Specified equation of state from heat capacity

[Dazed&Confused]

Homework Statement

The constant-volume heat capacity of a particular simple system is <br /> c_v = AT^3
where A is a constant. In addition the equation of state is known to be of the form
<br /> (v-v_0)p = B(T)<br />
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
<br /> {\frac{\partial S}{\partial v}}<br /> _U = \frac{B(T)}{T(v-v_0)}
and
<br /> {\frac{\partial S}{\partial T}}_v = AT^2<br />
I apply the first derivative to the second equation and vice versa. I equate and get
<br /> \frac{\partial}{\partial T} \left ( \frac{B(T)}{T} \right) \frac{1}{v-v_0} = 2A T {\frac{\partial T}{\partial v}}_u<br />
The rightmost term can be rewritten as
<br /> {\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
<br /> \frac{2}{T^2} \left [ -\frac{T B&#039;(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}<br />
which I solve for B(T) and get B(T) = ET with E a constant. Now
<br /> c_p = c_v + \frac{Tv\alpha^2}{\kappa_T}<br />
where \alpha is the isobaric compressability with temperature and \kappa_T is the isothermal compressability with pressure. Thus with
<br /> v = \frac{B(T)}{p} + v_0<br />
this should equal (I think) to
<br /> 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&#039;(T)^2}{B(T)}<br />
which in my case would be simply c_v + E. The answers give
<br /> c_v + (T^3/DT + E)<br />.
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
<br /> s = \frac{AT^3}{3} +f(v)
for some f. Then with the Maxwell relation
<br /> f&#039;(v) = \frac{B&#039;(T)}{v-v_0}<br />
which means that B&#039;(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.