Statistical Mechanics: Two systems reaching an equilibrium temperature

  • #1
tanaygupta2000
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Homework Statement:
System A has N particles and infinite energy levels 0, ε, 2ε, 3ε, ... is kept in thermal contact with system B having N particles and two energy levels 0 and ε, till they reach equilibrium. The total energy of system A is Nε and that of system is 3Nε/4. What is the sign of equilibrium temperature?
Relevant Equations:
Partition function, Z = ∑exp[-E(j)/kT]
Total energy, E = NkT^2 d[ln(Z)]/dT
First I found partition functions of both the systems and hence total energies of them using above formulas.
Z(A) = (1 - e-ε/kT)-1 and Z(B) = (1 + e-ε/kT)
Then I equated these values to the given values of total energies.
I got:
For System A, T(A) = ε/kln(2) > 0
For System B, T(B) = -ε/kln(3) < 0
Now how do I find the equilibrium temperature when these systems are kept in thermal contact?
Do I have to take the average values of T(A) and T(B) ?
 

Answers and Replies

  • #2
36,247
13,302
The equilibrium doesn't have to be at the average temperature (simple counterexample: Ice and a bit of hot water). Total energy is conserved, you can find the temperature as function of the energy of the individual systems. The equilibrium is at the place where the total energy stays the same but both systems have the same temperature.
The problem only asks about the sign, however. Can the first system achieve a negative temperature? If not, you don't need to calculate anything.
 
  • #3
tanaygupta2000
204
14
The equilibrium doesn't have to be at the average temperature (simple counterexample: Ice and a bit of hot water). Total energy is conserved, you can find the temperature as function of the energy of the individual systems. The equilibrium is at the place where the total energy stays the same but both systems have the same temperature.
The problem only asks about the sign, however. Can the first system achieve a negative temperature? If not, you don't need to calculate anything.
Since ε is positive, the temperature, being a function of ε, of first system is positive while that of second is negative. Now how do I proceed next for equilibrium ?
 
  • #4
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Initially the temperature of the first is positive and the second is negative, yes, but that isn't helping much.
Now how do I proceed next for equilibrium ?
I wrote two approaches in my previous post. The first one is to explicitly calculate the equilibrium, the second one is just to determine its sign without calculations (or with minimal calculations).
 
  • #5
tanaygupta2000
204
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The equilibrium is at the place where the total energy stays the same but both systems have the same temperature.
Okay the equilibrium temperature is the temperature when the energies of systems A and B are equal.
So I need to find the temperature where E = Nε + 3Nε/4.
 
  • #6
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Okay the equilibrium temperature is the temperature when the energies of systems A and B are equal.
Why?
That's certainly not true in general.

Stop guessing around. I posted two approaches that you can follow, they will both lead to the answer.
 
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  • #7
tanaygupta2000
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Why?
That's certainly not true in general.

Stop guessing around. I posted two approaches that you can follow, they will both lead to the answer.
Sorry sir I'm not getting your approaches.
I'm having two systems, one with a positive temperature and other with negative. I'm just asking for a simple formula for obtaining the equilibrium temperature of these two.
 
  • #8
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Let's break it down step by step:

> Total energy is conserved
Can you write that as an equation?

> you can find the temperature as function of the energy of the individual systems.
You found the temperature at one specific energy, you can find it as function of energy in general.

> The equilibrium is at the place where the total energy stays the same but both systems have the same temperature.
That's one equation, the other one comes from the total energy. That gives you two equations and two unknowns so you can solve the problem.
 
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  • #9
tanaygupta2000
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By applying conservation of energy,
Nεe-ε/kT/(1-e-ε/kT) + Nεe-ε/kT/(1+e-ε/kT) = Nε + 3Nε/4
I am getting
T(ε) = -ε/k ln(7/4)
Is this correct ?
 
Last edited:
  • #10
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The answer is not correct. If you plug it into the first term of the first equation you end up with a negative energy which doesn't make sense.
 
  • #11
tanaygupta2000
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The problem only asks about the sign, however. Can the first system achieve a negative temperature? If not, you don't need to calculate anything.
First system is having positive temperature.
Second system is having negative temperature.
I need to determine only the 'sign' of equilibrium temperature.
I'm not getting a way how to do it.
I've tried every way I know.
Please help it's my assignment !
Its the only part of the only question out of 20 questions I'm having confusion.
 
Last edited:
  • #12
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It's really a tiny step from what I hinted towards already: The first system has a positive temperature at any energy. It cannot have a negative temperature. That means once you show that you are done: Whatever the equilibrium energy distribution might be, it will have a positive temperature in both systems.
 

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