If 1 litre of N2 at `27^(@)C` and 760 mm Hg contains N molecules, 4 litres of `O_(2)` under the same conditions of temperature and pressure, shall contain

Options

A. N molecules
B. 2N molecules
C. `(N)/(4)` molecules
D. 4N molecules

Explanation

To solve the problem, we will use the ideal gas law and the relationship between volume and the number of molecules under constant temperature and pressure.

### 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
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

2. **Identify Given Values**:
- For nitrogen (\( N_2 \)):
- Volume (\( V_1 \)) = 1 L
- Temperature = \( 27^\circ C \) (which is \( 300 K \))
- Pressure = 760 mm Hg
- Number of molecules = \( N \)
- For oxygen (\( O_2 \)):
- Volume (\( V_2 \)) = 4 L
- Temperature and pressure are the same as for \( N_2 \).

3. **Use the Relationship Between Volume and Number of Molecules**: Since the temperature and pressure are constant, the volume is directly proportional to the number of molecules. This can be expressed as:
\[ \frac{V_1}{V_2} = \frac{N_1}{N_2} \]
where:
- \( N_1 \) = number of molecules in 1 L of \( N_2 \) = \( N \)
- \( N_2 \) = number of molecules in 4 L of \( O_2 \)

4. **Substitute the Known Values**:
\[ \frac{1 \text{ L}}{4 \text{ L}} = \frac{N}{N_2} \]

5. **Cross-Multiply to Solve for \( N_2 \)**:
\[ N_2 = 4N \]

6. **Conclusion**: Therefore, 4 liters of \( O_2 \) under the same conditions of temperature and pressure will contain \( 4N \) molecules.

### Final Answer:
The number of molecules in 4 liters of \( O_2 \) is \( 4N \).

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