According to this question, for some ideal gas

$$U(S,V,N) = \alpha e^{\frac{S}{N c_v}} V^{\frac{c_v-c_p}{c_v}} N^{\frac{c_p}{c_v}}$$

From this,

$$T = \frac{\partial U}{\partial S} = \frac{1}{Nc_v}U \implies TS = \frac{S}{Nc_v}U$$ $$-P = \frac{\partial U}{\partial V} = \frac{c_v-c_p}{c_v}\frac{1}{V}U\implies -PV = \frac{c_v-c_p}{c_v}U$$ $$\mu = \frac{\partial U}{\partial N} = \frac{c_p}{c_v}\frac{1}{N}U\implies \mu N = \frac{c_p}{c_v}U$$

So, on one hand

$$TS -PV+\mu N= \left(\frac{S}{Nc_v} + \frac{c_v-c_p}{c_v} + \frac{c_p}{c_v}\right)U = \left(\frac{S}{Nc_v} + 1 \right)U$$ On the other hand, by Euler's theorem on homogeneous functions, $$TS -PV+\mu N=U$$ which implies

$$\frac{S}{Nc_v} + 1 =1$$

that is, $$S=0$$

Where is the error?


1 Answer 1


In differentiation $\partial_N$ you didn't act on $\exp\left(\frac{S}{N c_v}\right)$.

If you differentiate properly you'll get: $$\mu = \frac{\partial U}{\partial N}=\frac{c_p}{c_v} \frac{U}{N} - \frac{S}{N c_v} \frac{U}{N},$$ where the second term cancels the problematic $\frac{S}{N c_v}$ term and yields the proper expression for the internal energy.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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