Determine Joule-Kelvin coefficient for gas given equations of state

[Like Tony Stark]

Homework Statement

The equations of state of a gas are $P=\frac{U}{V}$ and $T=3B(U²/NV)^{1/3}$. Determine $\alpha$ and $\mu$.

Relevant Equations

$U$: internal energy; $T$: temperature; $\mu$: Joule-Kelvin coefficient; $B$: positive constant; $V$: volume; $N$: number of moles; $\alpha$: coefficient of thermal expansion; $P$: pressure; $c_P$: heat capacity at constant pressure.

Hi

$\mu=\frac{\alpha TV–V}{N c_P}$. So, firstly, I have to calculate $\alpha$ and $c_P$. So $\alpha=\frac{1}{V} \frac{\partial V}{\partial T}$ at constant $P$. I can write $U=PV$, then I replace it in the equation of $T$, solve for $V$ and then I differentiate with respect to $T$.

Then, $c_P=\frac{T}{N} \frac{\partial S}{\partial T}$ at constant $P$. Do I have to find the fundamental equation for $S$ using Euler and Gibbs-Duhem relations, or is there an easier way?

 

[TSny]

Then, $c_P=\frac{T}{N} \frac{\partial S}{\partial T}$ at constant $P$. Do I have to find the fundamental equation for $S$ using Euler and Gibbs-Duhem relations, or is there an easier way?

Try using $c_P = \frac 1 N \left( \frac{dQ}{dT} \right)_P$ along with the first law.

 

[Like Tony Stark]

Try using $c_P = \frac 1 N \left( \frac{dQ}{dT} \right)_P$ along with the first law.

Thanks! I have arrived to $c_P=\frac{2T^2}{9B^3P}$ and $\alpha=\frac{NT^2}{9B^3P^2V}$. But when I replace this identities in the expression for $\mu$ I get $\mu=0$

 

[TSny]

Thanks! I have arrived to $c_P=\frac{2T^2}{9B^3P}$ and $\alpha=\frac{NT^2}{9B^3P^2V}$.

I believe these are correct. They will simplify nicely if you use $P=\frac{U}{V}$ and $T=3B(U^2/NV)^{1/3}$ to express $B^3$ in terms of $P$, $V$, $T$, and $N$.

But when I replace this identities in the expression for $\mu$ I get $\mu=0$

I don't get $\mu = 0$.