In a flask of volume V litres, 0.2 mol of oxygen 0.4 mol of nitrogen, 0.1 mole of `NH_(3)` and 0.3 mol of He are enclosed at `27^(@)C`. If the total pressure exerted by these non reacting gases is one atmosphere, the partial pressure exerted by nitrogen is (A) 0.1 atmosphere (B) 0.2 atmosphere (C)0.3 atmosphere (D)0.4 atmosphere

Options

A. 0.1 atmosphere
B. 0.2 atmosphere
C. 0.3 atmosphere
D. 0.4 atmosphere

Explanation

To find the partial pressure exerted by nitrogen in a flask containing a mixture of gases, we can follow these steps:

### Step 1: Identify the given data
- Moles of oxygen (O2) = 0.2 mol
- Moles of nitrogen (N2) = 0.4 mol
- Moles of ammonia (NH3) = 0.1 mol
- Moles of helium (He) = 0.3 mol
- Total pressure (P_total) = 1 atm

### Step 2: Calculate the total number of moles of gas
To find the total number of moles, we sum the moles of each gas:
\[ \text{Total moles} = \text{moles of O2} + \text{moles of N2} + \text{moles of NH3} + \text{moles of He} \]
\[ \text{Total moles} = 0.2 + 0.4 + 0.1 + 0.3 = 1.0 \text{ mol} \]

### Step 3: Calculate the mole fraction of nitrogen (N2)
The mole fraction (X) of a gas is calculated using the formula:
\[ X_{N2} = \frac{\text{moles of N2}}{\text{total moles}} \]
Substituting the values:
\[ X_{N2} = \frac{0.4}{1.0} = 0.4 \]

### Step 4: Use Dalton's Law of Partial Pressures
According to Dalton's Law, the partial pressure of a gas can be calculated as:
\[ P_{N2} = X_{N2} \times P_{\text{total}} \]
Substituting the values:
\[ P_{N2} = 0.4 \times 1 \text{ atm} = 0.4 \text{ atm} \]

### Final Answer
The partial pressure exerted by nitrogen is **0.4 atm**.

### Conclusion
The correct option is (D) 0.4 atmosphere.

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