1
$\begingroup$

So, $$dS = \frac{dE}{T}+\frac{P}{T}$$ For an ideal gas, the above relation reduces to, $$dS = \frac{C_VdE}{E}+\frac{nR}{V}$$ Integrating both sides, I can obtain, $$S(E,V) = C_V\ \ln{E}+nR\ \ln{V}+\textrm{constant}$$ That's fine for me, but I am struggling to find same relation from the partial derivative of $S(E,V)$. Using, $$\left ( \frac{\partial S }{\partial V} \right )_E = \frac{P}{T}= \frac{nR}{V}$$ Integrating both sides, I can obtain: $$S(E,V) = nR \ln{V}+f(E)$$ And using,$$\left ( \frac{\partial S }{\partial E} \right )_V = \frac{1}{T}= \frac{C_V}{E}$$ Again, integrating, I can obtain: $$S(E,V) = C_V \ln{E}+f(V)$$ So, comparing these two relations I can say: $$S(E,V) = nR \ln{V} + C_V\ln{E} $$ So, there is an extra constant term missing, which I can't produce with these two relations. I think that my conundrum is more due to Maths than the underlying physics in here. Would be very grateful if anyone can suggest, what's wrong I am doing in here. Thanks.

$\endgroup$
1
$\begingroup$

You can look at it like this: when evaluating $\left ( \frac{\partial S }{\partial E} \right )_V = \frac{C_V}{E}$ you should keep in mind what you already calculated: $$ \left ( \frac{\partial S }{\partial E} \right )_V = \left ( \frac{\partial\ (nR \ln{V}+f(E))}{\partial E} \right )_V = \left ( \frac{\partial\ nR \ln{V}}{\partial E} \right )_V + \left ( \frac{\partial f(E)}{\partial E} \right )_V = \left ( \frac{\partial f(E)}{\partial E} \right )_V = \frac{C_V}{E}$$ Because V is constant in the first partial derivative, the derivative is $0$. So after integrating you have information about $f(E)$ more so than about $S$. By integration you get: $$ f(E) = C_V \ln{E}+ \text{constant}$$ This constant can no longer depend on $V$ anymore because you defined $f$ to be only dependant on $E$. Filling in your value for $f$ in the equation you got earlier gives: $$ S(E,V) = nR \ln{V}+f(E) = nR \ln{V} + C_V \ln{E}+ \text{constant}$$

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

You have neglected constants of integration when comparing terms.

You have :

$$S(E,V)=nR\ln V +f(E)$$

and

$$S(E,V)=C_V\ln E +g(V)$$

You assume :

$$f(E)=C_V\ln E$$

but should say :

$$f(E)=C_V\ln E + K_1$$

for some constant $K_1$ and similarly for $g(V)$.

Doing that you get :

$$S(E,V)=nR\ln V + C_V\ln E + K_1$$

And hence your desired result.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ So, basically it should have been $S(E,V) = C_V \ln{E}+f(V)+K_1$ after integrating partial derivative? $\endgroup$ – Roshan Shrestha Apr 29 at 7:32
  • $\begingroup$ Sorry, I think I got your point now. $\endgroup$ – Roshan Shrestha Apr 29 at 7:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.