Thermodynamics

JEE Main level, single correct, medium

127 of 190

Question 127 of 190

The P−VP-V diagram of a diatomic ideal gas system going under cyclic process as shown in figure. The work done during an adiabatic process CDCD is (use γ=1.4)\gamma =1.4) :

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Answer: (A) –500J–500J

Adiabatic process is from CC to DD
WD=P2V2−P1V11−γWD=\dfrac{{P}_{2}{V}_{2}-{P}_{1}{V}_{1}}{1-\gamma }
=PDVD−PCVC1−γ=\dfrac{{P}_{D}{V}_{D}-{P}_{C}{V}_{C}}{1-\gamma }
=200(3)−(100)(4)1−1.4=\dfrac{200(3)-(100)(4)}{1-1.4}
=−500J=-500J

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