In a cubical vessel of side one metre a mixture of gases containing 22g of carbon dioxide, 8g of oxygen and 14g of nitrogen are taken. Find the pressure exerted by the mixture if the temperature is `30^(@)C`.

Explanation

To find the pressure exerted by the mixture of gases in a cubical vessel, 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 J/(mol·K))
- \( T \) = temperature (in Kelvin)

### Step-by-Step Solution:

1. **Calculate the Volume of the Vessel:**
The side of the cubical vessel is given as 1 meter. Therefore, the volume \( V \) is:
\[ V = \text{side}^3 = 1^3 = 1 \, \text{m}^3 \]

2. **Convert the Temperature to Kelvin:**
The temperature is given as \( 30^\circ C \). To convert this to Kelvin:
\[ T = 30 + 273 = 303 \, \text{K} \]

3. **Calculate the Number of Moles of Each Gas:**
- For Carbon Dioxide (\( CO_2 \)):
\[ n_{CO_2} = \frac{\text{mass}}{\text{molar mass}} = \frac{22 \, \text{g}}{44 \, \text{g/mol}} = 0.5 \, \text{mol} \]
- For Oxygen (\( O_2 \)):
\[ n_{O_2} = \frac{8 \, \text{g}}{32 \, \text{g/mol}} = 0.25 \, \text{mol} \]
- For Nitrogen (\( N_2 \)):
\[ n_{N_2} = \frac{14 \, \text{g}}{28 \, \text{g/mol}} = 0.5 \, \text{mol} \]

4. **Calculate the Total Number of Moles:**
Now, sum the number of moles of all gases:
\[ n = n_{CO_2} + n_{O_2} + n_{N_2} = 0.5 + 0.25 + 0.5 = 1.25 \, \text{mol} \]

5. **Substitute Values into the Ideal Gas Law:**
Now, substitute \( n \), \( R \), \( T \), and \( V \) into the ideal gas equation to find \( P \):
\[ P = \frac{nRT}{V} \]
Since \( V = 1 \, \text{m}^3 \):
\[ P = nRT = 1.25 \times 8.314 \times 303 \]

6. **Calculate the Pressure:**
Now, perform the calculation:
\[ P = 1.25 \times 8.314 \times 303 \approx 3.15 \times 10^3 \, \text{Pascals} \]

### Final Answer:
The pressure exerted by the mixture of gases is approximately:
\[ P \approx 3150 \, \text{Pascals} \]