The volume of gas A is twice than that of gas B. The compressibility factor of gas A is thrice than that of gas B at same temperature. The pressures of the gases for equal number of moles are:

Options

A. `P_A = 2P_(B)`
B. `P_(A)=3P_(B)`
C. `2P_(A)=3P_(B)`
D. `3P_(A)=2P_(B)`.

Explanation

To solve the problem, we need to relate the pressures of gases A and B using the given information about their volumes and compressibility factors. Let's break this down step by step.

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Volume of gas A (\(V_A\)) is twice that of gas B (\(V_B\)):
\[ V_A = 2V_B \]
- Compressibility factor of gas A (\(Z_A\)) is three times that of gas B (\(Z_B\)):
\[ Z_A = 3Z_B \]
- The number of moles of both gases is equal.

2. **Use the Ideal Gas Equation with Compressibility**:
The relationship for pressure in terms of the compressibility factor is given by:
\[ PV = ZnRT \]
For gas A:
\[ P_A V_A = Z_A n R T \]
For gas B:
\[ P_B V_B = Z_B n R T \]

3. **Substitute the Known Values**:
Substitute \(V_A\) and \(Z_A\) into the equation for gas A:
\[ P_A (2V_B) = (3Z_B) n R T \]
This simplifies to:
\[ 2P_A V_B = 3Z_B n R T \]

4. **Set Up the Equation for Gas B**:
From the equation for gas B:
\[ P_B V_B = Z_B n R T \]

5. **Divide the Two Equations**:
Now, divide the equation for gas A by the equation for gas B:
\[ \frac{2P_A V_B}{P_B V_B} = \frac{3Z_B n R T}{Z_B n R T} \]
The \(V_B\), \(n\), \(R\), and \(T\) cancel out:
\[ \frac{2P_A}{P_B} = 3 \]

6. **Rearrange to Find the Relationship Between Pressures**:
Rearranging gives:
\[ 2P_A = 3P_B \]
Therefore, we can express \(P_A\) in terms of \(P_B\):
\[ P_A = \frac{3}{2} P_B \]

### Conclusion:
The pressures of gases A and B are related by the equation \(2P_A = 3P_B\).