SL: Motivating the mathematical form of S

Thermodynamic Cycles

cycle

Entropy is a state function so, just like:

$\Delta U_{cycle} = 0$

so is

$\Delta S_{cycle} = 0$

The two steps that are not adiabatic yield:

$Q_H = Q_{12} = -W_{12} = nRT_H\ln \left (\frac{V_2}{V_1} \right )$

$Q_C = Q_{34} = -W_{34} = nRT_C\ln \left (\frac{V_4}{V_3} \right )$

The two adiabatic steps yield:

$\Delta U_{41} = W_{41} = \int_1^4 c_v dT = -\int_1^4 PdV$

Noting that $P=\frac{RT}{V}$, so:

$\int_1^4 \frac{c_v}{R} \frac{dT}{T} = -\int_1^4 \frac{dV}{V} = \ln \left (\frac{V_1}{V_4} \right )$

by similar logic:

$\int_2^3 \frac{c_v}{R} \frac{dT}{T} = -\int_2^3 \frac{dV}{V} = \ln \left (\frac{V_2}{V_3} \right )$

since $T_1=T_2$ and $T_3=T_4$ both $c_v$-related (i.e., left-hand-side) integrals are the same, so:

$\ln \left (\frac{V_2}{V_3} \right ) = \ln \left (\frac{V_1}{V_4} \right )$

rearranging:

$\ln \left (\frac{V_2}{V_1} \right ) = \ln \left (\frac{V_3}{V_4} \right )$

Combining this result with our non-adiabatic steps:

$Q_H = Q_{12} = -W_{12} = nRT_H\ln \left (\frac{V_2}{V_1} \right ) = nRT_H\ln \left (\frac{V_3}{V_4} \right )$

$Q_C = Q_{34} = -W_{34} = nRT_C\ln \left (\frac{V_4}{V_3} \right ) = -nRT_C\ln \left (\frac{V_3}{V_4} \right )$

Taking the ratio $\frac{Q_C}{Q_H}$, gives:

$\frac{Q_C}{Q_H} = \frac{-nRT_C\ln \left (\frac{V_3}{V_4} \right )}{nRT_H\ln \left (\frac{V_3}{V_4} \right )}$

$\frac{Q_C}{Q_H} = \frac{-T_C}{T_H}$

or

$\frac{|Q_C|}{Q_H} = \frac{T_C}{T_H}$

This tells us two interesting things:

Note:

This suggests that, despite the fact that $Q$ is a path function, the ratio of $\frac{Q}{T}$ (at least for reversible processes) is a state function.