# Combined law for an open system

1. Jan 29, 2008

### ehrenfest

1. The problem statement, all variables and given/known data
I am confused about this solution. The question explicitly says that the systems are allowed energy exchange, so why is energy held fixed in the equation after "therefore"?

2. Relevant equations

3. The attempt at a solution

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2. Jan 30, 2008

### siddharth

The solution does not claim that the energy of each system is held fixed. In the equation, the parameter $\zeta$ doesn't represent the the energy of the state. The relation,

$$\zeta_A = \zeta_B$$

is obtained from the fact that at equilibrium, the closed system is at its most probable state (ie, the entropy is maximum).

Refer to section 1.7 in "Elementary Statistical Physics" by Kittel or section 3.4 in Reif's statistical physics text for a discussion.

Last edited: Jan 30, 2008
3. Jan 30, 2008

### ehrenfest

The equation after "since" is always true, correct?

I am just confused about how they went from that to the next equation with the partial derivatives. It seems like they just divided each side by dN_A and then changed the total derivatives to partial derivatives and then added the subscripts V,E which I understand are redundant. I am just confused about how they changed the total derivatives to partial derivatives.

4. Jan 31, 2008

### siddharth

Yes, that's the combined law for an open system.

It's just a differential, right? For example, if S is a function of N,V,E, ie S=S(N,V,E) then

$$dS = \frac{\partial S}{\partial N} dN + \frac{\partial S}{\partial V} dV + \frac{\partial S}{\partial E} dE$$

and at constant V,E

$$dS = \frac{\partial S}{\partial N} dN$$

5. Jan 31, 2008

### ehrenfest

Why would the partial $$\left(\frac{\partial{E_A}}{\partial{N_A}}\right)_{V,E}$$ not be identically 0 because the subscript E means energy is not changing?

6. Jan 31, 2008

### kdv

The total energy does not change, but the energy of the two subsytems A and B may vary.

7. Jan 31, 2008

### ehrenfest

I don't see how that explains the equation after "therefore".

8. Jan 31, 2008

### siddharth

Why not? The notation is to tell you that V and E are held constant. I think it follows directly.

9. Apr 24, 2010

### cracketabhi

Re: thermodynamics

dudes... the question asked the minimum value of.... so . if u've got the answer after manipulation, u must have ASSUMED the minimum value condition in the course of the problem and thus the answer...

@ others - wat say?

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