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Statistical physics

  • Thread starter ehrenfest
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  • #1
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[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: [itex]dQ = dU -dW = dU +PdV[/itex] (does anyone know how to write the inexact differential d in latex?).
And we know that [itex]C_p \equiv \frac{dQ}{dT}_P[/itex]. 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
[tex]S(A,T) = -\frac{NkT}{A-b}+\frac{aN^2}{A^2} +f(T)[/tex] 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

  • #2
Mapes
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A good definition for [itex]c_P[/itex] is

[tex]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[/tex]

Recall that for an ideal gas [itex]dU=Nc_V\,dT[/itex].

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

[tex]Q=\int Nc_P\,dT[/tex]
 
  • #3
Mapes
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Regarding your second question: try inverting [itex]\left(\frac{\partial T}{\partial \mathcal{S}}\right)_A[/itex] and using

[tex]d\mathcal{S}=\left(\frac{\partial \mathcal{S}}{dT}\right)_A dT+\left(\frac{\partial \mathcal{S}}{dA}\right)_T dA[/tex]
 
  • #4
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Regarding your second question: try inverting [itex]\left(\frac{\partial T}{\partial \mathcal{S}}\right)_A[/itex] and using

[tex]d\mathcal{S}=\left(\frac{\partial \mathcal{S}}{dT}\right)_A dT+\left(\frac{\partial \mathcal{S}}{dA}\right)_T dA[/tex]
Is it in general true that

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

?
 
  • #5
Mapes
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In my experience, it always works in thermodynamics. Outside engineering it may be risky. Mathematicians, want to weigh in?
 
  • #6
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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.
 
  • #7
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anyone?
 
  • #8
Hurkyl
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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 [itex]\xi[/itex], and you have a relationship

[tex]S(\xi) = f( T(\xi), A(\xi) )[/tex]

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


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


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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  • #9
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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!
 

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