Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and the same velocity V. The mass of the 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 the pressure in A and B are found to be `DeltaP and 1.5 DeltaP` respectively. Then

Options

A. `4m_(A)=9m_(B)`
B. `3m_(A)=3m_(B)`
C. `3m_(A)=2m_(B)`
D. `9m_(A)=4m_(B)`

Explanation

To solve the problem step by step, we will analyze the situation involving two identical containers A and B containing the same ideal gas, and derive the relationship between the masses of the gases in each container based on the given conditions.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- Both containers A and B contain the same ideal gas at the same initial temperature \( T \) and volume \( V \).
- Let the mass of the gas in container A be \( m_A \) and in container B be \( m_B \).

2. **Applying the Ideal Gas Law**:
- For container A:
\[ P_A V = \frac{m_A}{M} RT \quad \text{(1)} \]
- For container B:
\[ P_B V = \frac{m_B}{M} RT \quad \text{(2)} \]
- Here, \( M \) is the molar mass of the gas, \( R \) is the universal gas constant, and \( P_A \) and \( P_B \) are the initial pressures in containers A and B, respectively.

3. **Final Conditions After Isothermal Expansion**:
- Both gases expand isothermally to a final volume of \( 2V \).
- For container A:
\[ P_A' (2V) = \frac{m_A}{M} RT \quad \text{(3)} \]
- For container B:
\[ P_B' (2V) = \frac{m_B}{M} RT \quad \text{(4)} \]
- Here, \( P_A' \) and \( P_B' \) are the final pressures in containers A and B, respectively.

4. **Calculating the Change in Pressure**:
- The change in pressure for container A is given as \( \Delta P = P_A' - P_A \).
- From equations (1) and (3):
\[ \Delta P = \frac{m_A}{MRT} \cdot \frac{1}{2V} - \frac{m_A}{MRT} \cdot \frac{1}{V} = \frac{m_A}{MRT} \left( \frac{1}{2V} - \frac{1}{V} \right) = -\frac{m_A}{2MRTV} \]
- Therefore, we have:
\[ \Delta P = -\frac{m_A}{2MRTV} \]

5. **For Container B**:
- The change in pressure for container B is given as \( 1.5 \Delta P \).
- From equations (2) and (4):
\[ 1.5 \Delta P = P_B' - P_B = \frac{m_B}{MRT} \cdot \frac{1}{2V} - \frac{m_B}{MRT} \cdot \frac{1}{V} = \frac{m_B}{MRT} \left( \frac{1}{2V} - \frac{1}{V} \right) = -\frac{m_B}{2MRTV} \]
- Therefore, we have:
\[ 1.5 \Delta P = -\frac{m_B}{2MRTV} \]

6. **Setting Up the Equation**:
- Since we have \( \Delta P \) for A and \( 1.5 \Delta P \) for B, we can set up the equation:
\[ -\frac{m_B}{2MRTV} = 1.5 \left( -\frac{m_A}{2MRTV} \right) \]
- Simplifying this gives:
\[ m_B = 1.5 m_A \]

7. **Final Relationship**:
- Rearranging gives:
\[ 3m_A = 2m_B \]

### Conclusion:
From the calculations, we find that the relationship between the masses of the gases in containers A and B is:
\[ 3m_A = 2m_B \]