The internal energy of a monoatomic gas is 3nRT. One mole of helium is kept in a cylinder having internal cross section area of 17 cm² and fitted with a light movable frictionless piston. The gas is heated slowly by suppling 126 J heat. If the temperature rises by 4 °C, then the piston will mov $e______cm.$
(atmospheric pressure = 10⁵ Pa)

Question

The internal energy of a monoatomic gas is 3nRT. One mole of helium is kept in a cylinder having internal cross section area of 17 cm² and fitted with a light movable frictionless piston. The gas is heated slowly by suppling 126 J heat. If the temperature rises by 4 °C, then the piston will mov $e______cm.$

(atmospheric pressure = 10⁵ Pa)

G1 Physics · Accepted Answer

Using the internal-energy relation stated in the question, ΔU = 3nRΔT. For one mole, n = 1, R = 25/3 J·mol⁻¹·K⁻¹, ΔT = 4 K. Therefore, ΔU = 3×1×(25/3)×4 = 100 J. From the first law of thermodynamics, ΔQ = ΔU + W. Hence, W = 126 - 100 = 26 J. The piston moves slowly against atmospheric pressure, so W = PΔV. Also, ΔV = AΔx. Thus, 26 = 10⁵ × 17×10⁻⁴ × Δx. Therefore, Δx = 26/170 m ≈ 0.153 m = 15.3 cm. The nearest listed option is 15.5 cm. Correct option: C.