A diatomic ideal gas is used in a Carnot engi | vedclass

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(C) For an adiabatic process,the relation between temperature and volume is given by $T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$.Given $V_1 = V$ and

Vedclass · Accepted Answer

$0.75$

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(C) For an adiabatic process,the relation between temperature and volume is given by $T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$.Given $V_1 = V$ and $V_2 = 32V$,we have $T_1 V^{\gamma - 1} = T_2 (32V)^{\gamma - 1}$.This simplifies to $T_1 = (32)^{\gamma - 1} T_2$.For a diatomic gas,the adiabatic index $\gamma = 7/5$,so $\gamma - 1 = 2/5$.Substituting this,$T_1 = (32)^{2/5} T_2 = (2^5)^{2/5} T_2 = 2^2 T_2 = 4 T_2$.The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.Substituting $T_1 = 4T_2$,we get $\eta = 1 - \frac{T_2}{4T_2} = 1 - 0.25 = 0.75$.

$A$ diatomic ideal gas is used in a Carnot engine as the working substance. If during the adiabatic expansion part of the cycle the volume of the gas increases from $V$ to $32V$,the efficiency of the engine is

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