91Ó°ÊÓ

1. The first container is at pressure p₁.
2. The molecule with mass is m₁.
3. The rms speed is $v_{rms1}$.
4. The second container is at pressure p₂ .
5. The molecule with mass is m₂.

The rms speed is $v_{avg2}=2v_{rms1}$.

By using equations of average speed of the ideal gas Eq. (19-31) and therms speed of theideal gas Eq. (19-34), find the ratioof the masses (m₁ /m₂)of their molecule.

${\mathbf{v}}_{\mathbf{avg}}\mathbf{=}\sqrt{\frac{\mathbf{8}\mathbf{RT}}{\mathbf{Ï€²Ñ}}}··············································\left(1\right)$

${\mathbf{v}}_{\mathbf{rms}}\mathbf{=}\sqrt{\frac{\mathbf{3}\mathbf{RT}}{\mathbf{M}}}··············································\left(2\right)$

where,$\mathbf{γ}\mathbf{=}{\mathbf{C}}_{\mathbf{p}}\mathbf{/}{\mathbf{C}}_{\mathbf{V}}$, Ris the gas constant, Tis the temperature, vis velocity and M is the molar mass.

Dividing Eq. (1) by Eq. (2),

$$\begin{gathered} \frac{v_{avg2}}{v_{rms1}}=\frac{\sqrt{\frac{8RT}{\mathrm{π}{M}_2}}}{\sqrt{\frac{3RT}{M_1}}} \\ =\sqrt{\frac{8M_1}{3\mathrm{π}{M}_2}}…………………………………………\left(3\right) \end{gathered}$$

For${\text{v}}_{avg2}=2{\text{v}}_{rms1}$,we have-

$\frac{m_1}{m_2}=\frac{M_1}{M_2}$

From Eq. (3),

$$\begin{gathered} \frac{m_1}{m_2}=\frac{M_1}{M_2} \\ \frac{m_1}{m_2}=\frac{3\mathrm{Ï€}}{8}{\left(\frac{v_{avg2}}{v_{rms1}}\right)}^2 \\ \frac{m_1}{m_2}=\frac{3\mathrm{Ï€}}{8}{\left(2\right)}^2 \\ \frac{m_1}{m_2}=\frac{3\mathrm{Ï€}}{2} \\ \frac{m_1}{m_2}=4.71 \end{gathered}$$

Hence, the ratio(m₁ /m₂) of the masses of their molecules is, 4.71 .