Two identical containers A and B having same volume of an ideal gas at same temperature have mass of the gas as `m_(1)` and `m_(2)` respectively and `2m_(1) = 3m_(2)`. The gas in each cylinder expands isomthermally to double of its volume. If change in pressure in A is `300 Pa`, then the change in pressure in B is

Options

A. 200 Pa
B. 300 Pa
C. 400 Pa
D. 500 Pa

Answer

To solve the problem, we will use the ideal gas law and the relationship between mass, pressure, and volume during isothermal expansion.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- We have two identical containers A and B with ideal gases.
- Mass of gas in container A is \( m_1 \) and in container B is \( m_2 \).
- The relationship between the masses is given as \( 2m_1 = 3m_2 \).

2. **Expressing Masses**:
- From the relationship \( 2m_1 = 3m_2 \), we can express \( m_2 \) in terms of \( m_1 \):
\[ m_2 = \frac{2}{3} m_1 \]

3. **Using the Ideal Gas Law**:
- The ideal gas law states that \( PV = nRT \), where \( n \) is the number of moles.
- The number of moles \( n \) can be expressed as \( n = \frac{m}{M} \), where \( M \) is the molar mass of the gas.
- For container A:
\[ P_1 V = \frac{m_1}{M} RT \implies P_1 = \frac{m_1 RT}{MV} \]
- For container B:
\[ P_2 V = \frac{m_2}{M} RT \implies P_2 = \frac{m_2 RT}{MV} \]

4. **Calculating the Initial Pressures**:
- Substitute \( m_2 = \frac{2}{3} m_1 \) into the equation for \( P_2 \):
\[ P_2 = \frac{\left(\frac{2}{3} m_1\right) RT}{MV} = \frac{2m_1 RT}{3MV} \]

5. **Calculating Change in Pressure for Container A**:
- The gas in container A expands isothermally to double its volume:
\[ P_1' = \frac{m_1 RT}{2MV} \]
- The change in pressure for container A is:
\[ \Delta P_A = P_1 - P_1' = \frac{m_1 RT}{MV} - \frac{m_1 RT}{2MV} = \frac{m_1 RT}{MV} \left(1 - \frac{1}{2}\right) = \frac{m_1 RT}{2MV} \]

6. **Given Change in Pressure for Container A**:
- We know \( \Delta P_A = 300 \, \text{Pa} \):
\[ \frac{m_1 RT}{2MV} = 300 \]

7. **Calculating Change in Pressure for Container B**:
- The gas in container B also expands isothermally to double its volume:
\[ P_2' = \frac{m_2 RT}{2MV} \]
- The change in pressure for container B is:
\[ \Delta P_B = P_2 - P_2' = \frac{2m_1 RT}{3MV} - \frac{m_2 RT}{2MV} \]
- Substitute \( m_2 = \frac{2}{3} m_1 \):
\[ \Delta P_B = \frac{2m_1 RT}{3MV} - \frac{\left(\frac{2}{3} m_1\right) RT}{2MV} = \frac{2m_1 RT}{3MV} - \frac{m_1 RT}{3MV} = \frac{m_1 RT}{3MV} \]

8. **Relating Changes in Pressure**:
- From the previous steps, we have:
\[ \Delta P_B = \frac{m_1 RT}{3MV} \]
- We can relate \( \Delta P_B \) to \( \Delta P_A \):
\[ \Delta P_B = \frac{1}{3} \Delta P_A = \frac{1}{3} \times 300 = 100 \, \text{Pa} \]

### Final Answer:
The change in pressure in container B is \( 100 \, \text{Pa} \).