# Derive the entropy of an ideal gas

1. Oct 22, 2006

### Lee

THe Question asks 'Derive the entropy of an ideal gas when its molar specific heat at constant volume is constant.'

So I've taken

$$\Delta S = \int_{S_0}^{S} dS = \int_{T_0}^{T} \frac{\partial_S} {\partial_V} dT + \int_{V_0}^{V} \frac{\partial_S}{\partial_V} dV$$

in this context what would be the next best step?

2. Oct 22, 2006

### Andrew Mason

If the specific heat remains constant at all temperatures, then it is possible to integrate from temperature 0 to T.

Since $dQ = TdS = dU + PdV = nC_vdT + PdV$ at constant volume $nC_vdT = TdS$

so:

$$\int_0^T dS = \int_0^T nC_v dT/T = S_T - S_0$$

If you let the entropy of the gas at 0 K be 0: $S_0 = 0$, then ST represents the entropy of the gas at temperature T.

AM