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(B) For a Carnot engine, the efficiency $\eta$ is given by $\eta = 1 - \frac{T_2}{T_1}$, where $T_1$ is the temperature of the source and $T_2$ is the

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$75$

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(B) For a Carnot engine, the efficiency $\eta$ is given by $\eta = 1 - \frac{T_2}{T_1}$, where $T_1$ is the temperature of the source and $T_2$ is the temperature of the sink.During the adiabatic expansion process, the relation between temperature and volume is $T V^{\gamma-1} = 	ext{constant}$.For a diatomic gas, the adiabatic index $\gamma = 1.4 = \frac{7}{5}$.Thus, $\gamma - 1 = 0.4 = \frac{2}{5}$.Given that the volume increases by a factor of $32$, i.e.,$V_2 = 32 V_1$.Using the adiabatic relation: $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.$\frac{T_2}{T_1} = \left( \frac{V_1}{V_2} ight)^{\gamma-1} = \left( \frac{1}{32} ight)^{2/5}$.$\frac{T_2}{T_1} = \left( (2^5)^{-1} ight)^{2/5} = (2^{-5})^{2/5} = 2^{-2} = \frac{1}{4} = 0.25$.Now, the efficiency $\eta = 1 - \frac{T_2}{T_1} = 1 - 0.25 = 0.75$.Converting to percentage, $\eta = 75 \%$.

$A$ Carnot engine uses diatomic gas as a working substance. During the adiabatic expansion part of the cycle, if the volume of the gas becomes $32$ times its initial volume, then the efficiency of the engine is (in $\%$)

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