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An ideal gas with constant heat capacity $\mathrm{C}_{\mathrm{V}}=\frac{3}{2} \mathrm{n} \mathrm{R}$ is made to carry out a cycle that is depicted by a triangle in the figure given below.

The following statement is true about the cycle-
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The following statement is true about the cycle-
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The efficiency is given by $1-\frac{1}{2} \frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{P}_{2} \mathrm{~V}_{2}}$
$$
\begin{array}{l}
\mathrm{C}_{\mathrm{V}}=\frac{3}{2} \mathrm{R}, \mathrm{C}_{\mathrm{P}}=\mathrm{C}_{\mathrm{V}}+\mathrm{R}=\frac{5 \mathrm{R}}{2} \\
\mathrm{f}=3 \\
\mathrm{~W}=\frac{1}{2}\left(\mathrm{~V}_{2}-\mathrm{V}_{1}\right)\left(\mathrm{P}_{2}-\mathrm{P}_{1}\right)
\end{array}
$$
For BA $\quad Q=n C p \Delta T=n\left(\frac{5}{2} R\right) \Delta T=\frac{5}{2}\left(P_{1} V_{1}-P_{2} V_{2}\right)$
For AC $\quad Q_{A C}=\frac{1}{2}\left(P_{1}+P_{2}\right)\left(V_{2}-V_{1}\right)+n C_{V} \Delta T$
now $n \mathrm{Cv} \Delta \mathrm{T}=\frac{3}{2}\left(\mathrm{P}_{2} \mathrm{~V}_{2}-\mathrm{P}_{1} \mathrm{~V}_{1}\right)$
now $\eta=\frac{W}{Q}=\frac{\frac{1}{2}\left(V_{2}-V_{1}\right)\left(P_{2}-P_{1}\right)}{\frac{1}{2} \times\left(P_{1}+P_{2}\right)\left(V_{2}-V_{1}\right)+\frac{3}{2}\left(P_{2} V_{2}-P_{1} V_{1}\right)}$
using formula for heat we can calculate heat absorbed in $\mathrm{AC}$.
\begin{array}{l}
\mathrm{C}_{\mathrm{V}}=\frac{3}{2} \mathrm{R}, \mathrm{C}_{\mathrm{P}}=\mathrm{C}_{\mathrm{V}}+\mathrm{R}=\frac{5 \mathrm{R}}{2} \\
\mathrm{f}=3 \\
\mathrm{~W}=\frac{1}{2}\left(\mathrm{~V}_{2}-\mathrm{V}_{1}\right)\left(\mathrm{P}_{2}-\mathrm{P}_{1}\right)
\end{array}
$$
For BA $\quad Q=n C p \Delta T=n\left(\frac{5}{2} R\right) \Delta T=\frac{5}{2}\left(P_{1} V_{1}-P_{2} V_{2}\right)$
For AC $\quad Q_{A C}=\frac{1}{2}\left(P_{1}+P_{2}\right)\left(V_{2}-V_{1}\right)+n C_{V} \Delta T$
now $n \mathrm{Cv} \Delta \mathrm{T}=\frac{3}{2}\left(\mathrm{P}_{2} \mathrm{~V}_{2}-\mathrm{P}_{1} \mathrm{~V}_{1}\right)$
now $\eta=\frac{W}{Q}=\frac{\frac{1}{2}\left(V_{2}-V_{1}\right)\left(P_{2}-P_{1}\right)}{\frac{1}{2} \times\left(P_{1}+P_{2}\right)\left(V_{2}-V_{1}\right)+\frac{3}{2}\left(P_{2} V_{2}-P_{1} V_{1}\right)}$
using formula for heat we can calculate heat absorbed in $\mathrm{AC}$.
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