The molecular mass of a gas having r.m.s. speed four times as that of another gas having molecular mass 32 is

The molecular mass of a gas having r.m.s. speed four times as that of another gas having molecular mass 32 is
  1. 2
  2. 4
  3. 26
  4. 32

Solution

$\begin{array}{ll} & \mathrm{v}_{\mathrm{rms}} \propto \frac{1}{\sqrt{\mathrm{M}}} \\ \therefore \quad & \frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_2}{\left(\mathrm{v}_{\mathrm{rms}}\right)_1}=\sqrt{\frac{\mathrm{M}_1}{\mathrm{M}_2}} \\ & \text { Given that }\left(\mathrm{v}_{\mathrm{rms}}\right)_2=4\left(\mathrm{v}_{\mathrm{rms}}\right)_1 \\ \therefore \quad & \sqrt{\frac{32}{\mathrm{M}_2}}=4 \\ \therefore \quad & \frac{32}{\mathrm{M}_2}=16 \\ \therefore \quad & \mathrm{M}_2=2\end{array}$ *

Asked in: MHT CET 2023 (14 May Shift 2)

Practice more Kinetic Theory of Gases questions on Aicharya