A vessel contains `28 g` of `N_(2)` and `32 g` of `O_(2)` at temperature `T = 1800 K` and pressure `2 atm`. What would be the pressure when `N_(2)` dissociates `30%` and `O_(2)` dissociates `50%` and temperature remains constant ?

Options

A. 2 atm
B. 1 atm
C. 2.8 atm
D. 1.4 atm

Answer

To find the pressure in the vessel after the dissociation of nitrogen and oxygen, we can follow these steps:

### Step 1: Calculate the initial number of moles of \(N_2\) and \(O_2\)

The molecular mass of \(N_2\) is approximately \(28 \, \text{g/mol}\) and that of \(O_2\) is \(32 \, \text{g/mol}\).

- For \(N_2\):
\[ n_{N_2} = \frac{\text{mass}}{\text{molecular mass}} = \frac{28 \, \text{g}}{28 \, \text{g/mol}} = 1 \, \text{mol} \]

- For \(O_2\):
\[ n_{O_2} = \frac{\text{mass}}{\text{molecular mass}} = \frac{32 \, \text{g}}{32 \, \text{g/mol}} = 1 \, \text{mol} \]

### Step 2: Determine the amount of dissociation

- \(30\%\) of \(N_2\) dissociates:
\[ \text{Dissociated moles of } N_2 = 0.30 \times 1 \, \text{mol} = 0.30 \, \text{mol} \]
This dissociation produces \(2 \times 0.30 = 0.60 \, \text{mol}\) of \(N\).

- \(50\%\) of \(O_2\) dissociates:
\[ \text{Dissociated moles of } O_2 = 0.50 \times 1 \, \text{mol} = 0.50 \, \text{mol} \]
This dissociation produces \(2 \times 0.50 = 1.00 \, \text{mol}\) of \(O\).

### Step 3: Calculate the total number of moles after dissociation

- Remaining moles of \(N_2\):
\[ n_{N_2 \, \text{after}} = 1 - 0.30 = 0.70 \, \text{mol} \]

- Remaining moles of \(O_2\):
\[ n_{O_2 \, \text{after}} = 1 - 0.50 = 0.50 \, \text{mol} \]

- Total moles after dissociation:
\[ n_{\text{total}} = n_{N_2 \, \text{after}} + n_{O_2 \, \text{after}} + \text{moles of } N + \text{moles of } O \]
\[ n_{\text{total}} = 0.70 + 0.50 + 0.60 + 1.00 = 2.80 \, \text{mol} \]

### Step 4: Use the ideal gas law to find the new pressure

Using the relation \(P_1V = n_1RT\) and \(P_2V = n_2RT\) (where \(V\) and \(R\) are constant), we can set up the following equation:

\[ \frac{P_1}{n_1} = \frac{P_2}{n_2} \]

Given:
- \(P_1 = 2 \, \text{atm}\)
- \(n_1 = 2 \, \text{mol}\) (initial total moles)
- \(n_2 = 2.80 \, \text{mol}\) (after dissociation)

Rearranging gives:
\[ P_2 = P_1 \times \frac{n_2}{n_1} = 2 \, \text{atm} \times \frac{2.80}{2} = 2.80 \, \text{atm} \]

### Final Answer

The pressure when \(N_2\) dissociates \(30\%\) and \(O_2\) dissociates \(50\%\) is \(P_2 = 2.80 \, \text{atm}\).

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