Question

A mixture of $2$ moles of helium gas $\left( \text{atomic mass = }4u \right)$, and $1$ mole of argon gas $\left( \text{atomic mass = }40u \right)$ is kept at $300K$ in a container. The ratio of their rms speeds $\left[ \dfrac{{{V}_{rms}}\left( \text{helium} \right)}{{{V}_{rms}}\left( \text{argon} \right)} \right]$, is close to:
$A)\text{ }2.24$
$B)\text{ }0.45$
$C)\text{ 0}\text{.32}$
$D)\text{ }3.16$

Answer 1

Hint: This problem can be solved by using the mathematical formula for the root mean square velocity of an ideal gas as a function of its atomic mass and its temperature. By applying the formula for both the gases, we can get the required ratio of the rms velocities.

Formula used:
${{V}_{rms}}=\sqrt{\dfrac{3RT}{M}}$

Complete step-by-step answer:
We can get the required ratio by applying the mathematical formula for the root mean square velocity of an ideal gas for both the cases.
The root mean square velocity ${{V}_{rms}}$ of an ideal gas at temperature $T$ is given by,
${{V}_{rms}}=\sqrt{\dfrac{3RT}{M}}$ --(1)
Where $M$ is the atomic mass of the gas and $R=8.314J.mo{{l}^{-1}}{{K}^{-1}}$ is the universal gas constant.
Hence, let us analyze the question.
Let the rms speed of helium gas be ${{V}_{rms}}\left( \text{helium} \right)$
The temperature of the gas is $T=300K$.
The atomic mass of helium gas is ${{M}_{\text{helium}}}=4u$.
Therefore, using (1), we get,
${{V}_{rms}}\left( \text{helium} \right)=\sqrt{\dfrac{3RT}{{{M}_{\text{helium}}}}}$ --(2)
Similarly,
Let the rms speed of argon gas be ${{V}_{rms}}\left( \text{argon} \right)$
The temperature of the gas is $T=300K$.
The atomic mass of helium gas is ${{M}_{\arg on}}=40u$.
Therefore, using (1), we get,
${{V}_{rms}}\left( \text{argon} \right)=\sqrt{\dfrac{3RT}{{{M}_{\text{argon}}}}}$ --(3)
Upon dividing (2) by (3), we will get the required ratio. Hence, let us do that.
$\dfrac{{{V}_{rms}}\left( \text{helium} \right)}{{{V}_{rms}}\left( \text{argon} \right)}=\dfrac{\sqrt{\dfrac{3RT}{{{M}_{\text{helium}}}}}}{\sqrt{\dfrac{3RT}{{{M}_{\text{argon}}}}}}=\sqrt{\dfrac{{{M}_{\text{argon}}}}{{{M}_{\text{helium}}}}}$
Therefore, using the above information, we get.
$\dfrac{{{V}_{rms}}\left( \text{helium} \right)}{{{V}_{rms}}\left( \text{argon} \right)}=\sqrt{\dfrac{{{M}_{\text{argon}}}}{{{M}_{\text{helium}}}}}=\sqrt{\dfrac{40u}{4u}}=\sqrt{10}\approx 3.16$
Hence, we have got the required ratio as $3.16:1$.
Hence, the correct option is $D)\text{ }3.16$.

Note: Students might get confused upon seeing the number of moles of the individual gases being mentioned. However they must remember the formula of root mean square velocity of an ideal gas and thus realize that the number of moles of the gases given in the question is immaterial to the question. Sometimes, questions contain extra information on purpose to waiver the confidence of the students.