A closed container contains equal number of moles of two gases X and Y at a total pressure of 710 mm of Hg . If gas X is removed from the mixture , the pressure will (A) become double (B) become half (C) remain same (d) become one - fourth

Options

A. become double
B. become half
C. remain same
D. become one - fourth .

Explanation

To solve the problem, we need to analyze the situation using the ideal gas law and the relationship between pressure, volume, and the number of moles of gas.

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- We have a closed container with equal moles of two gases, X and Y.
- Let the number of moles of gas X be \( n_X = a \) and the number of moles of gas Y be \( n_Y = a \).
- The total number of moles in the container is \( n_{total} = n_X + n_Y = a + a = 2a \).
- The total pressure in the container is given as \( P_{total} = 710 \, \text{mm of Hg} \).

2. **Using the Ideal Gas Law**:
- The ideal gas law states that \( PV = nRT \).
- Since the volume (V) and temperature (T) are constant in this closed container, the pressure (P) is directly proportional to the number of moles (n).
- Therefore, we can express the relationship as \( P \propto n \).

3. **Initial Pressure Calculation**:
- The initial pressure due to both gases is \( P_{total} = P_X + P_Y \).
- Since the moles of gases are equal, the contribution to the pressure from each gas can be considered equal.
- Thus, the pressure due to gas X and gas Y can be represented as:
\[ P_X = \frac{n_X}{n_{total}} \times P_{total} = \frac{a}{2a} \times 710 = \frac{710}{2} = 355 \, \text{mm of Hg} \]
\[ P_Y = \frac{n_Y}{n_{total}} \times P_{total} = \frac{a}{2a} \times 710 = \frac{710}{2} = 355 \, \text{mm of Hg} \]

4. **Removing Gas X**:
- When gas X is removed, the number of moles of gas Y remains \( n_Y = a \).
- The total number of moles now is \( n_{total}' = n_Y = a \).

5. **Final Pressure Calculation**:
- The new pressure \( P_{final} \) can be calculated using the same proportionality:
\[ P_{final} = \frac{n_Y}{n_{total}'} \times P_{total} = \frac{a}{a} \times 710 = 710 \, \text{mm of Hg} \]
- However, since gas X is removed, we need to find the new pressure in terms of the original total pressure:
\[ P_{final} = \frac{P_{total}}{2} = \frac{710}{2} = 355 \, \text{mm of Hg} \]

6. **Conclusion**:
- The pressure after removing gas X becomes half of the original pressure.
- Therefore, the correct answer is **(B) become half**.