Specified equation of state from heat capacity

[Dazed&Confused]

Homework Statement

The constant-volume heat capacity of a particular simple system is $$c_v = AT^3$$
where A is a constant. In addition the equation of state is known to be of the form
$$(v-v_0)p = B(T)$$
where $B(T)$ is an unspecified function of T. Evaluate the permissible functional form of $B(T)$.

Homework Equations

3. The Attempt at a Solution [/B]

So we have
$${\frac{\partial S}{\partial v}} _U = \frac{B(T)}{T(v-v_0)}$$
and
$${\frac{\partial S}{\partial T}}_v = AT^2$$
I apply the first derivative to the second equation and vice versa. I equate and get
$$\frac{\partial}{\partial T} \left ( \frac{B(T)}{T} \right) \frac{1}{v-v_0} = 2A T {\frac{\partial T}{\partial v}}_u$$
The rightmost term can be rewritten as
$${\frac{\partial T}{\partial v}}_u = -\frac{ {\frac{\partial u}{\partial v}}_T}{{\frac{\partial u}{\partial T}}_v} = -\frac{T \frac{\partial s}{\partial v}_T - p}{c_v}= -\frac{T \frac{\partial p}{\partial T}_v - p}{c_v}$$ so that
$$\frac{2}{T^2} \left [ -\frac{T B'(T)}{v-v_0} + \frac{B(T)}{v-v_0} \right] = \frac{\partial}{\partial T} \left ( \frac{B(T)}{T} \right) \frac{1}{v-v_0}$$
which I solve for $B(T)$ and get $B(T) = ET$ with $E$ a constant. Now
$$c_p = c_v + \frac{Tv\alpha^2}{\kappa_T}$$
where $\alpha$ is the isobaric compressability with temperature and $\kappa_T$ is the isothermal compressability with pressure. Thus with
$$v = \frac{B(T)}{p} + v_0$$
this should equal (I think) to
$$c_v + T \left(\frac{\partial V }{\partial T}_p\right)^2\left/\right. \left(\frac{\partial V}{\partial p}\right)_T =c_v+ T\frac{B'(T)^2}{B(T)}$$
which in my case would be simply $c_v + E$. The answers give
$$c_v + (T^3/DT + E)$$.
Now when solving for their $B(T)$ I get a very complicated expression. I do not see where my mistake lies, except I am not 100% sure if the two partial derivatives I had at the beginning commute.

 

[Chestermiller]

I think that it might be easier to start with the Maxwell relationship: $$\left(\frac{\partial S}{\partial v}\right)_T=\left(\frac{\partial p}{\partial T}\right)_v$$

 

[Dazed&Confused]

That's definitely a good idea! Integrating $c_v/T$ we find
$$s = \frac{AT^3}{3} +f(v)$$
for some f. Then with the Maxwell relation
$$f'(v) = \frac{B'(T)}{v-v_0}$$
which means that $B'(T) = E$ for some E. So this confirms what I found before (much more easily).

 

[Chestermiller]

I would have done it a little differently: $$dS=AT^2dT+\frac{B'(T)}{v-v_0}dv$$So,
$$\frac{\partial^2 S}{\partial T \partial v}=\frac{\partial^2 S}{\partial v \partial T}=\frac{\partial}{\partial v}(AT^2)=0=\frac{\partial }{\partial T}\left(\frac{B'(T)}{v-v_0}\right)=\frac{B''(T)}{v-v_0}$$So, $$B''(T) = 0$$So, B(T) is a linear function of T.

 

[Dazed&Confused]

Maybe it was too late for me to be posting before but my first method misses the extra constant you'd get.