JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Heat & Thermodynamics

One mole of an ideal diatomic gas expands from volume V to 2V isothermally at a temperature

2 7^{\circ} C

and does W joule of work. If the gas undergoes same magnitude of expansion adiabatically from

2 7^{\circ} C

doing the same amount of work W, then its final temperature will be (close to)

____

^{\circ} C

Answer: (B)

Solution

For the isothermal expansion at

T = 300 K

, the work is

W = n R T ln \frac{V _{2}}{V _{1}} = R \times 300 \times ln 2 = 300 R ln 2

. For an adiabatic expansion doing the same work,

W = \frac{n R ( T _{1} - T _{2} )}{γ - 1}

with

γ = 1.4

for a diatomic gas. Equating the two,

\frac{R ( 300 - T _{2} )}{0.4} = 300 R ln 2

, so

300 - T_{2} = 0.4 \times 300 \times 0.693 = 83.2

, giving

T_{2} = 216.8 K

. In Celsius this is

216.8 - 273 \approx - 5 6^{\circ} C

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Single correct option