The mean kinetic energy of the molecules of an ideal gas at 399 is ' E '. The temperature at which the mean kinetic energy of its molecules will be ' E/2 ', is (a) 336 (b) 276 (c) 123 (d) 63

Answer: (d). Mean KE ∝ T (kelvin). T₁ = 399 + 273 = 672 K; for E/2, T₂ = 336 K. 336 - 273 = 63° C So the answer is (d) 63° C.

The mean kinetic energy of the molecules of an ideal gas at 399 is '… | MHT CET 2025 (22 Apr, Shift 1)

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Pressure & Kinetic Energy of a Gas

The mean kinetic energy of the molecules of an ideal gas at $399^\circ\text{C}$ is ' E '. The temperature at which the mean kinetic energy of its molecules will be ' $E/2$ ', is

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Answer: (d)

Mean KE $\propto T$ (kelvin).
$T_1 = 399 + 273 = 672$ K; for $E/2$ , $T_2 = 336$ K.
$336 - 273 = 63^\circ$ C
So the answer is (d) $63^\circ$ C.

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