Determine the efficiency of a Carnot cycle if | vedclass

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(B) For an adiabatic process,the relationship between temperature and volume is given by $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.Rearranging this,we

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$1-\frac{1}{\sqrt{3}}$

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(B) For an adiabatic process,the relationship between temperature and volume is given by $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.Rearranging this,we get $\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma-1}$.Given that the final volume $V_2 = 3V_1$,we have $\frac{V_1}{V_2} = \frac{1}{3}$.Substituting the values $\gamma = 1.5$ and $\frac{V_1}{V_2} = \frac{1}{3}$:$\frac{T_2}{T_1} = \left(\frac{1}{3}\right)^{1.5-1} = \left(\frac{1}{3}\right)^{0.5} = \frac{1}{\sqrt{3}}$.The efficiency $\eta$ of a Carnot cycle is defined as $\eta = 1 - \frac{T_2}{T_1}$.Substituting the ratio,we get $\eta = 1 - \frac{1}{\sqrt{3}}$.

Determine the efficiency of a Carnot cycle if,in the adiabatic expansion,the volume becomes $3$ times its initial value and $\gamma = 1.5$.

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