# Calculating temperature from entropy

1. Mar 10, 2012

### SoggyBottoms

Consider N >> 1 particles that can either be in groundstate $\epsilon_0$ or excited state $\epsilon_1$ and are thermally isolated, so the internal energy is fixed at $U = (N - n) \epsilon_0 + n \epsilon_1$. We want to calculate the temperature of this system.

This is how I attempt it: First calculate the multiplicity. For two particles, the possible microstates are -2, 0, 0, 2 and for 3 particles it's -3, -1, 1, 3, etc, so that the multiplicity becomes:
$$\Omega(n) = \frac{N!}{\frac{N + n}{2}! \frac{N - n}{2}!}$$
The entropy is, using the Stirling approximation, $S = k_B \ln{\Omega(n)} = k_B(N \ln N - \frac{N + n}{2} \ln{\frac{N + n}{2}} - \frac{N - n}{2} \ln{\frac{N - n}{2}})$.

The temperature we can calculate from $T = \left(\frac{\partial S}{\partial U}\right)^{-1}$. This is where I don't know how to continue (not sure I am on the right track in the first place).