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 need to find the ratio of moles of Neon gas in box B to box A, given the conditions of the two boxes.

### Step-by-Step Solution:

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

2. **Use the Ideal Gas Law:**
The ideal gas law is given by:
\[ PV = nRT \]
Rearranging this for pressure gives:
\[ P = \frac{nRT}{V} \]
where \( n \) is the number of moles, \( R \) is the gas constant, \( T \) is the temperature, and \( V \) is the volume.

3. **Set Up the Pressure Equations:**
Since the two boxes are interconnected, the pressures in both boxes will be equal when equilibrium is reached:
\[ P_A = P_B \]
Using the ideal gas law for both boxes, we have:
\[ \frac{n_A R T_A}{V_A} = \frac{n_B R T_B}{V_B} \]

4. **Cancel Out the Gas Constant R:**
Since \( R \) is the same for both boxes, we can cancel it out:
\[ \frac{n_A T_A}{V_A} = \frac{n_B T_B}{V_B} \]

5. **Substitute the Volume Ratio:**
From the volume ratio \( V_A : V_B = 1 : 4 \), we can express \( V_B \) in terms of \( V_A \):
\[ V_B = 4 V_A \]
Substituting this into the equation gives:
\[ \frac{n_A T_A}{V_A} = \frac{n_B T_B}{4 V_A} \]
Simplifying this, we can cancel \( V_A \):
\[ n_A T_A = \frac{n_B T_B}{4} \]

6. **Rearranging for the Ratio of Moles:**
Rearranging the equation to find the ratio of moles:
\[ \frac{n_B}{n_A} = \frac{4 T_A}{T_B} \]

7. **Substituting the Temperatures:**
Substitute \( T_A = 300 \, K \) and \( T_B = 600 \, K \):
\[ \frac{n_B}{n_A} = \frac{4 \times 300}{600} = \frac{1200}{600} = 2 \]

8. **Final Ratio:**
Thus, the ratio of moles of Neon gas in box B to box A is:
\[ \frac{n_B}{n_A} = 2 : 1 \]

### Conclusion:
The ratio of moles of Ne gas in box B to box A is **2:1**.

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