SL: Calculating entropy changes for ideal gases

Entropy calculations for ideal gases

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

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


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

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 $T$:

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


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


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


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.


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


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.