JEE Main 2026 Jan 22 Shift 1, Physics Q16: Thermodynamic Processes

JEE Main2026Jan 22, Shift 1Physics

The volume of an ideal gas increases 8 times and temperature becomes $(1/4)^{t h}$ 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
$C O_{2}$
B
$O_{2}$
C
$N H_{3}$
D
He

Solution

Reversible adiabatic ( $Δ Q = 0$ ): $T V^{γ - 1} =$ constant.

Apply: $T \cdot V^{γ - 1} = \frac{T}{4} \cdot (8 V)^{γ - 1}$ , so $4 = 8^{γ - 1}$ .

Take logs: $lo g 4 = (γ - 1) lo g 8 \Rightarrow 2 = 3 (γ - 1) \Rightarrow γ = \frac{5}{3}$ .

$γ = 5/3$ corresponds to a monatomic gas, so the answer is He.

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JEE Main 2026 Jan 22 Shift 1, Physics Q16: Thermodynamic Processes

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