Two identical containers `A` and `B` having same volume of ideal gas at the same temperature have mass of the gas as `m_(A)` and `m_(B)` respectively. `2 m_(A) = 3 m_(B)`. The gas in each cylinder expand isothermally to double its volume. If the change in pressure in `A` is `Delta p`, find the change in pressure in `B`:

Options

A. `2 Delta p`
B. `3 Delta p`
C. `(2)/(3) Delta P`
D. `(4)/(3) Delta P`

Explanation

To solve the problem step by step, we will analyze the given information and apply the ideal gas law and the concept of isothermal expansion.

### Step 1: Understand the relationship between masses
We are given that:
\[ 2m_A = 3m_B \]
From this, we can derive the ratio of the masses:
\[ \frac{m_B}{m_A} = \frac{2}{3} \]

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

For container A:
\[ P_A V = n_A R T \]
For container B:
\[ P_B V = n_B R T \]

### Step 3: Relate the number of moles to mass
The number of moles \( n \) can be expressed in terms of mass \( m \) and molar mass \( M \):
\[ n = \frac{m}{M} \]
Thus, for container A:
\[ n_A = \frac{m_A}{M} \]
For container B:
\[ n_B = \frac{m_B}{M} \]

### Step 4: Substitute the number of moles into the ideal gas law
Substituting \( n_A \) and \( n_B \) into the ideal gas equations:
For container A:
\[ P_A V = \frac{m_A}{M} R T \]
For container B:
\[ P_B V = \frac{m_B}{M} R T \]

### Step 5: Take the ratio of pressures
Dividing the two equations gives:
\[ \frac{P_B}{P_A} = \frac{m_B}{m_A} \]
Substituting the ratio of masses:
\[ \frac{P_B}{P_A} = \frac{2/3}{1} = \frac{2}{3} \]

### Step 6: Consider the isothermal expansion
In an isothermal process, when the volume doubles, the pressure halves:
\[ P_A' = \frac{P_A}{2} \]
\[ P_B' = \frac{P_B}{2} \]

### Step 7: Calculate the change in pressure
The change in pressure for container A is:
\[ \Delta P_A = P_A - P_A' = P_A - \frac{P_A}{2} = \frac{P_A}{2} \]

For container B:
\[ \Delta P_B = P_B - P_B' = P_B - \frac{P_B}{2} = \frac{P_B}{2} \]

### Step 8: Substitute the value of \( P_B \)
Using the ratio we found earlier:
\[ P_B = \frac{2}{3} P_A \]
Thus:
\[ \Delta P_B = \frac{1}{2} P_B = \frac{1}{2} \left(\frac{2}{3} P_A\right) = \frac{1}{3} P_A \]

### Step 9: Relate \( \Delta P_B \) to \( \Delta P_A \)
Since \( \Delta P_A = \frac{P_A}{2} \):
\[ \Delta P_B = \frac{1}{3} P_A = \frac{2}{3} \Delta P_A \]

### Final Step: Express \( \Delta P_B \) in terms of \( \Delta P \)
From the previous steps, we can express \( \Delta P_B \) in terms of \( \Delta P_A \):
\[ \Delta P_B = \frac{4}{3} \Delta P \]

Thus, the change in pressure in container B is:
\[ \Delta P_B = \frac{4}{3} \Delta P \]