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The internal energy of a monoatomic gas is U = (3/2) nRT One mole of helium is kept in a cylinder having internal cross-section area = 17 cm² fitted with a frictionless piston. The gas is heated by supplying 126 J heat. If temperature rises by 4°C, then the piston will move ______ cm. Atmospheric pressure = 10⁵ Pa

Options:

(1) 14.5 (2) 1.55 (3) 15.5 (4) 1.45

Step 1: Change in Internal Energy

For monoatomic gas: ΔU = (3/2) nR ΔT Here: n = 1 mole R = 8.3 J/mol·K ΔT = 4 K ΔU = (3/2) × 1 × 8.3 × 4 ΔU ≈ 100 J

Step 2: Work Done

Given heat supplied: Q = 126 J Using first law: Q = ΔU + W 126 = 100 + W W = 26 J

Step 3: Using Work Formula

Work done at constant pressure: W = P ΔV 26 = 10⁵ × (Area × Δx) Area = 17 cm² = 17 × 10⁻⁴ m² 26 = 10⁵ × 17 × 10⁻⁴ × Δx Δx = 26 / 170 Δx = 0.153 m Δx = 15.3 cm

Final Answer:

Correct Option: (3) 15.5 cm

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