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  4. The temperature at which the velocity of oxygen will be half of that of hydrogen at NTP is

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Q. The temperature at which the velocity of oxygen will be half of that of hydrogen at NTP is

BHUBHU 2011

Solution:

$ {{v}_{{{O}_{2}}}}=\frac{1}{2}{{v}_{{{H}_{2}}}} $
$ \therefore $ $ \sqrt{\frac{3RT}{32}}=\frac{1}{2}\sqrt{\frac{3R\times 273}{2}} $
Or $ \frac{T}{32}=\frac{273}{8} $
$ \therefore $ $ T=4\times 273 $
$ T=1092K=1092-273 $
$ T=819{}^\circ C $