A vessel of volume `5000 cm^3` contains `(1/20)` moles of molecular nitrogen at 1800K. If `30%` of the molecules are now dissociated the pressure inside the vessel (in Pa) will be

Options

A. `1.49 xx 10^(5)`
B. `1.95 xx 10^5`
C. `2.25 xx 10^5`
D. `3.78 xx 10^5`

Answer

To solve the problem step by step, we will use the Ideal Gas Law, which is given by the equation:

\[ PV = nRT \]

Where:
- \( P \) = Pressure (in Pascals)
- \( V \) = Volume (in cubic meters)
- \( n \) = Number of moles of gas
- \( R \) = Universal gas constant (\( 8.314 \, \text{J/(mol K)} \))
- \( T \) = Temperature (in Kelvin)

### Step 1: Convert the volume from cm³ to m³
Given:
\[ V = 5000 \, \text{cm}^3 \]

To convert cm³ to m³:
\[ V = 5000 \, \text{cm}^3 \times 10^{-6} \, \text{m}^3/\text{cm}^3 = 5 \times 10^{-3} \, \text{m}^3 \]

### Step 2: Calculate the initial number of moles of nitrogen
Given:
\[ n = \frac{1}{20} \, \text{moles} \]

### Step 3: Determine the number of moles after dissociation
30% of the nitrogen molecules dissociate. Therefore, the number of moles that dissociate is:
\[ \text{Moles dissociated} = 0.3 \times n = 0.3 \times \frac{1}{20} = \frac{0.3}{20} = \frac{3}{200} \, \text{moles} \]

The dissociation of nitrogen gas (N₂) produces two nitrogen atoms (N):
\[ N_2 \rightarrow 2N \]

Thus, for every mole of \( N_2 \) that dissociates, we get 2 moles of \( N \). Therefore, the total number of moles after dissociation becomes:
\[ n_{\text{final}} = n + 2 \times \text{Moles dissociated} \]
\[ n_{\text{final}} = \frac{1}{20} + 2 \times \frac{3}{200} \]
\[ n_{\text{final}} = \frac{1}{20} + \frac{6}{200} \]
\[ n_{\text{final}} = \frac{10}{200} + \frac{6}{200} = \frac{16}{200} = \frac{4}{50} = \frac{2}{25} \, \text{moles} \]

### Step 4: Use the Ideal Gas Law to find the pressure
Now, we can use the Ideal Gas Law to find the pressure:
\[ P = \frac{nRT}{V} \]

Substituting the values:
- \( n = \frac{2}{25} \, \text{moles} \)
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 1800 \, \text{K} \)
- \( V = 5 \times 10^{-3} \, \text{m}^3 \)

Calculating:
\[ P = \frac{\left(\frac{2}{25}\right) \times 8.314 \times 1800}{5 \times 10^{-3}} \]

Calculating the numerator:
\[ \frac{2}{25} \times 8.314 \times 1800 = \frac{2 \times 8.314 \times 1800}{25} \]
\[ = \frac{29932.8}{25} = 1197.312 \]

Now, substituting back into the pressure equation:
\[ P = \frac{1197.312}{5 \times 10^{-3}} = 239462.4 \, \text{Pa} \]

### Final Result
The pressure inside the vessel after 30% of the nitrogen molecules have dissociated is approximately:
\[ P \approx 2.39 \times 10^5 \, \text{Pa} \]