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Q.

A flask contains hydrogen and helium in the ratio 2 : 1 by mass. The temperature of the mixture is 27^oC. Obtain the ratio of rms speeds of the molecules of the two gases. Molecular mass of hydrogen = 2, molecular mass of helium = 4.

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a

2 : 1

b

$\sqrt{2} : 1$

c

1 : 2

d

1 : 3

answer is B.

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Detailed Solution

Average KE of the molecule $= \frac{1}{2} m v_{r m s}^{2}$
For hydrogen, $\frac{1}{2} m_{H} v_{H}^{2} = (\frac{3}{2}) k T$ …(1)
For helium, $\frac{1}{2} m_{H e} v_{H e}^{2} = \frac{3}{2} k T$ …(2)
So, $\frac{v_{H}^{2}}{v_{H e}^{2}} = \frac{M_{H e}}{M_{H}} = \frac{4}{2} = 2$
Or, $\frac{v_{H}}{v_{H e}} = \sqrt{2} : 1$ .

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A flask contains hydrogen and helium in the ratio 2 : 1 by mass. The temperature of the mixture is 27oC. Obtain the ratio of rms speeds of the molecules of the two gases. Molecular mass of hydrogen = 2, molecular mass of helium = 4.