SL: Calculating entropy changes for ideal gases

Entropy calculations for ideal gases

For one mole of ideal gas undergoing a closed, reversible process:

$dU = \delta Q_{rev} + \delta W = \delta Q_{rev} - PdV$

Recall:

$dH = dU + d(PV) = dU + PdV + VdP$

Combining yields:

$dH = \delta Q_{rev} - PdV + PdV + VdP = \delta Q_{rev} + VdP$

$\delta Q_{rev} = dH - VdP$

Using $dH = c_PdT$ and $V = RT/P$ gives:

$\delta Q_{rev} = c_PdT - \frac{RTdP}{P}$

Dividing by $T$:

$dS = \frac{\delta Q_{rev}}{T} = \frac{c_PdT}{T} - \frac{RdP}{P}$

Integrating:

$\frac{\Delta S}{R} = \int_{T_1}^{T_2} \frac{c_P}{R}\frac{dT}{T} - \ln\frac{P_2}{P_1}$

Alternatively:

$\frac{\Delta S}{R} = \int_{T_1}^{T_2} \frac{c_V}{R}\frac{dT}{T} + \ln\frac{V_2}{V_1}$ (Why?)

Notation:

Because $\Delta S$ is a state function, which is calculated from the reversible path anyway (for the system), and this equation depends only on state functions, this equation is general for ideal gases.

Outcome:

Develop hypothetical reversible paths between two states in order to calculate entropy changes

Outcome:

Use the heat capacity to calculate entropy changes

Test Yourself:

Use these expressions to calculate the entropy changes for each step of a carnot cycle operated between $P_1 = 2bar, P_2 = 1 bar, P_3=0.05bar, and P_4=0.1bar$ and $T_1 = 1000K, T_2= 1000K, T_3 = 300K, and T_4 = 300K$. (Assume $c_P = (5/2)R$.) Plot the results on a TS diagram.