Thermodynamics

JEE Main level, single correct, medium

21 of 190

Question 21 of 190

The internal energy of a monoatomic gas is 3 nRT. One mole of helium is kept in a cylinder having internal cross section area of 17 cm217 \mathrm{~cm}^{2} and fitted with a light movable frictionless piston. The gas is heated slowly by suppling 126 J heat. If the temperature rises by 4∘C4^{\circ} \mathrm{C}, then the piston will move ____\_\_\_\_ cm.
(atmospheric pressure =105 Pa=10^{5} \mathrm{~Pa})

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Answer: (B) 15.5

The question states the internal energy of the gas is U=3nRTU = 3nRT.
The change in internal energy is ΔU=3nRΔT\Delta U = 3nR\Delta T.
Given: n=1n = 1 mole, ΔT=4∘C=4\Delta T = 4^{\circ}C = 4 K, and R≈8.314R \approx 8.314 J/mol·K.
ΔU=3×1×8.314×4=99.768\Delta U = 3 \times 1 \times 8.314 \times 4 = 99.768 J.
According to the first law of thermodynamics, ΔQ=ΔU+W\Delta Q = \Delta U + W.
Given heat supplied ΔQ=126\Delta Q = 126 J.
Work done by the gas W=ΔQ−ΔU=126−99.768=26.232W = \Delta Q - \Delta U = 126 - 99.768 = 26.232 J.
Since the piston is light and movable, the process occurs at constant atmospheric pressure P=105P = 10^5 Pa.
Work done W=PΔV=P(A×d)W = P \Delta V = P(A \times d), where AA is the cross-section area and dd is the displacement of the piston.
A=17 cm2=17×10−4 m2A = 17 \text{ cm}^2 = 17 \times 10^{-4} \text{ m}^2.
26.232=105×(17×10−4)×d26.232 = 10^5 \times (17 \times 10^{-4}) \times d.
26.232=170×d26.232 = 170 \times d.
d=26.232170≈0.1543d = \frac{26.232}{170} \approx 0.1543 m.
Converting to cm: d≈15.43d \approx 15.43 cm.
Rounding to the nearest option, we get 15.515.5 cm.

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