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. 1. `4m_(A)=9m_(B)`
B. 2. `2m_(A)=3m_(B)`
C. 3. `3m_(A)=2m_(B)`
D. 4. `9m_(A)=4m_(B)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Initial Conditions
We have two identical containers A and B containing the same ideal gas at the same temperature. The gas in container A has a mass \( m_A \) and the gas in container B has a mass \( m_B \). Both gases are allowed to expand isothermally to a final volume of \( 2V \).

**Hint:** Identify that both containers have the same initial conditions and are undergoing the same process.

### Step 2: Use the Ideal Gas Law
The ideal gas law states that:
\[ PV = nRT \]
Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = temperature

Since the temperature is constant and the gas is ideal, we can express the initial pressures in terms of the number of moles and the volume.

### Step 3: Express the Number of Moles
The number of moles of gas in each container can be expressed as:
- For container A:
\[ n_A = \frac{m_A}{M} \]
- For container B:
\[ n_B = \frac{m_B}{M} \]
Where \( M \) is the molar mass of the gas.

### Step 4: Write the Initial Pressures
Using the ideal gas law, we can write the initial pressures for both containers:
- For container A:
\[ P_A = \frac{n_A RT}{V} = \frac{m_A RT}{MV} \]
- For container B:
\[ P_B = \frac{n_B RT}{V} = \frac{m_B RT}{MV} \]

### Step 5: Consider the Changes in Pressure
After the isothermal expansion, the pressures change:
- For container A, the change in pressure is \( \Delta P \):
\[ P_{A, final} = P_A + \Delta P \]
- For container B, the change in pressure is \( 1.5 \Delta P \):
\[ P_{B, final} = P_B + 1.5 \Delta P \]

### Step 6: Apply the Ideal Gas Law After Expansion
After expansion, we can write:
- For container A:
\[ P_A + \Delta P = \frac{n_A RT}{2V} \]
- For container B:
\[ P_B + 1.5 \Delta P = \frac{n_B RT}{2V} \]

### Step 7: Substitute and Rearrange
Substituting the expressions for \( P_A \) and \( P_B \):
1. For container A:
\[ \frac{m_A RT}{MV} + \Delta P = \frac{m_A RT}{2MV} \]
Rearranging gives:
\[ \Delta P = \frac{m_A RT}{2MV} - \frac{m_A RT}{MV} = -\frac{m_A RT}{2MV} \]

2. For container B:
\[ \frac{m_B RT}{MV} + 1.5 \Delta P = \frac{m_B RT}{2MV} \]
Rearranging gives:
\[ 1.5 \Delta P = \frac{m_B RT}{2MV} - \frac{m_B RT}{MV} = -\frac{m_B RT}{2MV} \]

### Step 8: Equate the Two Expressions for \( \Delta P \)
Since both expressions for \( \Delta P \) must be equal:
\[ -\frac{m_A RT}{2MV} = -\frac{m_B RT}{3MV} \]

### Step 9: Simplify the Equation
Cancelling common terms gives:
\[ \frac{m_A}{2} = \frac{m_B}{3} \]
Cross-multiplying yields:
\[ 3m_A = 2m_B \]

### Step 10: Find the Relationship Between Masses
Thus, we can express the relationship between the masses:
\[ \frac{m_A}{m_B} = \frac{2}{3} \]

### Final Answer
The relationship between the masses of the gases in containers A and B is:
\[ \frac{m_A}{m_B} = \frac{2}{3} \]

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