How Does the Third Law of Thermodynamics Apply in Calculating Molar Entropy?

[WarDieS]

Homework Statement

They gave me a system with $u=\frac{3}{2} pv$ and $u^{1/2}=bTv^{1/3}$
So i know that $p=\frac{2b^{2}T^{2}}{3v^{1/3}}$

And it says i have to use the third law of thermodynamics to obtain the integrating factor

Homework Equations

$\lim_{T\to0}S=0$

$\left(\frac{\partial u}{\partial v}\right)_{T}=T\left(\frac{\partial s}{\partial v}\right)_{T}-p$

The Attempt at a Solution

So what i did is

$\left(\frac{\partial u}{\partial v}\right)_{T}=T\left(\frac{\partial s}{\partial v}\right)_{T}-p\Rightarrow\left(\frac{\partial s}{\partial v}\right)_{T}=\frac{1}{T}\left(\frac{\partial u}{\partial v}\right)_{T}+\frac{p}{T}$

So we end with
$ds=\frac{4b^{2}T}{3v^{1/3}}dv$
so ..
$s=2b^{2}Tv^{2/3}+f(T)$

I know the solution is just $s=2b^{2}Tv^{2/3}$

But i don't know how to use the third law ot obtain the integrating factor and obtain the molar entropy without the uknown function of TI also tried using the differential function of $S(T,v)$ like this
$ds=\underbrace{\left(\frac{\partial s}{\partial T}\right)_{v}}_{c_{v}/T}dT+\left(\frac{\partial s}{\partial v}\right)_{T}dv$

But i don't know if $c_{v}$ its constant, also even if so, it still wrong.

I don't know how to really solve this, please help, and THX ! :D

 

[WarDieS]

All right i just solved it, so what i did is:

$dS=\frac{dU}{T}+\frac{p}{T}dV$

knowing that

$\frac{p}{T}=\frac{2}{3}\frac{b^{2}T}{v^{1/3}}$

and

$du=\frac{2 b^2 T^2}{3 v^{1/3}}dv$

so

$ds=\frac{2b^{2}T}{3v^{1/3}}dv$

to get finally

$s=2b^2 T v^{2/3}$

Still didn't use the third law to obtain the integrating factor, i am puzzled why i should use it