Two moles of helium gas are taken over the cycle ABCD, as shown in the P-T diagram. The work done on the gas in taking it from D to A is: (a) +414 R (b) -690 R (c) +690 R (d) -414 R

Answer: (a). D to A: T = 300 K constant (isothermal), P rises from 1 × 10⁵ to 2 × 10⁵ Pa, so the gas is compressed. Work on gas = nRTln (PA)/(PD) = 2 × R × 300 × ln 2. = 600R × 0.69 ≈ 414R So the answer is (a) +414R.

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.

The work done on the gas in taking it from $D$ to $A$ is:

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Answer: (a)

D to A: $T = 300$ K constant (isothermal), $P$ rises from $1 \times 10^{5}$ to $2 \times 10^{5}$ Pa, so the gas is compressed.
Work on gas $= nRT\ln\frac{P_A}{P_D} = 2 \times R \times 300 \times \ln 2$ .
$= 600R \times 0.69 \approx 414R$
So the answer is (a) $+414R$ .

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