Question

The gas having average speed four times as that of $ SO_2 $ (molecular mass $ 64 $ ) is

• A. $ He $ (molecular mass $ 4 $ )
• B. $ O_2 $ (molecular mass $ 32 $ )
• C. $ H_2 $ (molecular mass $ 2 $ )
• D. $ CH_4 $ (molecular mass $ 16 $ )

Answer 1

Answer: (A)

Average speed $ (v_{av}) $ of gas molecules is inversely proportional to the square root of the absolute temperature (T) of the gas
$ v_{av} = \sqrt{\frac{8RT}{\pi M}} $
where R is gas constant and M the molecular weight.
Given, $ \, \, \, v_1 = v, M_1 = 64, v_2 = 4v $
$ \therefore \frac{v_1}{v_2} = \sqrt{\frac{M_2}{M_1}} $
$ \frac{v}{4v} = \sqrt{\frac{M_2}{64}} $
$ \Rightarrow M_2 = \frac{64}{16} = 4 $
Hence, the gas is helium (molecular mass 4).