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← JEE Advanced 2026 · Paper 1
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JEE Advanced 2026 · Paper 1, Chemistry question 1

An ideal gas (0.5 mol), initially at 2 bar pressure, is compressed at a constant temperature of 600 K in two steps: first, against a constant external pressure of PP bar (2<P<8)(2<P<8), and then against constant external pressure of 8 bar. At each step, the compression is stopped only when the pressure of the gas becomes equal to the external pressure. The total work done on the gas in these steps is WW. Considering all possible values of PP (2<P<8)(2<P<8) and taking the gas constant as RR (in J K−1^{-1} mol−1^{-1}), the minimum value of ∣W∣|W| (in J) is

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Answer and worked solution

Official answer: B

Each step is an irreversible isothermal compression against a constant pressure, ending when the gas pressure equals it. Use V=nRTpV=\dfrac{nRT}{p} with nRT=0.5×600R=300RnRT=0.5\times600R=300R.

Work on the gas.
W=P(V0−V1)+8(V1−V2)=nRT[P(12−1P)+8(1P−18)]=nRT(P2+8P−2).W=P\left(V_0-V_1\right)+8\left(V_1-V_2\right)=nRT\left[P\left(\frac12-\frac1P\right)+8\left(\frac1P-\frac18\right)\right]=nRT\left(\frac P2+\frac8P-2\right).

Minimum. P2+8P\frac P2+\frac8P is smallest at P=4P=4, where it equals 4. So
∣W∣min⁡=300R×2=600R.|W|_{\min}=300R\times2=600R.

Topics:First law and enthalpy
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