JEE Main 2026 Apr 8 Shift 2, Physics Q10: Laws of Thermodynamics

JEE Main2026Apr 8, Shift 2Physics

One mole of diatomic gas having rotational modes only is kept is a cylinder with a piston system. The cross-section area of the cylinder is $4$ cm $^{2}$ . The gas is heated slowly to raise the temperature by $1.2^{\circ}$ 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)

A
$24.8$
B
$25$
C
$15.04$
D
$29.98$

Solution

For a diatomic gas with translational $+$ rotational modes only, $f = 5$ . The process is isobaric (piston free, atmospheric pressure constant).

Work done by gas: $W = P Δ V = (1 0^{5}) (4 \times 1 0^{- 4}) (25 \times 1 0^{- 3}) = 1.0$ J.

Change in internal energy: $Δ U = \frac{f}{2} n R Δ T = \frac{5}{2} (1) (8.3) (1.2) = 24.9$ J.

Heat supplied: $Q = Δ U + W = 24.9 + 1.0 \approx 25.9$ J, closest to the listed $25$ J.

Note: The given numbers do not satisfy $P Δ V = n R Δ T$ exactly (state data is inconsistent), so the closest available option, $25$ J, is the intended answer.

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JEE Main 2026 Apr 8 Shift 2, Physics Q10: Laws of Thermodynamics

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