JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Heat & Thermodynamics

The temperature of a gas is

- 7 8^{\circ} C

and the average translational kinetic energy of its molecules is K . The temperature at which the average translational kinetic energy of the molecules of the same gas becomes 2 K is :

(A) $- 3 9^{\circ} C$
(B) $11 7^{\circ} C$ ✓ Correct
(C) $12 7^{\circ} C$
(D) $- 7 8^{\circ} C$

Answer: (B)

Solution

Average translational KE per molecule is

KE = \frac{3}{2} k_{B} T

, so

KE \propto T

(in kelvin).

Given

T_{i} = - 7 8^{\circ} C = 195 K

corresponds to KE

= K

. For KE

= 2 K

, temperature must double:

T_{f} = 2 \times 195 = 390 K = 11 7^{\circ} C

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Single correct option

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