A diatomic gas is used in a Carnot heat engin | vedclass

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(C) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.Given $\eta = 80\% = 0.8$,we have $0.8 = 1 - \frac{T_2}{T_1}$,wh

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$(\frac{1}{5})^{\frac{5}{2}}$

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(C) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.Given $\eta = 80\% = 0.8$,we have $0.8 = 1 - \frac{T_2}{T_1}$,which implies $\frac{T_2}{T_1} = 0.2 = \frac{1}{5}$.For an adiabatic process,the relation between temperature and volume is $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.Rearranging this,we get $\frac{T_1}{T_2} = (\frac{V_2}{V_1})^{\gamma-1}$.Since $\frac{T_1}{T_2} = 5$,we have $5 = (\frac{V_2}{V_1})^{\gamma-1}$.For a diatomic gas,the adiabatic index $\gamma = 1.4 = \frac{7}{5}$,so $\gamma - 1 = \frac{2}{5}$.Thus,$5 = (\frac{V_2}{V_1})^{2/5}$.Taking the power of $5/2$ on both sides,we get $(\frac{V_2}{V_1}) = 5^{5/2}$.Therefore,the ratio of initial volume to final volume is $\frac{V_1}{V_2} = (\frac{1}{5})^{5/2}$.

$A$ diatomic gas is used in a Carnot heat engine. If the efficiency of the given Carnot heat engine is $80\%$,then find the ratio of the initial volume to the final volume of the gas during adiabatic expansion.

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