JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Heat & Thermodynamics

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) $CO_{2}$
(B) $O_{2}$
(C) $NH_{3}$
(D)He✓ Correct

Answer: (D)

Solution

With no heat exchange the change is adiabatic, so

T V^{γ - 1}

is constant:

T_{i} V_{i}^{γ - 1} = \frac{T _{i}}{4} (8 V_{i})^{γ - 1}

. This gives

8^{γ - 1} = 4

, i.e.

2^{3 (γ - 1)} = 2^{2}

, so

γ - 1 = \frac{2}{3}

and

γ = \frac{5}{3}

. A ratio of specific heats of

\frac{5}{3}

corresponds to a monatomic gas, which among the choices is helium.

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Single correct option

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