The volume of an ideal gas increases 8 times and temperature becomes (1/4) th of initial temperature during a reversible change. If there is no exchange of heat in this process (Δ Q = 0) then identify the gas from the following options (Assuming the gases given in the options are ideal gases): (a) NH₃ (b) O₂ (c) CO₂ (d) He

Answer: (d). Δ Q = 0, reversible: TVγ-1 = const. T/4(8V)γ-1 = TVγ-1 ⇒ 8γ-1 = 4 ⇒ 3(γ - 1) = 2 γ = 5/3, which is a monoatomic gas. So the answer is (d) He.

The volume of an ideal gas increases 8 times and temperature becomes… | JEE Main 2026 (22 Jan, Shift 1)

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Thermodynamic Processes

The volume of an ideal gas increases 8 times and temperature becomes $(1/4)$ th of initial temperature during a reversible change. If there is no exchange of heat in this process $(\Delta Q = 0)$ then identify the gas from the following options (Assuming the gases given in the options are ideal gases):

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Answer: (d)

$\Delta Q = 0$ , reversible: $TV^{\gamma-1} = \text{const}$ .
$\frac{T}{4}(8V)^{\gamma-1} = TV^{\gamma-1} \Rightarrow 8^{\gamma-1} = 4 \Rightarrow 3(\gamma - 1) = 2$
$\gamma = \frac{5}{3}$ , which is a monoatomic gas.
So the answer is (d) He.

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