Ratio of Masses in Ideal Gas Mixture Cylinders

To find the ratio of the masses $\left(\frac{m_1}{m_2}\right)$, we apply the ideal gas equation $PV = nRT$ to both cylinders.

Let the molar mass of $\text{He}$ be $M_{\text{He}} = M$. Given that the molar mass of $\text{Ar}$ is $10$ times that of $\text{He}$, we have $M_{\text{Ar}} = 10M$.

Cylinder 1:

• Mass of $\text{He} = m_1$
• Mass of $\text{Ar} = m_2$

The total number of moles of gas in the first cylinder, $n_1$, is: $n_1 = n_{\text{He}} + n_{\text{Ar}} = \frac{m_1}{M_{\text{He}}} + \frac{m_2}{M_{\text{Ar}}} = \frac{m_1}{M} + \frac{m_2}{10M} = \frac{10m_1 + m_2}{10M}$

Using the ideal gas equation for Cylinder 1: $P_1 V = n_1 R T$ $P_1 V = \left(\frac{10m_1 + m_2}{10M}\right) RT \quad \text{--- (1)}$

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Cylinder 2:

• Mass of $\text{He} = m_2$
• Mass of $\text{Ar} = m_1$

The total number of moles of gas in the second cylinder, $n_2$, is: $n_2 = n_{\text{He}} + n_{\text{Ar}} = \frac{m_2}{M_{\text{He}}} + \frac{m_1}{M_{\text{Ar}}} = \frac{m_2}{M} + \frac{m_1}{10M} = \frac{10m_2 + m_1}{10M}$

Using the ideal gas equation for Cylinder 2: $P_2 V = n_2 R T$ $P_2 V = \left(\frac{10m_2 + m_1}{10M}\right) RT \quad \text{--- (2)}$

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Ratio of Pressures:

Given that the volumes ($V$) and temperatures ($T$) are equal for both cylinders, and the external pressure applied to the first cylinder is $5$ times that on the second cylinder ($P_1 = 5P_2$), we divide equation (1) by equation (2):

$\frac{P_1 V}{P_2 V} = \frac{n_1 RT}{n_2 RT}$

$\frac{5P_2}{P_2} = \frac{n_1}{n_2}$

$5 = \frac{\frac{10m_1 + m_2}{10M}}{\frac{10m_2 + m_1}{10M}}$

$5 = \frac{10m_1 + m_2}{10m_2 + m_1}$

Cross-multiplying to solve for $\frac{m_1}{m_2}$:

$5(10m_2 + m_1) = 10m_1 + m_2$

$50m_2 + 5m_1 = 10m_1 + m_2$

$50m_2 - m_2 = 10m_1 - 5m_1$

$49m_2 = 5m_1$

$\frac{m_1}{m_2} = \frac{49}{5} = 9.8$