(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

They gave me a system with [itex]u=\frac{3}{2} pv[/itex] and [itex]u^{1/2}=bTv^{1/3}[/itex]

So i know that [itex]p=\frac{2b^{2}T^{2}}{3v^{1/3}}[/itex]

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

2. Relevant equations

[itex]\lim_{T\to0}S=0[/itex]

[itex]\left(\frac{\partial u}{\partial v}\right)_{T}=T\left(\frac{\partial s}{\partial v}\right)_{T}-p[/itex]

3. The attempt at a solution

So what i did is

[itex]\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}[/itex]

So we end with

[itex]ds=\frac{4b^{2}T}{3v^{1/3}}dv[/itex]

so ..

[itex]s=2b^{2}Tv^{2/3}+f(T)[/itex]

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

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 T

I also tried using the differential function of [itex]S(T,v)[/itex] like this

[itex]ds=\underbrace{\left(\frac{\partial s}{\partial T}\right)_{v}}_{c_{v}/T}dT+\left(\frac{\partial s}{\partial v}\right)_{T}dv[/itex]

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

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

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# Homework Help: Calculate the molar entropy

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