Thermodynamics

JEE Main level, single correct, medium

155 of 190

Question 155 of 190

An engine operates by taking nnmoles of an ideal gas through the cycle ABCDAABCDA shown in figure. The thermal efficiency of the engine is:

(Take Cv=1.5R{C}_{v}=1.5R, whereRR is gas constant)

Show answer and solution

Answer: (B) 0.150.15

Work done in cycle ==area under closed curve

W=P0V0W={P}_{0}{V}_{0}

Heat given =QAB+QBC={Q}_{AB}+{Q}_{BC}

=nCVΔTAB+nCpΔTBC=n{C}_{V}\Delta {T}_{AB}+n{C}_{p}\Delta {T}_{BC}

=32(nRTB−nRTA)+52(nRTC−nRTB)=\dfrac{3}{2}\left(nR{T}_{B}-nR{T}_{A}\right)+\dfrac{5}{2}\left(nR{T}_{C}-nR{T}_{B}\right)

=32(2P0V0−P0V0)+52(4P0V0−2P0V)=\dfrac{3}{2}\left(2{P}_{0}{V}_{0}-{P}_{0}{V}_{0}\right)+\dfrac{5}{2}\left(4{P}_{0}{V}_{0}-2{P}_{0}V\right)

=132P0V0=\dfrac{13}{2}{P}_{0}{V}_{0}

Thermal efficiency, η=WQabsorbed=213=0.15\eta =\dfrac{W}{{Q}_{absorbed}}=\dfrac{2}{13}=0.15

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