JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Heat & Thermodynamics

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 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^{\circ} C

, then the piston will move ____ cm.
(atmospheric pressure

= 1 0^{5} Pa

)

Answer: (C)

Solution

With the internal energy taken as

U = 3 n R T

, the change is

Δ U = 3 n R Δ T = 3 \times 1 \times R \times 4

; using

R \approx 8.3 J mo l^{- 1} K^{- 1}

this is about

100 J

. The light frictionless piston keeps the pressure equal to atmospheric, so by the first law the work done by the gas is

W = Q - Δ U = 126 - 100 = 26 J

. For this isobaric expansion

W = P A Δ x

, hence

Δ x = \frac{W}{P A} = \frac{26}{1 0 ^{5} \times 17 \times 1 0 ^{- 4}} = \frac{26}{170} \approx 0.153 m \approx 15.5 cm

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Single correct option