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Question

Physics Question on Thermodynamics

An engine operates by taking nn moles of an ideal gas through the cycle ABCDAABCDA shown in figure. The thermal efficiency of the engine is : (Take Cv=1.5R,C_v = 1.5 \, R, where RR is gas constant)

A

0.24

B

0.15

C

0.32

D

0.08

Answer

0.15

Explanation

Solution

w=P0V0w = P_{0}V_{0}
Heat given = QAB=QBCQ_{AB} = Q_{BC}
=nCVdTAB+nCPdTBC= nC_{V}dT_{AB} + nC_{P}dT_{BC}
=32(nRTBnRTA)+52(nRTCnRTB)=\frac{3}{2}\left(nRT_{B}-nRT_{A}\right)+\frac{5}{2} \left(nRT_{C} - nRT_{B}\right)
=32(2P0V0P0V0)+52(4P0V02P0V)= \frac{3}{2}\left(2P_{0}V_{0} - P_{0}V_{0}\right)+\frac{5}{2}\left(4P_{0}V_{0} - 2P_{0}V\right)
=132P0V0= \frac{13}{2}P_{0}V_{0}
n=wQgiven=213=0.15n = \frac{w}{Qgiven} = \frac{2}{13} = 0.15