Question 9

Let the molar mass of He be $M$. Then, the molar mass of Ar is $10M$.

For the first cylinder:

Moles of He, $n_{He1} = \frac{m_1}{M}$

Moles of Ar, $n_{Ar1} = \frac{m_2}{10M}$

Total moles $n_1 = \frac{m_1}{M} + \frac{m_2}{10M} = \frac{10m_1 + m_2}{10M}$

For the second cylinder:

Moles of He, $n_{He2} = \frac{m_2}{M}$

Moles of Ar, $n_{Ar2} = \frac{m_1}{10M}$

Total moles $n_2 = \frac{m_2}{M} + \frac{m_1}{10M} = \frac{10m_2 + m_1}{10M}$

Using the ideal gas equation $PV = nRT$, at constant $V$ and $T$, $P \propto n$.

Given $P_1 = 5P_2$, we have $n_1 = 5n_2$.

$\frac{10m_1 + m_2}{10M} = 5 \left( \frac{10m_2 + m_1}{10M} \right)$

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

$5m_1 = 49m_2$

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