A vessel of volume 5 litres contains `1.4 g` of `N_2` and `0.4 g` of He at `1500 K`. If `30 %` of the nitrogen molecules are dissociated into atoms then find the gas pressure.

Explanation

To solve the problem step by step, we will follow the given information and apply the ideal gas law.

### Step 1: Calculate the number of moles of Nitrogen (N₂)
The molecular mass of Nitrogen (N₂) is 28 g/mol. The mass of Nitrogen given is 1.4 g.

\[ n_{N_2} = \frac{\text{mass}}{\text{molecular mass}} = \frac{1.4 \, \text{g}}{28 \, \text{g/mol}} = 0.05 \, \text{mol} \]

### Step 2: Calculate the number of moles of Helium (He)
The molecular mass of Helium (He) is 4 g/mol. The mass of Helium given is 0.4 g.

\[ n_{He} = \frac{\text{mass}}{\text{molecular mass}} = \frac{0.4 \, \text{g}}{4 \, \text{g/mol}} = 0.1 \, \text{mol} \]

### Step 3: Determine the dissociation of Nitrogen
30% of the Nitrogen molecules dissociate into atoms. Therefore, the number of moles that dissociate is:

\[ \text{Dissociated moles} = 0.30 \times n_{N_2} = 0.30 \times 0.05 = 0.015 \, \text{mol} \]

### Step 4: Calculate the moles of Nitrogen remaining as molecules
The remaining moles of Nitrogen as molecules after dissociation:

\[ \text{Remaining moles of } N_2 = n_{N_2} - \text{Dissociated moles} = 0.05 - 0.015 = 0.035 \, \text{mol} \]

### Step 5: Calculate the total number of moles after dissociation
Each dissociated molecule of Nitrogen produces 2 atoms. Therefore, the number of moles of Nitrogen atoms produced is:

\[ \text{Moles of } N \text{ atoms} = 2 \times \text{Dissociated moles} = 2 \times 0.015 = 0.030 \, \text{mol} \]

Now, the total number of moles of Nitrogen (atoms + molecules) is:

\[ n_{N_{total}} = \text{Remaining moles of } N_2 + \text{Moles of } N \text{ atoms} = 0.035 + 0.030 = 0.065 \, \text{mol} \]

### Step 6: Calculate the total number of moles in the system
Now, we add the moles of Helium:

\[ n_{total} = n_{N_{total}} + n_{He} = 0.065 + 0.1 = 0.165 \, \text{mol} \]

### Step 7: Apply the Ideal Gas Law to find the pressure
Using the ideal gas equation \( PV = nRT \), we can solve for pressure \( P \):

\[ P = \frac{nRT}{V} \]

Where:
- \( n = 0.165 \, \text{mol} \)
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 1500 \, \text{K} \)
- \( V = 5 \, \text{liters} = 5 \times 10^{-3} \, \text{m}^3 \)

Substituting the values:

\[ P = \frac{0.165 \times 8.314 \times 1500}{5 \times 10^{-3}} \]

Calculating this gives:

\[ P = \frac{0.165 \times 8.314 \times 1500}{0.005} = \frac{2050.497}{0.005} = 410099.4 \, \text{Pa} \approx 4.1 \times 10^5 \, \text{Pa} \]

### Final Answer
The pressure in the vessel is approximately \( 4.1 \times 10^5 \, \text{Pa} \).

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