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 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) 10.96 (c) 15.04 (d) 29.98

Answer: (b). Work against the atmosphere: W = P A x = 10⁵ × 4×10⁻⁴ × 0.025 = 1 J. "Rotational modes only" is meant to count the energy in the 2 rotational degrees of freedom: U = 2 × 1/2RT = RT per mole. Δ U = nRΔ T = 1 × 8.3 × 1.2 = 9.96 J Q = Δ U + W = 9.96 + 1 = 10.96 J So the answer is (b) 10.96 J.

One mole of diatomic gas having rotational modes only is kept in a… | JEE Main 2026 (8 Apr, Shift 2)

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Thermodynamic Processes

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

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Answer: (b)

Work against the atmosphere: $W = P A x = 10^5 \times 4\times10^{-4} \times 0.025 = 1\text{ J}$ .
"Rotational modes only" is meant to count the energy in the 2 rotational degrees of freedom: $U = 2 \times \frac{1}{2}RT = RT$ per mole.
$\Delta U = nR\Delta T = 1 \times 8.3 \times 1.2 = 9.96\text{ J}$
$Q = \Delta U + W = 9.96 + 1 = 10.96\text{ J}$
So the answer is (b) 10.96 J.

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Thermodynamic Processes question from JEE Main 2026 (8 Apr, Shift 2)

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