# Does $U(S,V,N) = \alpha e^{\frac{S}{N c_v}} V^{\frac{c_v-c_p}{c_v}} N^{\frac{c_p}{c_v}}$ really imply zero entropy?

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?

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.