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(C) For a diatomic gas,the adiabatic index $\gamma = 1.4 = 7/5$.In a Carnot cycle,the adiabatic expansion occurs between temperatures $T_1$ (source) a

Vedclass · Accepted Answer

$0.75$

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(C) For a diatomic gas,the adiabatic index $\gamma = 1.4 = 7/5$.In a Carnot cycle,the adiabatic expansion occurs between temperatures $T_1$ (source) and $T_2$ (sink).The relation for adiabatic process is $T V^{\gamma - 1} = 	ext{constant}$.Thus,$T_1 V_B^{\gamma - 1} = T_2 V_C^{\gamma - 1}$,where $V_B = V$ and $V_C = 32V$.$\frac{T_2}{T_1} = \left( \frac{V_B}{V_C} ight)^{\gamma - 1} = \left( \frac{V}{32V} ight)^{1.4 - 1} = \left( \frac{1}{32} ight)^{0.4} = \left( \frac{1}{32} ight)^{2/5}$.Since $32 = 2^5$,we have $\frac{T_2}{T_1} = (2^{-5})^{2/5} = 2^{-2} = 1/4 = 0.25$.The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.$\eta = 1 - 0.25 = 0.75$.

$A$ diatomic ideal gas is used as the working substance in a Carnot engine. During the adiabatic expansion process,if the volume of the gas increases from $V$ to $32V$,what is the efficiency of the engine?

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