Thermodynamics of the free and independent electron gas

[amjad-sh]

Homework Statement

(a) Deduce from the thermodynamics identities:

$c_v=(\frac {\partial u}{\partial T})_n=T(\frac{\partial s}{\partial T})_n$
and equations: $f(ε)=\frac {1}{e^{ε-μ/k_BT}+1}$
and $u=\int \frac{1}{4π^3}ε(k)f(ε(k)) \, {d \vec k}$
and from the third law of thermodynamics(s→0 asT→0)
that the entropy density is given by:
$s=-K_B\int \frac{1}{4π^3}[f\ln f +(1-f)\ln(1-f)] \, {d \vec k}$

Homework Equations

The Attempt at a Solution

[/B]I tried to calculate $(\frac {\partial u}{\partial T})_n$ first

I used $\frac{\partial u}{\partial T}=\frac{\partial u}{\partial \beta}\frac{\partial \beta}{\partial T}$
where$\frac{\partial \beta}{\partial T}= \frac{-1}{K_BT^2}$
now $\frac{\partial u}{\partial \beta}=\frac{\partial }{\partial \beta}\int \frac{1}{4π^3}ε(k)\frac {1}{e^{ε-μ/k_BT}+1} \, {d \vec k}$
then$\frac{\partial u}{\partial \beta}=\int \frac{1}{4π^3}ε\frac {(ε-μ)e^{(ε-μ)\beta}}{(e^{(ε-μ)/k_BT}+1)^2} \, {d \vec k}$
$\frac{\partial u}{\partial T}=-K_B\int \frac{1}{4π^3}ε\frac {(\beta)^2(ε-μ)e^{(ε-μ)\beta}}{(e^{(ε-μ)/k_BT}+1)^2} \, {d \vec k}$

I reached to this part and then couldn't continue further.

 

[BigJDubs]

So I thought of using the second equation $f(ε)=\frac {1}{e^{ε-μ/k_BT}+1}$ which can be written as$f(ε)=\frac {1}{e^{(ε-μ)\beta)+1}$Now using the first equation, we have $c_v=(\frac {\partial u}{\partial T})_n=T(\frac{\partial s}{\partial T})_n=K_B\int \frac{1}{4π^3}[f\ln f +(1-f)\ln(1-f)] \, {d \vec k}$Is this the correct approach? Thank you