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JEE Main 2024 (9 Apr, Shift 2) · Pressure & Kinetic Energy of a Gas

Pressure & Kinetic Energy of a Gas question from JEE Main 2024 (9 Apr, Shift 2)

Q13JEE Main 2024 · 9 APR · SHIFT 2Easy

Pressure & Kinetic Energy of a Gas

The temperature of a gas is $-78^\circ\text{C}$ 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 $2K$ is:

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Answer: (d)

Average KE $\propto T$ (kelvin).
$T_1 = -78 + 273 = 195$ K; for $2K$ , $T_2 = 390$ K.
$390 - 273 = 117^\circ$ C
So the answer is (d) $117^\circ$ C.

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