Question

Two cylinders, both fitted with frictionless pistons, are filled with mixtures of He and Ar gases. In the first cylinder, the masses of He and Ar are \(m_1\) and \(m_2\), respectively. In the second cylinder, the masses of He and Ar are \(m_2\) and \(m_1\), respectively. The molar mass of Ar is \(10\) times the molar mass of He. The external pressure applied by the piston on the first cylinder needs to be \(5\) times that on the second cylinder so that the volume of the gas mixtures in both the cylinders are equal at the same temperature. Assuming He and Ar behave like ideal gases, the value of \(\left(\dfrac{m_1}{m_2}\right)\) is ____.

Hint:

For ideal gases at same temperature: \[ V \propto \frac{n}{P} \] Always:
• first calculate total moles
• then apply ideal gas proportionality
• carefully substitute pressure ratios Also remember: \[ n = \frac{m}{M} \] where \(m\) is mass and \(M\) is molar mass.

Answer 1

Step 1: Understanding the Concept:
This problem centers on the Ideal Gas Equation and the behavior of gas mixtures.
According to the Ideal Gas Law, the volume of a gas or a mixture of gases is determined by the total number of moles, the temperature, and the pressure: \[ PV = n_{\text{total}}RT \] Since the gases (Helium and Argon) are both noble gases and are assumed to behave ideally, we can calculate the total moles in each cylinder by summing the moles of each individual component.
The core relationship given here is that the volume (\(V\)) and temperature (\(T\)) are identical for both systems.
This implies that the ratio of the total number of moles to the pressure must be the same for both cylinders.
Step 2: Key Formula or Approach:
Let \(M_{\text{He}}\) be the molar mass of Helium and \(M_{\text{Ar}}\) be the molar mass of Argon.
We are given: \(M_{\text{Ar}} = 10 \cdot M_{\text{He}}\).
For Cylinder 1: \[ n_1 = \frac{m_1}{M_{\text{He}}} + \frac{m_2}{M_{\text{Ar}}} = \frac{m_1}{M_{\text{He}}} + \frac{m_2}{10M_{\text{He}}} = \frac{10m_1 + m_2}{10M_{\text{He}}} \] For Cylinder 2: \[ n_2 = \frac{m_2}{M_{\text{He}}} + \frac{m_1}{M_{\text{Ar}}} = \frac{m_2}{M_{\text{He}}} + \frac{m_1}{10M_{\text{He}}} = \frac{10m_2 + m_1}{10M_{\text{He}}} \] From the Ideal Gas Law: \( V = \frac{nRT}{P} \).
Given \(V_1 = V_2\) and \(T_1 = T_2\), we have: \[ \frac{n_1}{P_1} = \frac{n_2}{P_2} \] Step 3: Detailed Explanation:
We are given the pressure condition: \(P_1 = 5P_2\).
Substituting the values of \(n_1, n_2,\) and \(P_1\) into the equality: \[ \frac{\left( \frac{10m_1 + m_2}{10M_{\text{He}}} \right)}{5P_2} = \frac{\left( \frac{10m_2 + m_1}{10M_{\text{He}}} \right)}{P_2} \] We can cancel out the common terms \(10M_{\text{He}}\) and \(P_2\) from both sides: \[ \frac{10m_1 + m_2}{5} = 10m_2 + m_1 \] Now, multiply both sides by 5 to clear the fraction: \[ 10m_1 + m_2 = 5(10m_2 + m_1) \] \[ 10m_1 + m_2 = 50m_2 + 5m_1 \] Rearrange the terms to group \(m_1\) on one side and \(m_2\) on the other: \[ 10m_1 - 5m_1 = 50m_2 - m_2 \] \[ 5m_1 = 49m_2 \] The ratio \((m_1/m_2)\) is therefore: \[ \frac{m_1}{m_2} = \frac{49}{5} = 9.8 \]
Step 4: Final Answer:
By equating the volume-to-temperature ratios for both ideal gas mixtures, we find that the ratio of the masses \(m_1/m_2\) is exactly 9.8.