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The molecules of a given mass of a gas have rms velocity of $200$ ms $^{-1}$ at $27^{\circ}C$ and $1.0 \times 10^{5}$ Nm $^{-2}$ pressure. When the temperature and pressure of the gas are, respectively, $127^{\circ}C$ and $0.05 \times 10^{5}$ Nm $^{-2},$ the rms velocity of its molecules in ms $^{-1}$ is

medium

Kinetic Theory

2016

physics

$\frac{100 \sqrt{2}}{3}$

$\frac{100}{3}$

Explanation

To solve this problem, we will use the formula for the root mean square (rms) velocity of gas molecules: $\\ v_{rms} = \sqrt{\frac{3RT}{M}} \\$

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$100 \sqrt{2}$

$\frac{400}{\sqrt{3}}$

However, since we are comparing the same gas under different conditions, we can use the relation:

\\ \frac{v_{rms1}}{v_{rms2}} = \sqrt{\frac{T_1}{T_2}} \\

where

v_{rms1}

and

v_{rms2}

are the rms velocities at temperatures

T_1

and

T_2

respectively.

\\

Given:

\\ v_{rms1} = 200

ms

^{-1} \\ T_1 = 27^{\circ}C = 27 + 273 = 300

K

\\ T_2 = 127^{\circ}C = 127 + 273 = 400

K

\\

Substitute these values into the equation:

\\ \frac{200}{v_{rms2}} = \sqrt{\frac{300}{400}} \\

Simplify the square root:

\\ \frac{200}{v_{rms2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \\

Rearrange to solve for

v_{rms2}: \\ v_{rms2} = 200 \times \frac{2}{\sqrt{3}} \\ v_{rms2} = \frac{400}{\sqrt{3}} \\

Therefore, the rms velocity of the gas molecules at the new conditions is

\frac{400}{\sqrt{3}}

ms

^{-1}. \\

This corresponds to Option 4.