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