JEE Main 2025 · 24 January · evening · Single correct · +4 / −1
Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : In an insulated container, a gas is adiabatically shrunk to half of its initial volume. The temperature of the gas decreases.
Reason (R) : Free expansion of an ideal gas is an irreversible and an adiabatic process.
In the light of the above statements, choose the correct answer from the options given below :
- (a) Both (A) and (R) are true and (R) is the correct explanation of (A)
- (b) Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
- (c) (A) is false but (R) is true
- (d) (A) is true but (R) is false
Answer and solution
Answer (c) (A) is false but (R) is true
Solution
Physics sir · 4 steps
Squeezing a gas with no heat escaping heats it up, so the assertion has the direction backwards.
- First law. In an adiabatic process Q = 0, so Δ U = Q - W = -W. When the gas is compressed, the work done by the gas is negative, so Δ U is positive.
TV^(γ-1) = constant ⇒ T₂ = T₁(V₁/V₂)^(γ-1) = T₁ × 2^(γ-1)
- Check Assertion. Since γ > 1, 2^(γ-1) > 1, so T₂ > T₁. The work done on the gas raises its internal energy and its temperature. (A) is false.
- Check Reason. In free expansion the gas rushes into vacuum inside an insulated container: Q = 0, so it is adiabatic. It happens suddenly, not through equilibrium states, and cannot be retraced, so it is irreversible. (R) is true.
- So. (A) is false and (R) is true, which is option (c).
The idea to keep. Adiabatic compression heats a gas and adiabatic expansion cools it, because all the work goes into internal energy. Free expansion is the odd one out: adiabatic, irreversible, and Δ T = 0 for an ideal gas.
Answer from the NTA final answer key. Solution by PrepWiser.