JEE Main 2025 · 4 April · morning · Single correct · +4 / −1
One mole of an ideal gas expands isothermally and reversibly from 10 dm³ to 20 dm³ at 300 K . Δ U, q and work done in the process respectively are
Given: R=8.3 J K⁻¹ mol⁻¹
ln 10=2.3
log 2=0.30
log 3=0.48
- (a) 0,21.84 kJ,-1.726 J
- (b) 0,21.84 kJ, 21.84 kJ
- (c) 0,1.718 kJ,-1.718 kJ
- (d) 0,-17.18 kJ, 1.718 J
Answer and solution
Answer (c) 0,1.718 kJ,-1.718 kJ
Solution
Chemistry ma’am · 4 steps
Isothermal ideal gas means Δ U = 0, so whatever work the gas does, it takes back exactly as heat.
- Internal energy. For an ideal gas, U depends only on temperature. The temperature stays at 300 K, so Δ U = 0.
w = -2.303 nRT log V₂/V₁
- Put in numbers. w = -2.303 × 1 × 8.3 × 300 × log 2. With the given data, 2.303 log 2 ≈ 2.3 × 0.30 = 0.69, so w = -2490 × 0.69 = -1718 J = -1.718 kJ. The sign is negative because the gas does work on the surroundings as it expands.
- Heat from first law. Δ U = q + w = 0, so q = -w = +1.718 kJ. The gas absorbs heat to keep its temperature constant.
- So. Δ U = 0, q = 1.718 kJ and w = -1.718 kJ, which is option (c).
The idea to keep. In any isothermal process of an ideal gas, Δ U = 0, so q = -w. Heat and work always have equal size and opposite signs. Use that to reject options fast.
Answer from the NTA final answer key. Solution by PrepWiser.