A mixture of 2 moles of helium gas (atomic mass $=4 \mathrm{amu}$ ) and 1 mole of argon gas (atomic mass $=40 \mathrm{amu}$ ) is kept at $300 \mathrm{~K}$ in a container. The ratio of the rms speeds $\left(\frac{v_{\mathrm{rms}}( ext { helium })}{v_{\mathrm{rms}}(\operatorname{argon})} ight)$ is

Question

A mixture of 2 moles of helium gas (atomic mass $=4 \mathrm{amu}$ ) and 1 mole of argon gas (atomic mass $=40 \mathrm{amu}$ ) is kept at $300 \mathrm{~K}$ in a container. The ratio of the rms speeds $\left(\frac{v_{\mathrm{rms}}(	ext { helium })}{v_{\mathrm{rms}}(\operatorname{argon})}ight)$ is

Aicharya · Accepted Answer

Using $V_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}} \Rightarrow V_{r m s} \propto \frac{1}{\sqrt{M}}$ $\frac{v_{	ext {rms (helium) }}}{v_{	ext {rms (argon) }}}=\sqrt{\frac{M_{	ext {argon }}}{M_{	ext {helium }}}}=\sqrt{\frac{40}{4}}=\sqrt{10} \approx 3.16$

A mixture of 2 moles of helium gas (atomic mass $=4 \mathrm{amu}$ ) and 1 mole of argon gas (atomic mass…

Solution