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Derive the equation to find efficiency of Carnot engine cycle.

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It is defined as ratio of net mechanical work done per cycle by the gas to the amount of heat energy absorbed per cycle from the source.

In isothermal process P1V1 = P2V2    …(1)

As B(V2, P2) and C(V3, P3) lie on same adiabatic

P2\(V_{2}^{\gamma}\) = P3\(V_{3}^{\gamma}\)      …(2)

Again C and D lie on same isothermal 

P3V3 = P4V4      …(3)

Finally D and A lie on same adiabatic

P4\(V_{4}^{\gamma}\) = P1\(V_{1}^{\gamma}\)      …(4)

Multiplying (1), (2), (3), (4),

\(V_{2}^{\gamma-1}\)\(V_{4}^{\gamma-1}\)=\(V_{1}^{\gamma-1}\)\(V_{3}^{\gamma-1}\)

(V2V4) = (V1V3)

V2V4 = V1V3

\(\frac{V_2}{V_1}\)=\(\frac{V_3}{V_4}\)

Loge\(\frac{V_2}{V_1}\)=Loge\(\frac{V_3}{V_4}\)

Dividing (4) by (2),

\(\frac{Q_2}{Q_1}\)=\(\frac{KT_2log_e\frac{V_3}{V_4}}{KT_1log_e\frac{V_2}{V_1}}\)

=\(\frac{T_2}{T_1}\)

\(\frac{Q_2}{Q_1}\)=\(\frac{T_2}{T_1}\)

η=1-\(\frac{T_2}{T_1}\)

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