Issue deriving the third law of thermodynamics

[laser1]

Homework Statement

Given $\frac{1}{kT}=\frac{d\ln(\Omega)}{dE}$, derive an expression for entropy

Relevant Equations

$dU=dQ+dW$

Okay, so we have that $$dU = \left( \frac{\partial U}{\partial V} \right)_S dV + \left( \frac{\partial U}{\partial S} \right)_V dS$$ And comparing that to the first law, we get that $$T=\left(\frac{\partial U}{\partial S}\right)_V$$. Comparing expressions of $T$, $$\left(\frac{dE}{dk\ln(\Omega)}\right)=\left(\frac{\partial U}{\partial S}\right)_V$$, it ALMOST seems like $S=k\ln(\Omega)$. But I have one doubt about the constant volume... why isn't this an issue?

 

[ergospherical]

This is a weirdly stated question. What exactly are you being asked to prove? Typically the entropy is defined by $S(E) = k \log \Omega(E)$ and the temperature is defined by $\tfrac{1}{T} = \tfrac{\partial S}{\partial E}$.

 

[laser1]

This is a weirdly stated question. What exactly are you being asked to prove? Typically the entropy is defined by $S(E) = k \log \Omega(E)$ and the temperature is defined by $\tfrac{1}{T} = \tfrac{\partial S}{\partial E}$.

This is from Blundell and Blundell.

 

[Philip Koeck]

This is from Blundell and Blundell.

View attachment 352047

That's only one of many possible definitions of entropy.
If you add any function of V alone to the right hand side of (14.36) it still works out.

 

[laser1]

That's only one of many possible definitions of entropy.
If you add any function of V alone to the right hand side of (14.36) it still works out.

I thought $S=K_B\ln\Omega$ though. If $S=B(V)\cdot K_B\ln\Omega$ instead where $B$ is a function of $V$, surely they are not the same.

 

[Philip Koeck]

I thought $S=K_B\ln\Omega$ though. If $S=B(V)\cdot K_B\ln\Omega$ instead where $B$ is a function of $V$, surely they are not the same.

Add, not multiply!

 

[Philip Koeck]

That's only one of many possible definitions of entropy.
If you add any function of V alone to the right hand side of (14.36) it still works out.

@laser1, does that solve your inititial problem?

 

[laser1]

@laser1, does that solve your inititial problem?

yeah thanks, basically having V constant changes nothing as it gets wiped out when the derivative is taken

 

[Philip Koeck]

This is from Blundell and Blundell.

View attachment 352047

I'm wondering whether (14.35) ( or (4.7) ) in Blundell is accurate.
Maybe there should be a partial derivative there too.
Have you checked how they come up with (4.7)?

Also I don't think the homework from the first post should ask for a derivation.
What Blundell gives is just a justification. They find an expression that's compatible with classical thermodynamics, but they don't derive the only possible expression.

 

[laser1]

I'm wondering whether (14.35) ( or (4.7) ) in Blundell is accurate.
Maybe there should be a partial derivative there too.
Have you checked how they come up with (4.7)?

Also I don't think the homework from the first post should ask for a derivation.
What Blundell gives is just a justification. They find an expression that's compatible with classical thermodynamics, but they don't derive the only possible expression.

I have attached the part of the book. I mean, the two systems are fixed, so perhaps we can say constant volume?

Regarding the derivation part, this was the first time in the book that $SK_B\ln(\Omega)$ was mentioned! btw, as for the question, I did not find it in the book/lecture notes, I made it up myself. Which is perhaps why when seeing it for the first time and being familiar with $SK_B\ln(\Omega)$, one would be confused.

 

[Philip Koeck]

, as for the question, I did not find it in the book/lecture notes, I made it up myself. Which is perhaps why when seeing it for the first time and being familiar with $SK_B\ln(\Omega)$, one would be confused.

I see. I wouldn't ask for a derivation. I don't think you can derive statistical entropy from something else.

 

[Philip Koeck]

I have attached the part of the book. I mean, the two systems are fixed, so perhaps we can say constant volume?

I agree. They state that the total energy is fixed, which includes that the system is not doing any work on the surroundings. So I would just assume the simplest, i.e. both volumes constant.
I think they should use partial derivatives at constant volume in the whole argument.