What is the thermal efficiency of an engine that operates by taking n moles of diatomic ideal gas through the cycle \({\bf{1}} \to {\bf{2}} \to {\bf{3}} \to {\bf{4}} \to {\bf{1}}\) shown in Fig. P20.38?

The initial pressure for diatomic gas is\({P_1} = {P_0}\)

The initial volume for diatomic gas is\({V_1} = {V_0}\)

The final pressure for diatomic gas is\({P_2} = 2{P_0}\)

The final volume for diatomic gas is\({V_2} = 2{V_0}\)

The number of moles of diatomic gas is \(n\).

The thermal efficiency for the engine is calculated from the ratio of net work done in cycle to the heat input of cycle.

The net work done in the cycle is given as:

\(W = \left( {{P_2} - {P_1}} \right)\left( {{V_2} - {V_1}} \right)\)

Substitute all the values in the above equation.

\(\begin{aligned}W = \left( {2{P_0} - {P_0}} \right)\left( {2{V_0} - {V_0}} \right)\\W = {P_0}{V_0}\end{aligned}\)

The heat input of the engine is given as:

\(\begin{aligned}Q = \frac{{nR\left( {{T_2} - {T_1}} \right)}}{{\gamma - 1}} + \frac{{\gamma nR\left( {{T_3} - {T_2}} \right)}}{{\gamma - 1}}\\Q = \frac{{nR}}{{\gamma - 1}}\left( {{T_2} - {T_1} + \gamma \left( {{T_3} - {T_2}} \right)} \right)\\Q = \frac{{nR}}{{\gamma - 1}}\left( {\gamma \left( {{T_3} - 2{T_2}} \right) - {T_1}} \right)\end{aligned}\)

Here,\(R\)is the universal gas constant and\(\gamma \)is the specific heat ratio for diatomic gas.

The thermal efficiency of engine is given as:

\(\eta = \frac{W}{Q}\)

Substitute all the values in the above equation.

\(\begin{aligned}\eta = \frac{{{P_0}{V_0}}}{{\left( {\frac{{nR}}{{\gamma - 1}}\left( {\gamma \left( {{T_3} - 2{T_2}} \right) - {T_1}} \right)} \right)}}\\\eta = \frac{{\left( {\gamma - 1} \right){P_0}{V_0}}}{{nR\left( {\gamma \left( {{T_3} - 2{T_2}} \right) - {T_1}} \right)}}\end{aligned}\)

Therefore, the thermal efficiency of engine is \(\frac{{\left( {\gamma - 1} \right){P_0}{V_0}}}{{nR\left( {\gamma \left( {{T_3} - 2{T_2}} \right) - {T_1}} \right)}}\).