Statistical Physics: Solving C_P & Problem 4

In summary, the conversation is about solving problems in statistical physics involving the calculation of C_P and finding an explicit expression for dQ. It also involves finding the integral of Nc_P and discussing the validity of a mathematical relationship. The participants consider the use of the first law of thermodynamics and the concept of functions to solve the problems.
  • #1
ehrenfest
2,020
1
[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: [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)!

Homework Equations





The Attempt at a Solution

 
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  • #2
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
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
Mapes said:
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
In my experience, it always works in thermodynamics. Outside engineering it may be risky. Mathematicians, want to weigh in?
 
  • #6
Mapes said:
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
anyone?
 
  • #8
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
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!
 

FAQ: Statistical Physics: Solving C_P & Problem 4

What is statistical physics?

Statistical physics is a branch of physics that applies statistical methods and concepts to study the behavior of large systems of particles, such as atoms and molecules. It aims to understand and predict the macroscopic properties of these systems based on the microscopic behavior of their constituent particles.

What is C_P in statistical physics?

C_P, or specific heat capacity, is a thermodynamic quantity that measures the amount of heat required to raise the temperature of a material by one degree. In statistical physics, it is used to describe the relationship between the internal energy of a system and its temperature.

How is C_P calculated in statistical physics?

C_P can be calculated using the formula C_P = (dQ/dT)_P, where dQ is the change in heat and dT is the change in temperature at constant pressure. This quantity can also be obtained from the slope of the temperature dependence of the internal energy of a system.

What is Problem 4 in statistical physics?

Problem 4 is a specific problem in statistical physics that involves finding the specific heat capacity of a system of particles. It may involve using mathematical models and equations to analyze the behavior of the system and make predictions about its properties.

What are some real-world applications of statistical physics?

Statistical physics has many practical applications, including in fields such as materials science, chemistry, and engineering. It is used to understand and predict the behavior of materials, such as the thermal conductivity of a substance or the phase transitions of a material. It is also applied in fields like biophysics to study the behavior of complex biological systems.

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