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JEE Main 2026 · 8 Apr · Shift 2, Physics question 35

One mole of diatomic gas having rotational modes only is kept in a cylinder with a piston system. The cross-section area of the cylinder is 4 cm². The gas is heated slowly to raise the temperature by 1.2 °C during which the piston moves by 25 mm. The amount of heat supplied to the gas is ________ J.

(Atmospheric pressure = 100 kPa, R = 8.3 J/mol. K) (Neglect mass of the piston)

Choose one option
Answer and worked solution

Official answer: A

Key: both (A) 24.8 and (B) 25 accepted.

Internal energy. A diatomic gas with rotation has Cv=52RC_v=\tfrac52R:
ΔU=1×52×8.3×1.2=24.9 J≈25 J.\Delta U=1\times\tfrac52\times8.3\times1.2=24.9\text{ J}\approx25\text{ J}.

Work. At atmospheric pressure the piston does P A d=105×4×10−4×0.025=1P\,A\,d=10^5\times4\times10^{-4}\times0.025=1 J of work. That would make Q=ΔU+W≈25.9Q=\Delta U+W\approx25.9 J.

Note. The data are inconsistent: heating 1 mol by 1.2 K at constant pressure needs ΔV=nRΔT/P≈10−4\Delta V=nR\Delta T/P\approx10^{-4} m³, ten times the stated A d=10−5A\,d=10^{-5} m³. The accepted answers match ΔU\Delta U alone.

Topics:First law and internal energySpecific heats of gases
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