fraggedmemory said:
I'm not sure about the sentence "all variables and given/known data".
fraggedmemory said:
Use the Maxwell relations and the Euler chain relation to express (ds/dt)p in terms of the heat capacity Cv = (du/dt)v. The expansion coefficient alpha = 1/v (dv/dt)p, and the isothermal compressibility Kt = -1/v (dV/dp)T. Hint. Assume that S= S(p,V)
Homework Equations
dQ(rev) = Tds
The maxwell relations
Euler Chain relation
In fact I succeed in order to obtain a relation between:
[tex]\left(\frac{\partial s}{\partial T}\right)_p[/tex]
and [tex]c_v, \alpha, k_T, T, v[/tex] just following the mapes's hint to consider S as a function of T and V. Where minuscle letter for extensive quantity means: "this quantity is molar", and all transformation are intended to involve a costant number of molecules.
In fact:
[tex]\left(\frac{\partial s}{\partial T}\right)_p = \left(\frac{\partial s}{\partial T}\right)_v + \left(\frac{\partial s}{\partial v}\right)_T \left(\frac{\partial v}{\partial T}\right)_T[/tex]
The Maxwell's relation following from [tex]d(-p dv - sdT)=0[/tex] tell us:
[tex]\left(\frac{\partial s}{\partial v}\right)_p= \left(\frac{\partial p}{\partial T}\right)_v[/tex]
Now the Euler's chain rule give us the link between the first derivative in second addend, the compressibility and the thermal expansion coefficient. In fact:
[tex]\left(\frac{\partial p}{\partial T}\right)_v \left(\frac{\partial T}{\partial v}\right)_p\left(\frac{\partial v}{\partial p}\right)_T = -1[/tex]
and:
[tex]\left(\frac{\partial T}{\partial v}\right)_p = 1/\left(\frac{\partial v}{\partial T}\right)_p[/tex]
so that:
[tex]\left(\frac{\partial p}{\partial T}\right)_v = \frac{\alpha}{k_T}[/tex]
In order to complete the derivation we need to use the given alternative definition of specific heat:
[tex]c_v = T \left(\frac{\partial s}{\partial T}\right)_v = \left(\frac{\partial u}{\partial T}\right)_v[/tex]
Which follow from:
[tex]T dS = dU + p dV[/tex]
So obtaining:
[tex]\left(\frac{\partial s}{\partial T}\right)_p = \frac{c_v}{T} + \frac{\alpha^2 v}{k_T}[/tex]
is this what was required?
fraggedmemory said:
The Attempt at a Solution
Alright, my attempts at this involved trying find common partial derivatives from the information already given. I couldn't find anything. But then looking at the hint I thought that there might be a way to express the
change in entropy with respect to pressure and volume. I get this ds = (dU + PdV)/T assuming constant pressure. I am really not sure what I am suppose to do. I especially don't get what the expansion coefficient and thermal compressibility has to do with anything, but that might be because I can't see the big picture with this problem.
A step by step explanation would be greatly appreciated.