56 g of nitrogen and 96 g of oxygen are mixed isothermaly and at a total pressure of 10 atm. The partial pressures of oxygen and nitrogen (in atm) are respectively :

Options

A. 4,6
B. 5,5
C. 2,8
D. 6,4

Explanation

To solve the problem of finding the partial pressures of nitrogen (N₂) and oxygen (O₂) when 56 g of nitrogen and 96 g of oxygen are mixed isothermally at a total pressure of 10 atm, we can follow these steps:

### Step 1: Calculate the number of moles of nitrogen (N₂)
The molar mass of nitrogen (N₂) is 28 g/mol.

\[ \text{Moles of } N_2 = \frac{\text{Mass of } N_2}{\text{Molar mass of } N_2} = \frac{56 \text{ g}}{28 \text{ g/mol}} = 2 \text{ moles} \]

### Step 2: Calculate the number of moles of oxygen (O₂)
The molar mass of oxygen (O₂) is 32 g/mol.

\[ \text{Moles of } O_2 = \frac{\text{Mass of } O_2}{\text{Molar mass of } O_2} = \frac{96 \text{ g}}{32 \text{ g/mol}} = 3 \text{ moles} \]

### Step 3: Calculate the total number of moles in the mixture
Now, we can find the total number of moles in the gas mixture.

\[ \text{Total moles} = \text{Moles of } N_2 + \text{Moles of } O_2 = 2 + 3 = 5 \text{ moles} \]

### Step 4: Calculate the partial pressure of oxygen (O₂)
Using Dalton's Law of Partial Pressures, the partial pressure of a gas is given by:

\[ P_{O_2} = \left(\frac{\text{Moles of } O_2}{\text{Total moles}}\right) \times P_{\text{total}} \]

Substituting the values we have:

\[ P_{O_2} = \left(\frac{3}{5}\right) \times 10 \text{ atm} = 6 \text{ atm} \]

### Step 5: Calculate the partial pressure of nitrogen (N₂)
Similarly, we can calculate the partial pressure of nitrogen:

\[ P_{N_2} = \left(\frac{\text{Moles of } N_2}{\text{Total moles}}\right) \times P_{\text{total}} \]

Substituting the values:

\[ P_{N_2} = \left(\frac{2}{5}\right) \times 10 \text{ atm} = 4 \text{ atm} \]

### Final Answer
The partial pressures of oxygen and nitrogen are:
- Partial pressure of O₂ = 6 atm
- Partial pressure of N₂ = 4 atm