Question

Two containers are at the same temperature. The first contains a gas with pressure $P_1$, molecular mass $M_1$ and rms speed $v_{rms{}_1}$. The second contains gas with pressure $15P_1$, molecular mass $m_2$, and average speed $V_{avg2}=2.0v_{rms{}_1}$. What is the ratio $m_1:m_2$?

• A. $3:2$
• B. $2:\pi$
• C. $3\pi :2$
• D. $4:3\pi$

Answer 1

The correct option is C

$3\pi :2$

For the second gas,

$\frac{v_{rm{s}_2}}{\left(v_2\right)}=\sqrt{\frac{3K_{B}T}{m}}\times \sqrt{\frac{\pi m}{8K_{B}T}}=\sqrt{\frac{3\pi }{8}}$

$v_{rm{s}_2}=\sqrt{\frac{3\pi }{8}}\left(v_2\right).$

But, it is given, that $v_2=2.0v_{rm{s}_1}$

$v_{rm{s}_2}=\sqrt{\frac{3\pi }{8}}\times 2\times v_{rm{s}_1}.$

Gases have the same average kinetic energies at the same temperature,

$\frac{m_1}{2}{v}_{rm{s}_1}^2=\frac{m_2}{2}{v}_{rm{s}_2}^2$

$\Rightarrow \frac{m_1}{m_2}=\frac{\frac{3\pi }{8}\times 4\times v_{rm{s}_1}^2}{v_{rm{s}_1}^2}$

$\Rightarrow (m_1:m_2)=(3\pi :2).$