An engine operates by taking $n\,moles$ of an ideal gas through the cycle $ABCDA$ shown in figure. The thermal efficiency of the engine is : (Take $C_v =1 .5\, R$, where $R$ is gas constant)
JEE MAIN 2017, Diffcult
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$Work - done\left( W \right) = {P_0}{V_0}$
According to principle of calorimetry
Heat given$ = {Q_{AB}} = {Q_{BC}}$
$ = n{C_V}d{T_{AB}} + n{C_p}d{T_{BC}}$
$ = \frac{3}{2}\left( {nR{T_B} - nR{T_A}} \right) + \frac{5}{2}\left( {nR{T_C} - nR{T_B}} \right)$
$ = \frac{3}{2}\left( {2{P_0}{V_0} - {P_0}{V_0}} \right) + \frac{5}{2}\left( {4{P_0}{V_0} - 2{P_{.0}}{V_0}} \right)$
$ = \frac{{13}}{2}{P_0}{V_0}$
Thermal efficiency of engine $\left( \eta \right)$
$ = \frac{W}{{{Q_{given}}}} = \frac{2}{{13}} = 0.15$
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