91Ó°ÊÓ

• 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,

$V_{avg}=\sqrt{\frac{8RT}{\mathrm{Ï€}M}}$

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

$V_{avg}=\sqrt{\frac{8RT}{\mathrm{Ï€}M}}$

For the second container, molecular mass is.

$v_{avg2}=\sqrt{\frac{8RT}{\mathrm{π}{m}_2}}$……. (2)

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

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

$\frac{v_{avg2}}{v_{rms1}}=\sqrt{\frac{8m_1}{3\mathrm{π}{m}_2}}$……. (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

$$\begin{gathered} \frac{8m_1}{3\mathrm{Ï€}{m}_2}=4 \\ \begin{array}{c}\frac{m_1}{m_2}=\frac{3\mathrm{Ï€}}{2}\\ =4.7\end{array} \end{gathered}$$

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