Question

A mixture of $2$ moles of helium gas $\text{(atomic mass = 4u)}$, and $1$ mole of argon gas $\text{(atomic mass = 40u)}$ is kept at $300\text{K}$ in a container. The ratio of their $RMS$ speeds $\left[\frac{v_{\text{rms}}\left(\text{helium}\right)}{v_{\text{rms}}\left(\text{argon}\right)}\right]$ is close to:

• A. $0.45$
• B. $2.24$
• C. $3.16$
• D. $0.32$

Answer 1

The correct option is C $3.16$

The formula for $RMS$ speed is given by,

$v_{rms}=\sqrt{\frac{3RT}{M}}$

As, $T$ is same for both the gases,

$\Rightarrow \frac{v_{\text{rms(He)}}}{v_{\text{rms(Ar)}}}=\sqrt{\frac{M_{\text{Ar}}}{M_{\text{He}}}}=\sqrt{\frac{40}{4}}$

$\Rightarrow \frac{v_{\text{rms(He)}}}{v_{\text{rms(Ar)}}}=3.16$

Hence, option $\left(A\right)$ is correct.