Statistical physiics Problem 5.5

  • Thread starter ehrenfest
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In summary, the equations state that the internal energy is equal to the work done on the system minus the change in potential energy. The work done is equal to the force applied times the distance traveled.
  • #1
ehrenfest
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Homework Statement


http://ocw.mit.edu/NR/rdonlyres/Physics/8-044Spring-2004/85482B93-6A5E-4E2F-ABD2-E34AC245396C/0/ps5.pdf

I am stuck on Problem 5 part a. They say that the relevant state variables are H,M,T, and U. Obviously the first law of thermodynamics still holds: dU = dW+dQ (does anyone know how to make inexact differentials in latex)? But does dW = -PdV here? P and V were not among the state variables they talked about so does that really make sense? How do I proceed?


Homework Equations





The Attempt at a Solution

 
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  • #2
anyone?
 
  • #3
[itex]dW=H\,dM[/itex] and [itex]U=TS+MH[/itex]. Work terms always consist of a generalized force (an intensive quantity) and a generalized displacement (an extensive quantity). Examples: force x distance, magnetic field x magnetization, electric field x polarization, surface energy x area, stress x strain, etc.

In this problem [itex]P\,dV[/itex] work is evidently assumed to be negligible compared to [itex]M\,dH[/itex] work. You can tell because the problem states that there are only two independent variables (recall our https://www.physicsforums.com/showthread.php?p=1645661#post1645661").
 
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  • #4
OK, I see why [itex]dU = \delta Q +HdM[/itex]. But why is U = TS+MH true? I am trying to express [itex]C_M \equiv \left(\frac{\delta Q}{dT} \right)_M[/tex] as a derivative of the internal energy. Can you give me a hint how to do that?
 
  • #5
From what you've written, it looks like you can conclude that [itex]
C_M \equiv \left(\frac{\partial U}{\partial T} \right)_M
[/itex].

In general, [itex]
U=TS-PV+\sum\mu_i N_i +FL+ MH+EP+\gamma A+\sigma V\epsilon\dots
[/itex] where the terms represent the work terms I listed above. This is called the Euler form of the fundamental relation, if you want to find more information about it. Callen's Thermodynamics is a good reference.
 

1. What is Statistical Physics Problem 5.5?

Statistical Physics Problem 5.5 is a specific problem that falls under the field of statistical physics, which uses statistical methods to describe and analyze the behavior of large systems of particles. Problem 5.5 is just one example of a problem that can be solved using statistical physics principles.

2. What is the purpose of solving Statistical Physics Problem 5.5?

The purpose of solving Statistical Physics Problem 5.5 is to gain a deeper understanding of how statistical methods can be applied to complex systems, and to see how these methods can be used to make predictions about the behavior of these systems.

3. What concepts are involved in solving Statistical Physics Problem 5.5?

Solving Statistical Physics Problem 5.5 involves concepts such as statistical mechanics, thermodynamics, and probability theory. These concepts are used to analyze the behavior of particles in a system and make predictions about their behavior.

4. Is Statistical Physics Problem 5.5 a difficult problem to solve?

The difficulty of solving Statistical Physics Problem 5.5 depends on the individual's knowledge and understanding of the concepts involved. For someone with a strong background in statistical physics, it may not be difficult. However, for someone with limited knowledge in this field, it may be more challenging.

5. How can the solution to Statistical Physics Problem 5.5 be applied in real life?

The solution to Statistical Physics Problem 5.5 may have real-life applications in fields such as materials science, engineering, and thermodynamics. By understanding the behavior of particles in a system, predictions can be made about how these systems will behave in different conditions, which can be useful in designing and improving various technologies.

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