# $S(E,V)$ for an ideal gas

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.

## 2 Answers

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: $$S(E,V) = nR \ln{V}+f(E) = nR \ln{V} + C_V \ln{E}+ \text{constant}$$

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.

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