An engine operates by taking a monatomic ideal gas through the cycle shown in the figure. The percen

Question

An engine operates by taking a monatomic ideal gas through the cycle shown in the figure. The percentage efficiency of the engine is close to $\qquad$ .\ \

Professor PC Thomas Classes · Answer

(19) : Work done $\Delta W=$ Area of loop\ $=\left(3 P_{0}-P_{0}ight)\left(2 V_{0}-V_{0}ight)=2 P_{0} V_{0}$\ $Q_{A B}=n C_{V}\left(T_{B}-T_{A}ight)=\frac{3}{2} n R\left(T_{B}-T_{A}ight)$\ $=\frac{3}{2}\left(3 P_{0}-P_{0}ight) V_{0}=3 P_{0} V_{0}$\ $Q_{B C}=n C_{p}\left(T_{C}-T_{B}ight)=\frac{5}{2} n R\left(T_{C}-T_{B}ight)$\ $=\frac{5}{2}\left(3 P_{0}ight)\left(2 V_{0}-V_{0}ight)$ $$ =\frac{15}{2} P_{0} V_{0}=7.5 P_{0} V_{0} $$ Efficiency, $\eta=\frac{	ext { Work done }(\Delta W)}{	ext { Heat supplied }(\Delta Q)} 	imes 100$ $$ \begin{aligned} \eta & =\frac{\Delta W}{\Delta Q_{A B}+\Delta Q_{B C}} 	imes 100 \ & =\frac{2 P_{0} V_{0}}{3 P_{0} V_{0}+7.5 P_{0} V_{0}} 	imes 100=\frac{20}{105}=19 \% \end{aligned} $$