Calculate the efficiency of a heat engine in a gas whose ratio of specific heat is - while it is being taken through a cycle as shown in the indicator diagram-

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Question

Calculate the efficiency of a heat engine in a gas whose ratio of specific heat is $γ$ while it is being taken through a cycle as shown in the indicator diagram.

1 -

\frac{(T_{4} - T_{1})}{γ (T_{3} - T_{2})}

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1 -

\frac{γ (T_{4} - T_{1})}{(T_{3} - T_{2})}

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1 -

\frac{(T_{3} - T_{2})}{γ (T_{4} - T_{1})}

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\frac{γ (T_{3} - T_{2})}{(T_{4} - T_{1})}

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Solution

The correct option is A 1 - $\frac{(T_{4} - T_{1})}{γ (T_{3} - T_{2})}$
BC is an isobaric process and temperature from T $_{2}$ to T $_{3}$
Heat absorbed $= Q_{1} = C_{p} (T_{3} - T_{2})$
CD is an adiabatic process. So no heat is absorbed or rejected.
DA is an isochoric process.
So heat rejected $= Q_{2} = C_{v} (T_{4} - T_{1})$ and AB is an adiabatic process. So, again, no heat is absorbed or rejected during AB.
Net amount of heat converted to work $= Q_{2} - Q_{1} = C_{p} (T_{3} - T_{2}) - C_{v} (T_{4} - T_{1})$
So, efficiency $η = \frac{Work done}{Heat absorbed} = \frac{Q_{2} - Q_{1}}{Q_{1}} = \frac{C_{P} (T_{3} - T_{2}) - C_{v} (T_{4} - T_{1})}{C_{P} (T_{3} - T_{2})}$
$= 1 - \frac{C_{v}}{C_{p}} . \frac{(T_{4} - T_{1})}{T_{3} - T_{2}}$ $= 1 - \frac{1}{γ} . \frac{(T_{4} - T_{1})}{T_{3} - T_{2}}$

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Calculate the efficiency of a heat engine in a gas whose ratio of specific heat is γ while it is being taken through a cycle as shown in the indicator diagram.

Calculate the efficiency of a heat engine in a gas whose ratio of specific heat is $γ$ while it is being taken through a cycle as shown in the indicator diagram.