Question

Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and volume \(V\). The masses are \(m_A\) and \(m_B\). On isothermal expansion to \(2V\), pressure changes are \(\Delta p\) and \(1.5\Delta p\) respectively. The relation between masses is:

• A. \( \frac{m_A}{m_B} = \frac{4}{9} \)
• B. \( \frac{m_A}{m_B} = \frac{2}{3} \)
• C. \( \frac{m_A}{m_B} = \frac{3}{2} \)
• D. \( \frac{m_A}{m_B} = \frac{9}{4} \)

Hint:

In isothermal expansion to \(2V\), pressure drop is always half of initial pressure.

Answer 1

The Correct Option is C

Concept: For isothermal process: \[ pV = nRT \Rightarrow p \propto n \propto m \]

Step 1: Pressure change relation.
\[ \Delta p = p_i - p_f = p_i - \frac{p_i}{2} = \frac{p_i}{2} \] Thus: \[ \Delta p \propto p_i \propto m \]

Step 2: Use given data.
For A: \[ \Delta p_A \propto m_A \] For B: \[ \Delta p_B = 1.5\Delta p_A \propto m_B \] \[ \Rightarrow \frac{m_B}{m_A} = \frac{3}{2} \Rightarrow \frac{m_A}{m_B} = \frac{3}{2} \]