At which temperature the r.m.s. velocity of a hydrogen molecule equal to that of an oxygen molecule at 47 ° C ?

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TestHub · Accepted Answer

The r.m.s velocity of the molecule of a gas of molecular weight $M$ at Kelvin temperature $T$ is given by, $V_{r m s} = \sqrt{\frac{3 R T}{M}}$ .

Let $M_{O}$ and $M_{H}$ are molecular weight of oxygen and hydrogen and $T_{O}$ and $T_{H}$ the corresponding Kelvin temperature at which $V_{r . m . s}$ is same for both,

$V_{r . m . s} = \sqrt{\frac{3 R T_{O}}{M_{O}}} = \sqrt{\frac{3 R T_{H}}{M_{H}}}$ . Hence $\frac{T_{O}}{M_{O}} = \frac{T_{H}}{M_{H}}$

$T_{O} = 273 ° C + 47 ° C = 320 K$ ,

$M_{O} = 32$ ,

$M_{H} = 2 \Rightarrow T_{H} = \frac{2}{32} \times 320 = 20 K$

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