Two moles of helium gas are taken over the cycle ABCD, as shown in the P-T diagram. Assuming the gas to be ideal the work done on the gas in taking it from A to B is: (a) 300 R (b) 400 R (c) 500 R (d) 200 R

Answer: (b). A to B: P constant at 2 × 10⁵ Pa, T from 300 K to 500 K (isobaric). Work by gas = nRΔ T = 2 × R × 200 = 400R. Work done on the gas = -400R, magnitude 400R. So the answer is (b) 400R.

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

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P-V Diagrams & Cyclic Processes

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

Assuming the gas to be ideal the work done on the gas in taking it from $A$ to $B$ is:

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Answer: (b)

A to B: $P$ constant at $2 \times 10^{5}$ Pa, $T$ from 300 K to 500 K (isobaric).
Work by gas $= nR\Delta T = 2 \times R \times 200 = 400R$ .
Work done on the gas $= -400R$ , magnitude $400R$ .
So the answer is (b) $400R$ .

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