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`

Answer

To solve the problem, we need to analyze the situation involving two identical containers A and B containing the same ideal gas. Let's break down the steps to find the relationship between the masses of the gases in the two containers.

### Step 1: Understand the Initial Conditions
Both containers A and B contain the same ideal gas at the same temperature and velocity. The masses of the gases in containers A and B are denoted as \( m_A \) and \( m_B \), respectively.

### Step 2: Write the Initial Pressure Equations
The pressure of an ideal gas can be expressed using the ideal gas law:
\[ P = \frac{nRT}{V} \]
where:
- \( n \) is the number of moles,
- \( R \) is the gas constant,
- \( T \) is the temperature,
- \( V \) is the volume.

Since the number of moles \( n \) can be expressed as \( n = \frac{m}{M} \) (where \( m \) is the mass of the gas and \( M \) is the molar mass), we can write the initial pressures for both containers:

For container A:
\[ P_A = \frac{m_A RT}{MV} \]

For container B:
\[ P_B = \frac{m_B RT}{MV} \]

### Step 3: Write the Final Pressure Equations After Isothermal Expansion
After the gases expand isothermally to a final volume of \( 2V \), the pressures become:

For container A:
\[ P_A' = \frac{m_A RT}{M(2V)} = \frac{m_A RT}{2MV} \]

For container B:
\[ P_B' = \frac{m_B RT}{M(2V)} = \frac{m_B RT}{2MV} \]

### Step 4: Calculate the Change in Pressure for Both Containers
The change in pressure for container A is:
\[ \Delta P_A = P_A - P_A' = \frac{m_A RT}{MV} - \frac{m_A RT}{2MV} = \frac{m_A RT}{MV} - \frac{m_A RT}{2MV} = \frac{m_A RT}{2MV} \]

The change in pressure for container B is:
\[ \Delta P_B = P_B - P_B' = \frac{m_B RT}{MV} - \frac{m_B RT}{2MV} = \frac{m_B RT}{MV} - \frac{m_B RT}{2MV} = \frac{m_B RT}{2MV} \]

### Step 5: Relate the Changes in Pressure
According to the problem, we have:
\[ \Delta P_A = \Delta P \quad \text{and} \quad \Delta P_B = 1.5 \Delta P \]

Substituting the expressions for \( \Delta P_A \) and \( \Delta P_B \):
\[ \frac{m_A RT}{2MV} = \Delta P \quad \text{(1)} \]
\[ \frac{m_B RT}{2MV} = 1.5 \Delta P \quad \text{(2)} \]

### Step 6: Substitute Equation (1) into Equation (2)
From equation (1), we can express \( \Delta P \):
\[ \Delta P = \frac{m_A RT}{2MV} \]

Substituting this into equation (2):
\[ \frac{m_B RT}{2MV} = 1.5 \left(\frac{m_A RT}{2MV}\right) \]

### Step 7: Simplify the Equation
Cancelling \( RT/2MV \) from both sides gives:
\[ m_B = 1.5 m_A \]

### Step 8: Express the Relationship Between \( m_A \) and \( m_B \)
Rearranging gives:
\[ m_A = \frac{2}{3} m_B \]

### Conclusion
Thus, the relationship between the masses of the gases in the two containers is:
\[ 3m_A = 2m_B \]