Two boxes A and B having their volume ratio `1 : 4` and filled with Ne are inter connected through a narrow tube of negligible volume. Box A is kept at 300 K and box B at 600 K. The ratio of mole of Ne gas in box B to box A is ___________?

Explanation

To solve the problem, we will use the ideal gas law and the given information about the two boxes containing neon gas.

### Step-by-Step Solution:

1. **Understand the Ideal Gas Law**:
The ideal gas law is given by the equation \( PV = nRT \), where:
- \( P \) = Pressure
- \( V \) = Volume
- \( n \) = Number of moles
- \( R \) = Ideal gas constant
- \( T \) = Temperature in Kelvin

2. **Identify the Given Information**:
- Volume ratio of boxes A and B: \( V_A : V_B = 1 : 4 \)
- Temperature of box A: \( T_A = 300 \, K \)
- Temperature of box B: \( T_B = 600 \, K \)

3. **Establish the Relationship**:
Since the boxes are interconnected, the pressure in both boxes is the same. Therefore, we can write:
\[ \frac{n_A}{V_A T_A} = \frac{n_B}{V_B T_B} \]
Rearranging gives:
\[ \frac{n_B}{n_A} = \frac{V_B}{V_A} \cdot \frac{T_A}{T_B} \]

4. **Substitute the Known Values**:
- From the volume ratio \( V_A : V_B = 1 : 4 \), we have \( \frac{V_B}{V_A} = 4 \).
- Substitute the temperatures and volume ratio into the equation:
\[ \frac{n_B}{n_A} = 4 \cdot \frac{300}{600} \]

5. **Calculate the Ratio**:
\[ \frac{n_B}{n_A} = 4 \cdot \frac{1}{2} = 2 \]

6. **Final Result**:
The ratio of moles of Ne gas in box B to box A is:
\[ \frac{n_B}{n_A} = 2 : 1 \]