If the root mean square velocity of the molecules of hydrogen at NTP is $1.84$ km/s. Calculate the root mean square velocity of the oxygen molecule at NTP, molecule weight of hydrogen and oxygen are $2$ and $32$ respectively.

1.47

km/s

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0.94

km/s

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1.84

km/s

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0.47

km/s

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Solution

The correct option is B $0.47$ km/s
rms velocity of $M_{2}$ at $11 T P = V_{1} = 1.84 k m / s$
as, $V_{1} = \sqrt{\frac{3 R T}{M_{1}}} \to (1)$ $M_{1} =$ molar man of $H_{2} = 0.002$ kg/moL
and that of oxygen $(O_{2})$
$V_{2} = \sqrt{\frac{3 R T}{M_{2}}} \to (2)$ $M_{2} = 0.032$ kg/moL
so dividing $(1) \div (2)$ ,
$\frac{V_{1}}{V_{2}} = \sqrt{\frac{M_{2}}{M_{1}}} \Rightarrow V_{2} = V_{1} \sqrt{\frac{M_{1}}{M_{2}}}$
$\Rightarrow V_{2} = 1.84 \times \sqrt{\frac{2}{32}} = 1.84 \times \frac{1}{4} = 0.46$ km/s.

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