A closed vessel contains `8 g` of oxygen and ` 7 g` of nitrogen. The total pressure is 10 atm at a given temperature. If now oxygen is absorbed by introducting a suitable absorbent, the pressure of the remaining gas in atm will be

Options

A. 2
B. 10
C. 4
D. 5

Explanation

To solve the problem, we will use the principles of Dalton's Law of Partial Pressures and the Ideal Gas Law. Here’s a step-by-step solution:

### Step 1: Calculate the number of moles of oxygen and nitrogen.
- **Given:**
- Mass of oxygen (O₂) = 8 g
- Molecular mass of oxygen = 32 g/mol
- Mass of nitrogen (N₂) = 7 g
- Molecular mass of nitrogen = 28 g/mol

- **Calculating moles:**
\[ \text{Moles of O}_2 = \frac{\text{Mass of O}_2}{\text{Molecular mass of O}_2} = \frac{8 \, \text{g}}{32 \, \text{g/mol}} = 0.25 \, \text{mol} \]
\[ \text{Moles of N}_2 = \frac{\text{Mass of N}_2}{\text{Molecular mass of N}_2} = \frac{7 \, \text{g}}{28 \, \text{g/mol}} = 0.25 \, \text{mol} \]

### Step 2: Calculate the total number of moles in the mixture.
- **Total moles:**
\[ \text{Total moles} = \text{Moles of O}_2 + \text{Moles of N}_2 = 0.25 \, \text{mol} + 0.25 \, \text{mol} = 0.5 \, \text{mol} \]

### Step 3: Use Dalton's Law to find the partial pressures.
- **Total pressure (P_total) = 10 atm**
- **Using Dalton's Law:**
\[ P_{\text{total}} = P_{\text{O}_2} + P_{\text{N}_2} \]

- **Partial pressure of each gas:**
\[ P_{\text{O}_2} = \frac{\text{Moles of O}_2}{\text{Total moles}} \times P_{\text{total}} = \frac{0.25}{0.5} \times 10 \, \text{atm} = 5 \, \text{atm} \]
\[ P_{\text{N}_2} = \frac{\text{Moles of N}_2}{\text{Total moles}} \times P_{\text{total}} = \frac{0.25}{0.5} \times 10 \, \text{atm} = 5 \, \text{atm} \]

### Step 4: Calculate the pressure after oxygen is absorbed.
- **After oxygen is absorbed, only nitrogen remains.**
- **Thus, the pressure of the remaining gas (nitrogen) will be:**
\[ P_{\text{N}_2} = 5 \, \text{atm} \]

### Final Answer:
The pressure of the remaining gas (nitrogen) after oxygen is absorbed will be **5 atm**.

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