The temperature at which the mean kinetic energy of the molecules of gas is ((1)/(4))^ th of the mean kinetic energy of its molecules at 147^ ^ C is

Correct answer: A. 105 K The mean kinetic energy of gas molecules is directly proportional to the absolute temperature of the gas, E T . Let the initial temperature be T_1 = 147^ ^ C = 147 + 273 = 420 K . Let the final temperature be T_2 . Given that E_2 = (1)/(4)E_1 . Therefore, T_2 = (1)/(4)T_1 = (1)/(4) × 420 K = 105 K . Answer: 105 K

MHT CET 202626 AprilMorning Shift

Exam: MHT CET
Year: 2026
Date: 26 April
Shift: Morning Shift
Subject: Physics
Chapter: Kinetic Theory of Gases

MHT CET 2026 · 26 April · Morning Shift · Physics · Kinetic Theory of Gases

The temperature at which the mean kinetic energy of the molecules...

The temperature at which the mean kinetic energy of the molecules of gas is

\left(\dfrac{1}{4}\right)^{th}

of the mean kinetic energy of its molecules at

147^{}^{\circ}\text{C}

Options

A $105$ K
B $-147$ K
C $173^{}^{\circ}\text{C}$
D $-105^{}^{\circ}\text{C}$

Correct answer

A. $105$ K

Step-by-step solution

The mean kinetic energy of gas molecules is directly proportional to the absolute temperature of the gas,

E \propto T

.

Let the initial temperature be

T_1 = 147^{{}^{\circ}}\text{C} = 147 + 273 = 420\text{ K}

.

Let the final temperature be

T_2

.

Given that

E_2 = \dfrac{1}{4}E_1

.

Therefore,

T_2 = \dfrac{1}{4}T_1 = \dfrac{1}{4} \times 420\text{ K} = 105\text{ K}

.

Answer:

105

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