Derive the entropy of an ideal gas

[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?

 

[Andrew Mason]

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?

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 S_T represents the entropy of the gas at temperature T.

AM

 

[kyle8921]

The next step would be to use the definition of entropy, which is given by:

S = \frac{Q}{T}

Where Q is the heat added to the system and T is the temperature. In this case, since the molar specific heat at constant volume is constant, we can use the relation:

C_v = \frac{\partial_Q}{\partial_T}

where C_v is the molar specific heat at constant volume.

We can then rearrange the above equation to get:

dQ = C_v dT

Substituting this into the equation for entropy, we get:

dS = \frac{C_v}{T}dT

We can then integrate both sides from the initial state (S_0) to the final state (S) to get:

\Delta S = \int_{S_0}^{S} dS = \int_{T_0}^{T} \frac{C_v}{T}dT

This gives us the entropy change for an ideal gas when its molar specific heat at constant volume is constant.