# Statistical physics

[SOLVED] statistical physics

1. Homework Statement
http://ocw.mit.edu/NR/rdonlyres/Physics/8-044Spring-2004/85482B93-6A5E-4E2F-ABD2-E34AC245396C/0/ps5.pdf [Broken]
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)!
2. Homework Equations

3. The Attempt at a Solution

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## Answers and Replies

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

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

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

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.

anyone?

Hurkyl
Staff Emeritus
Science Advisor
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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.

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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!