Statistical Physics: Solving C_P & Problem 4

[ehrenfest]

[SOLVED] statistical physics

Homework Statement

http://ocw.mit.edu/NR/rdonlyres/Physics/8-044Spring-2004/85482B93-6A5E-4E2F-ABD2-E34AC245396C/0/ps5.pdf
I am working on number 3 part a.
I am trying to calculate C_P.
From the first law of thermodynamics: dQ = dU -dW = dU +PdV (does anyone know how to write the inexact differential d in latex?).
And we know that C_p \equiv \frac{dQ}{dT}_P. But I don't see how to get an explicit expression for dQ. Should I expand dU and dV in terms of the other independent variables or what? What variables should I choose to be independent?

EDIT: I actually need help with Problem 4 also. I can integrate (dS/dA)_T w.r.t A and get that
S(A,T) = -\frac{NkT}{A-b}+\frac{aN^2}{A^2} +f(T) but then I have no idea how to find f(T)!

Homework Equations

The Attempt at a Solution

 

[Mapes]

A good definition for c_P is

c_P=\frac{1}{N}\left(\frac{\partial H}{\partial T}\right)_P=\frac{1}{N}\left[\frac{\partial (U+PV)}{\partial T}\right]_P

Recall that for an ideal gas dU=Nc_V\,dT.

Once you find c_P you should be able to integrate your equation for \delta Q as

Q=\int Nc_P\,dT

 

[Mapes]

Regarding your second question: try inverting \left(\frac{\partial T}{\partial \mathcal{S}}\right)_A and using

d\mathcal{S}=\left(\frac{\partial \mathcal{S}}{dT}\right)_A dT+\left(\frac{\partial \mathcal{S}}{dA}\right)_T dA

 

[ehrenfest]

Regarding your second question: try inverting \left(\frac{\partial T}{\partial \mathcal{S}}\right)_A and using

d\mathcal{S}=\left(\frac{\partial \mathcal{S}}{dT}\right)_A dT+\left(\frac{\partial \mathcal{S}}{dA}\right)_T dA

Is it in general true that

1/\left(\frac{\partial T}{\partial \mathcal{S}}\right)_A = \left(\frac{\partial \mathcal{S}}{\partial T} \right)_A

?

 

[Mapes]

In my experience, it always works in thermodynamics. Outside engineering it may be risky. Mathematicians, want to weigh in?

 

[ehrenfest]

In my experience, it always works in thermodynamics. Outside engineering it may be risky. Mathematicians, want to weigh in?

Yes, it would really help me if a mathematician posted exactly when that is true.

 

[ehrenfest]

anyone?

 

[Hurkyl]

I assume that notation means you're "Treating S (resp. T) as a function of A and T (resp. S), and differentiating, holding A as constant"?

Or more precisely, S, T, and A are functions of your state \xi, and you have a relationship

S(\xi) = f( T(\xi), A(\xi) )

and you're interested in f_1(T(\xi), A(\xi)), the partial derivative of this function with respect to the first place, evaluated at (T(\xi), A(\xi)?


Well, for any particular value of A, this is just ordinary, one variable calculus -- let f_a denote the function defined by f_a(x) = f(x, a). If f_a is invertible, then it's easy to find a relationship: just differentiate the identity x = f_a( f_a^{-1}(x)).


For a more geometric flavor, if restricting to a subspace where A is constant means that the differentials dS and dT are proportional (i.e. dS = f dT for some f), then it's just a matter of algebra to express dT in terms of dS where possible.

 

[ehrenfest]

So it is in general true (as long as we assume differentiability of the function and its inverse and do not divide by 0)! Yay!