# An engine operates by taking a monatomic ideal

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

Solution:

$\mathrm{W}_{\mathrm{ABCDA}}=2 \mathrm{P}_{0} \mathrm{~V}_{0}$

$\mathrm{Q}_{\text {in }}=\mathrm{Q}_{\mathrm{AB}}+\mathrm{Q}_{\mathrm{BC}}$

$\mathrm{Q}_{\mathrm{AB}}=\mathrm{nC}\left(\mathrm{T}_{\mathrm{B}}-\mathrm{T}_{\mathrm{A}}\right)$

$=\frac{\mathrm{n} 3 \mathrm{R}}{2}\left(\mathrm{~T}_{\mathrm{B}}-\mathrm{T}_{\mathrm{A}}\right)$

$=\frac{3}{2}\left(\mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}-\mathrm{P}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}\right)$

$=\frac{3}{2}\left(3 P_{B} V_{0}=P_{0} V_{0}\right)=3 P_{0} V_{0}$

$Q_{B C}=n C_{P}\left(T_{C}-T_{B}\right)$

$=\frac{n 5 R}{2}\left(T_{C}-T_{B}\right)$

$=\frac{5}{2}\left(\mathrm{P}_{\mathrm{C}} \mathrm{V}_{\mathrm{C}}-\mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}\right)$

$=\frac{5}{2}\left(6 \mathrm{P}_{0} \mathrm{~V}_{0}-3 \mathrm{P}_{0} \mathrm{~V}_{0}\right)=\frac{15}{2} \mathrm{P}_{0} \mathrm{~V}_{0}$

$\eta=\frac{W}{Q_{i n}} \times 100=\frac{2 P_{0} V_{0}}{3 P_{0} V_{0}+\frac{15}{2} P_{0} V_{0}} \times 100$

$\eta=\frac{400}{21}=19.04 \approx 19$

$\eta=19$