JEE Main 2022 · 28 June · morning · Single correct · +4 / −1
Given below are two statements :
Statement I : When μ amount of an ideal gas undergoes adiabatic change from state (P₁, V₁, T₁) to state (P₂, V₂, T₂), then work done is W = (μ R(T₂ - T₁))/(1 - γ), where γ = Cₚ/Cᵥ and R = universal gas constant.
Statement II : In the above case, when work is done on the gas, the temperature of the gas would rise.
Choose the correct answer from the options given below :
- (a) Both Statement I and Statement II are true.
- (b) Both Statement I and Statement II are false.
- (c) Statement I is true but Statement II is false.
- (d) Statement I is false but Statement II is true.
Answer and solution
Answer (a) Both Statement I and Statement II are true.
Solution
Physics sir · 4 steps
In an adiabatic change Q = 0, so the work and the change in internal energy are locked together.
- First law. In an adiabatic process no heat flows, so Q = 0. Then Q = Δ U + W gives W = -Δ U, where W is the work done by the gas.
W = -μ Cᵥ (T₂ - T₁) = -(μ R (T₂ - T₁))/(γ - 1) = (μ R (T₂ - T₁))/(1 - γ)
- Statement I. This uses Cᵥ = R/(γ - 1). The result is exactly the expression given, taken as work done by the gas, so Statement I is true.
- Statement II. If work is done on the gas, W < 0, so Δ U = -W > 0. For an ideal gas U depends only on T, so the temperature rises. Statement II is true.
- So. Both statements are true, which is option (a).
The idea to keep. Adiabatic means Q = 0, so all the work goes into or out of internal energy. Compress a gas adiabatically and it heats up; let it expand and it cools.
Answer from the NTA final answer key. Solution by PrepWiser.