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 \(4 \mathrm{~cm}^{2}\). The gas is heated slowly to raise the temperature by \(1.2 { }^{\circ} \mathrm{C}\) during which the piston moves by 25 mm. The amount of heat supplied to the gas is \_\_\_\_ J. (Atmospheric pressure \(=100 \mathrm{kPa}\), \(\mathrm{R}=8.3 \mathrm{~J} / \mathrm{mol} . \mathrm{K}\)) (Neglect mass of the piston)

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Heat & Thermodynamics

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)

Answer: (B)

With the piston free against the atmosphere the process is isobaric, so

Q = Δ U + W

. Work done:

W = P Δ V = P A d = (1 0^{5}) (4 \times 1 0^{- 4}) (0.025) = 1 J

. Internal energy of one mole of a diatomic gas (translational + rotational,

C_{V} = \frac{5}{2} R

Δ U = n C_{V} Δ T = \frac{5}{2} (8.3) (1.2) = 24.9 J

. Thus

Q = 24.9 + 1 = 25.9 \approx 25 J

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Single correct option