JEE Main 2026 Jan 23 Shift 2, Physics Q14: Thermodynamic Processes

JEE Main2026Jan 23, Shift 2Physics

One mole of an ideal diatomic gas expands from volume $V$ to $2 V$ isothermally at a temperature $27^{\circ} C$ and does $W$ joule of work. If the gas undergoes same magnitude of expansion adiabatically from $27^{\circ} C$ doing the same amount of work $W$ , then its final temperature will be (close to) ______ $^{\circ} C$ .

A
$- 189$
B
$- 56$
C
$- 30$
D
$- 117$

Solution

Isothermal work: $W_{i so} = n R T ln (V_{2} / V_{1}) = 1 \cdot R \cdot 300 \cdot ln 2 \approx 300 R \cdot 0.693$ J.

Adiabatic work for diatomic gas ( $γ = 1.4$ ):

$W_{a d i} = \frac{n R ( T _{1} - T _{2} )}{γ - 1} = \frac{R ( 300 - T _{f} )}{0.4}$ .

Setting $W_{a d i} = W_{i so}$ :

$\frac{300 - T _{f}}{0.4} = 300 \cdot 0.693$

$300 - T_{f} = 120 \cdot 0.693 \approx 83.16$

$T_{f} \approx 216.84$ K $\approx - 56.3^{\circ} C$ .

Closest option: $- 56$ .

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JEE Main 2026 Jan 23 Shift 2, Physics Q14: Thermodynamic Processes

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