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

Concept: For ideal gases, \[ PV = nRT \] At constant temperature, \[ V \propto \frac{n}{P} \] Thus, if two gaseous systems have equal volume at the same temperature, \[ \frac{n_1}{P_1} = \frac{n_2}{P_2} \] The number of moles is calculated using: \[ n = \frac{\text{mass}}{\text{molar mass}} \] We are also given: \[ M_{\mathrm{Ar}} = 10M_{\mathrm{He}} \] This relation will help simplify the mole expressions.

Step 1: Calculating total moles in the first cylinder. Let molar mass of He be: \[ M \] Then molar mass of Ar is: \[ 10M \] In the first cylinder:
• Mass of He \(= m_1\)
• Mass of Ar \(= m_2\) Hence moles of He: \[ \frac{m_1}{M} \] and moles of Ar: \[ \frac{m_2}{10M} \] Therefore total moles in first cylinder are: \[ n_1 = \frac{m_1}{M} + \frac{m_2}{10M} \] Taking common denominator: \[ n_1 = \frac{10m_1 + m_2}{10M} \]

Step 2: Calculating total moles in the second cylinder. In the second cylinder:
• Mass of He \(= m_2\)
• Mass of Ar \(= m_1\) Thus moles of He: \[ \frac{m_2}{M} \] and moles of Ar: \[ \frac{m_1}{10M} \] Hence total moles are: \[ n_2 = \frac{m_2}{M} + \frac{m_1}{10M} \] \[ n_2 = \frac{10m_2 + m_1}{10M} \]

Step 3: Using the equal volume condition. The problem states: \[ P_1 = 5P_2 \] Also the volumes are equal and temperature is same. Using: \[ \frac{n_1}{P_1} = \frac{n_2}{P_2} \] Substitute \(P_1 = 5P_2\): \[ \frac{n_1}{5P_2} = \frac{n_2}{P_2} \] Cancel \(P_2\): \[ \frac{n_1}{5} = n_2 \] Thus, \[ n_1 = 5n_2 \]

Step 4: Substituting mole expressions. Using expressions for \(n_1\) and \(n_2\): \[ \frac{10m_1 + m_2}{10M} = 5\left( \frac{10m_2 + m_1}{10M} \right) \] Multiply both sides by \(10M\): \[ 10m_1 + m_2 = 5(10m_2 + m_1) \] Expand: \[ 10m_1 + m_2 = 50m_2 + 5m_1 \] Rearranging: \[ 10m_1 - 5m_1 = 50m_2 - m_2 \] \[ 5m_1 = 49m_2 \] \[ \frac{m_1}{m_2} = \frac{49}{5} \] But this contradicts the expected physical simplification. Let us carefully interpret the pressure statement again. The pressure on the first cylinder must be \(5\) times the second cylinder: \[ P_1 = 5P_2 \] For equal volume at same temperature: \[ \frac{n_1RT}{P_1} = \frac{n_2RT}{P_2} \] Thus, \[ \frac{n_1}{5P_2} = \frac{n_2}{P_2} \] \[ n_1 = 5n_2 \] Substituting correctly: \[ 10m_1 + m_2 = 5(10m_2 + m_1) \] \[ 10m_1 + m_2 = 50m_2 + 5m_1 \] \[ 5m_1 = 49m_2 \] Thus mathematically: \[ \boxed{\frac{m_1}{m_2} = \frac{49}{5}} \] Final Answer: \[ \boxed{\frac{49}{5}} \]