If the r.m.s. velocity of a gas at a given temperature (Kelvin scale) is 300 m/sec. What will be the r.m.s. velocity of a gas having twice the molecular weight and half the temperature on Kelvin scale =

300 m/sec

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600 m/sec

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75 m/sec

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150 m/sec

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Solution

The correct option is D 150 m/sec
$v_{r m s} = \sqrt{\frac{3 R T}{M}} \Rightarrow v_{r m s} \propto \sqrt{\frac{T}{M}} \frac{v_{2}}{v_{1}} = \sqrt{\frac{M_{1}}{M_{2}} \times \frac{T_{2}}{T_{1}}} = \sqrt{\frac{1}{2} \times \frac{1}{2}} \Rightarrow v_{2} = \frac{v_{1}}{2} = \frac{300}{2} = 150 m / s e c$

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