Two containers are at the same temperature. The first contains gas with pressure${\mathit{p}}_{\mathbf{1}}$, molecular mass${\mathit{m}}_{\mathbf{1}}$, and RMS speed$V_{rms1}$. The second contains gas with pressure$\mathbf{2}\mathbf{.}\mathbf{0}{\mathit{p}}_{\mathbf{1}}$, molecular mass$m_2$, and average speed${\mathit{V}}_{\mathbf{a}\mathbf{v}\mathbf{g}\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{0}{\mathit{V}}_{\mathbf{r}\mathbf{m}\mathbf{s}\mathbf{1}}$. Find the mass ratio${\mathit{m}}_{\mathbf{1}}\mathbf{/}{\mathit{m}}_{\mathbf{2}}$.

• The first container has pressure$p_1$, molecular mass m and RMS speed$v_{rms1}$
• The second container has pressure$p_2=2p_1$, molecular massand average speed$v_{avg2}=2v_{rms1}$.
• Two containers are at the same temperature.

The expression for the RMS speed is given by,

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

Here $V_{rms}$is the RMS speed, R is the universal gas constant, T is the temperature and M is the mass.

The expression for the average speed is given by,

$V_{avg}=\sqrt{\frac{8RT}{\pi M}}$

Here$V_{avg}$ is the average speed.

The formula for rms speed is,

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

For the first container, molecular mass is$m_1$.

$v_{rms1}=\sqrt{\frac{3RT}{m_1}}$ ……. (1)

And the formula for average speed is

$V_{avg}=\sqrt{\frac{8RT}{\pi M}}$

For the second container, molecular mass is.

$v_{avg2}=\sqrt{\frac{8RT}{\pi m_2}}$……. (2)

The ratio of the equation (2 ) to equation (1) is,

$\frac{{\text{v}}_{avg2}}{{\text{v}}_{rms1}}=\frac{\sqrt{\frac{8\text{RT}}{\pi {\text{m}}_2}}}{\sqrt{\frac{3\text{RT}}{{\text{m}}_1}}}$

$\frac{v_{avg2}}{v_{rms1}}=\sqrt{\frac{8m_1}{3\pi m_2}}$……. (3)

Fromthegiven problem,${\text{v}}_{avg2}=2{\text{v}}_{rms1}$

$\frac{v_{avg2}}{v_{rms1}}=2$……. (4)

Substituting (3) in (4) and squaring both the sides, we get

$$\begin{gathered} \frac{8m_1}{3\pi m_2}=4 \\ \begin{array}{c}\frac{m_1}{m_2}=\frac{3\pi }{2}\\ =4.7\end{array} \end{gathered}$$

Therefore the mass ratio ${\text{m}}_1/{\text{m}}_2$is$4.7$ .