Two moles of helium gas are taken over the cycle ABCD, as shown in the P-T diagram. The net work done on the gas in the cycle ABCDA is: (a) 276 R (b) 1076 R (c) 1904 R (d) zero

Answer: (a). Work done on gas in each step ( n = 2, ln 2 = 0.69): AB isobaric heating: -nRΔ T = -400R. BC isothermal at 500 K, P halves (expansion): -nRTln 2 = -690R. CD isobaric cooling: +400R. DA isothermal at 300 K, P doubles: +414R. Net work on gas = -400R - 690R + 400R + 414R = -276R, magnitude 276R (the gas does 276R of net work). So the answer is (a) 276R.

Two moles of helium gas are taken over the cycle ABCD, as shown in… | JEE Main 2009 (AIEEE)

Q86JEE Main 2009 · AIEEEMedium

P-V Diagrams & Cyclic Processes

Two moles of helium gas are taken over the cycle $ABCD$ , as shown in the $P\text{-}T$ diagram.

The net work done on the gas in the cycle $ABCDA$ is:

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Answer: (a)

Work done on gas in each step ( $n = 2$ , $\ln 2 = 0.69$ ): AB isobaric heating: $-nR\Delta T = -400R$ .
BC isothermal at 500 K, $P$ halves (expansion): $-nRT\ln 2 = -690R$ .
CD isobaric cooling: $+400R$ . DA isothermal at 300 K, $P$ doubles: $+414R$ .
Net work on gas $= -400R - 690R + 400R + 414R = -276R$ , magnitude $276R$ (the gas does $276R$ of net work).
So the answer is (a) $276R$ .

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