Two containers $A and B$ contain ideal gases, helium and oxygen respectively. Volume of both containers are equal, and pressure is also equal. Container $A$ has twice the number of molecules than container $B .$ If $V_{A} and V_{B}$ represent the rms speed of gases in containers $A and B$ respectively, then

\frac{V_{A}}{V_{B}} = \sqrt{2}

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\frac{V_{A}}{V_{B}} = 4

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\frac{V_{A}}{V_{B}} = 2

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\frac{V_{A}}{V_{B}} = \sqrt{8}

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Solution

The correct option is C $\frac{V_{A}}{V_{B}} = 2$
As we know that the ideal gas is given by : $P V = n R T$

$F o r A : P V = 2 n R T_{A} \dots (1)$
$F o r B : P V = n R T_{B} \dots (2)$

Given that pressure and volume are same therefore,
from equations (1) and (2)khj
$T_{B} = 2 T_{A}$

RMS speed of gas is given by :

$V_{r m s} = \sqrt{\frac{3 R T}{M}}$

$\frac{V_{A}}{V_{B}} = \frac{\sqrt{\frac{3 R T_{A}}{4}}}{\sqrt{\frac{3 R T_{B}}{32}}}$

$\frac{V_{A}}{V_{B}} = \frac{\sqrt{\frac{3 R T_{A}}{4}}}{\sqrt{\frac{3 R T_{A}}{32}}} = 2$

$\frac{V_{A}}{V_{B}} = 2$

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