JEE Main 2025 · 3 April · evening · Single correct · +4 / −1
Pressure of an ideal gas, contained in a closed vessel, is increased by 0.4 % when heated by 1° C. Its initial temperature must be:
- (a) 2500 K
- (b) 25° C
- (c) 250° C
- (d) 250 K
Answer and solution
Answer (d) 250 K
Solution
Physics sir · 4 steps
Closed vessel means fixed volume, so the fractional rise in pressure equals Δ T / T in kelvin.
- Constant volume. The vessel is closed, so V and the amount of gas do not change. From PV = nRT, pressure is proportional to absolute temperature: P ∝ T.
(Δ P)/P = (Δ T)/T
- Put in numbers. A rise of 1°C is the same as a rise of 1 K, so Δ T = 1 K. And 0.4% means (Δ P)/P = 0.4/100 = 0.004.
- Solve for T. 0.004 = 1/T, so T = 1/0.004 = 250 K. The proportionality P ∝ T holds only on the kelvin scale, so this T is in kelvin.
- So. The initial temperature is 250 K, option (d).
The idea to keep. For a gas at fixed volume, the percentage change in pressure equals the percentage change in absolute temperature. The T you get is always in kelvin.
Answer from the NTA final answer key. Solution by PrepWiser.