Equal molecules of `N_2` and `O_2` are kept in a closed container at pressure P. If `N_2` is removed from the system, then the pressure of the container will be

Options

A. P
B. P//2
C. `P/4`
D. 2P

Answer

To solve the problem, we need to understand how the pressure in a closed container changes when one of the gases is removed. We will use the ideal gas law, which states that pressure (P), volume (V), number of moles (n), and temperature (T) are related by the equation:

\[ PV = nRT \]

Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles of gas
- \( R \) = universal gas constant
- \( T \) = temperature

### Step-by-Step Solution:

1. **Initial Conditions**:
- Let the number of moles of \( N_2 \) be \( n \) and the number of moles of \( O_2 \) also be \( n \) (since they are equal).
- The total number of moles in the container initially is \( n + n = 2n \).
- The initial pressure in the container is given as \( P \).

2. **Using the Ideal Gas Law**:
- The initial pressure can be expressed using the ideal gas law:
\[ P = \frac{(2n)RT}{V} \]

3. **Removing \( N_2 \)**:
- When \( N_2 \) is removed, the number of moles of \( O_2 \) remains \( n \) and the number of moles of \( N_2 \) becomes \( 0 \).
- The total number of moles left in the container is now \( n \) (only \( O_2 \)).

4. **New Pressure Calculation**:
- The pressure of the container after removing \( N_2 \) can be calculated using the ideal gas law again:
\[ P' = \frac{(n)RT}{V} \]

5. **Relating New Pressure to Initial Pressure**:
- From the initial condition, we know that:
\[ P = \frac{(2n)RT}{V} \]
- Therefore, we can express \( P' \) in terms of \( P \):
\[ P' = \frac{(n)RT}{V} = \frac{1}{2} \cdot \frac{(2n)RT}{V} = \frac{P}{2} \]

6. **Final Result**:
- Thus, the pressure of the container after removing \( N_2 \) will be:
\[ P' = \frac{P}{2} \]

### Summary:
If \( N_2 \) is removed from the system, the pressure of the container will be half of the initial pressure, which is \( \frac{P}{2} \).