\left(K_{B}T\right)}^{\frac{5}{2}}$

Step-by-step solution

Step 1. Given information

If $T<T_c$,

$C_V=5.0306{\left(\frac{2\pi m}{h^2}\right)}^{\frac{3}{2}}V{\left(K_{B}T\right)}^{\frac{3}{2}}{K}_B$

$U=\frac{2}{\sqrt{\pi }}\left(\frac{2\pi m}{h^2}\right)V·{\left(K_{B}T\right)}^{\frac{5}{2}}(1.7833)$

$=2.0122{\left(\frac{2\pi m}{h^2}\right)}^{\frac{3}{2}}V·{\left(K_{B}T\right)}^{\frac{5}{2}}$

Here,

$h$= Planck's constant,

$K_B$= Boltzmann's constant,

$V$= volume of the box,

$T$= temperature,

$m$= mass of the particle,

Step 2. To find entropy 

We have,

$S={\int }_0^{T}5.0306{\left(\frac{2\pi m}{h^2}\right)}^{\frac{3}{2}}V·K_B^{\frac{5}{2}}\frac{T^{\text{'}\frac{3}{2}}}{T^{\text{'}}}d{T}^{\text{'}}$

$=5.0306{\left(\frac{2\pi m}{h^2}\right)}^{\frac{3}{2}}V·K_B^{\frac{5}{2}}{\int }_0^T{T}^{\text{'}\frac{1}{2}}d{T}^{\text{'}}$

$={\left.5.0306{\left(\frac{2\pi m}{h^2}\right)}^{\frac{3}{2}}V·K_B^{\frac{5}{2}}\frac{T^{\frac{3}{2}}}{\frac{3}{2}}\right|}_0^T$

$=5.0306·\frac{2}{3}·{\left(\frac{2\pi m}{h^2}\right)}^{\frac{3}{2}}V·{\left(K_{B}T\right)}^{\frac{3}{2}}{K}_B$

$=5.0306·\frac{2}{3}·{\left(\frac{2\pi m}{h^2}\right)}^{\frac{3}{2}}V·{\left(K_{B}T\right)}^{\frac{3}{2}}{K}_B$

$S=3.3537{\left(\frac{2\pi m}{h^2}\right)}^{\frac{3}{2}}V·{\left(K_{B}T\right)}^{\frac{3}{2}}{K}_B$

Step 3. To find the Helmholtz energy 

We have,

$F=U-TS$

$=2.0122{\left(\frac{2\pi m}{h^2}\right)}^{\frac{3}{2}}V·{\left(K_{B}T\right)}^{\frac{5}{2}}-T\left(5.03056\times \frac{2}{3}{\left(\frac{2\pi m}{h^2}\right)}^{\frac{3}{2}}V·{\left(K_{B}T\right)}^{\frac{3}{2}}{K}_B\right)$

$={\left(\frac{2\pi m}{h^2}\right)}^{\frac{3}{2}}V·{\left(K_{B}T\right)}^{\frac{5}{2}}[2.0122-3.3537]$

$=-1.3415{\left(\frac{2\pi m{K}_{B}T}{h^2}\right)}^{\frac{3}{2}}V·\left(K_{B}T\right)$

$\therefore F=-1.3415{\left(\frac{2\pi m}{h^2}\right)}^{\frac{3}{2}}V·{\left(K_{B}T\right)}^{\frac{5}{2}}$

Step 4. To find the pressure of a Bose gas

We have,

$P=-{\left(\frac{\partial F}{\partial V}\right)}_{N,T}$

$=-\frac{\partial }{\partial V}\left[-1.3415{\left(\frac{2\pi m}{h^2}\right)}^{\frac{3}{2}}V·{\left(K_{B}T\right)}^{\frac{5}{2}}\right]$$=1.3415{\left(\frac{2\pi m}{h^2}\right)}^{\frac{3}{2}}{\left(K_{B}T\right)}^{\frac{5}{2}}\left(1\right)$$\therefore P=1.3415{\left(\frac{2\pi m}{h^2}\right)}^{\frac{3}{2}}{\left(K_{B}T\right)}^{\frac{5}{2}}$

Step 5. Examining the expression of pressure

We get to know that pressure is independent of volume and a function of temperature '$T$' only as for condensing gas.

Further reduction in the volume would condense more particles in ground state in the limit$T<T_c$