Question

Two identical containers A and B with friction-less pistons contain the same ideal gas at the same temperature and the same volume, \(V\). The mass of gas in A is \(m_{A}\) and that in B is \(m_{B}\). The gas in each cylinder is now allowed to expand isothermally to the same final volume \(2V\). The changes in pressure in A and B are found to be \(\Delta p\) and \(3\Delta p\) respectively. Then

• A. \(m_{A}=m_{B}\)
• B. \(3m_{A}=2m_{B}\)
• C. \(3m_{A}=m_{B}\)
• D. \(3m_{A}=4m_{B}\)

Hint:

In an isothermal expansion to twice the volume the pressure halves, so the pressure drop is proportional to the mass of gas.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For an ideal gas, \(pV=\frac{m}{M}RT\), so the initial pressure \(p=\frac{mRT}{MV}\) is proportional to the mass \(m\).

Step 2: Isothermal expansion
For expansion from \(V\) to \(2V\) at constant \(T\), \(p_{f}=\frac{p}{2}\). So \(\Delta p=\frac{p}{2}=\frac{mRT}{2MV}\).

Step 3: Calculation
\(\Delta p\propto m\). For A, \(\Delta p=\frac{m_{A}RT}{2MV}\) and for B, \(3\Delta p=\frac{m_{B}RT}{2MV}\).

Step 4: Ratio
Dividing, \(3=\frac{m_{B}}{m_{A}}\), that is \(3m_{A}=m_{B}\), which is option (C).

Final Answer:
The masses satisfy \(3m_{A}=m_{B}\). \[ \boxed{\text{(C) }3m_{A}=m_{B}} \]