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BITSAT 2021

Exam: BITSAT
Year: 2021
Subject: Physics
Chapter: Kinetic Theory of Gases

BITSAT 2021 · Physics · Kinetic Theory of Gases

The temperature of an ideal gas is increased from 120 K to 480 K ....

The temperature of an ideal gas is increased from

120 \mathrm{ K}

to

480 \mathrm{ K}

. If at

120 \mathrm{ K}

, the root mean square velocity of the gas molecules is

v

, at

480 \mathrm{ K}

it becomes

Options

A $4 u$
B $2 u$
C $\frac{u}{2}$
D $\frac{u}{4}$

Correct answer

B. $2 u$

Step-by-step solution

(

\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{\sqrt{\frac{3 \mathrm{RT}_{1}}{\mathrm{M}}}}{\sqrt{\frac{3 \mathrm{RT}_{2}}{\mathrm{M}}}}=\sqrt{\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}} \therefore \mathrm{v}_{2}=

2

\mathrm{v}_{1}

)

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