2 moles each of hydrogen, carbon dioxide and chlorine are mixed in a close vessel of volume 3 litres and temperature `0^@ C`. Calculate the pressure exerted by the mixture. (`R=8.31 J mol^(-1) K^(-1)` )

Explanation

To calculate the pressure exerted by a mixture of gases in a closed vessel, we can 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
- \( R \) = ideal gas constant (8.31 J/mol·K)
- \( T \) = temperature (in Kelvin)

### Step-by-Step Solution:

1. **Convert Volume to Cubic Meters**:
The volume given is 3 liters. We need to convert this to cubic meters:
\[ V = 3 \, \text{liters} = 3 \times 10^{-3} \, \text{m}^3 \]

2. **Convert Temperature to Kelvin**:
The temperature is given as \( 0^\circ C \). We convert this to Kelvin:
\[ T = 0 + 273 = 273 \, \text{K} \]

3. **Calculate Total Moles of Gas**:
We have 2 moles of hydrogen, 2 moles of carbon dioxide, and 2 moles of chlorine. Therefore, the total number of moles \( n \) is:
\[ n = 2 + 2 + 2 = 6 \, \text{moles} \]

4. **Use the Ideal Gas Law to Calculate Pressure**:
Rearranging the ideal gas law to solve for pressure \( P \):
\[ P = \frac{nRT}{V} \]
Substituting the known values:
\[ P = \frac{6 \, \text{moles} \times 8.31 \, \text{J/mol·K} \times 273 \, \text{K}}{3 \times 10^{-3} \, \text{m}^3} \]

5. **Calculate the Pressure**:
Now, performing the calculation:
\[ P = \frac{6 \times 8.31 \times 273}{3 \times 10^{-3}} \]
\[ P = \frac{13661.58}{0.003} = 4.55 \times 10^6 \, \text{Pa} \]

6. **Final Result**:
The pressure exerted by the mixture is approximately:
\[ P \approx 4.55 \times 10^6 \, \text{Pa} \, \text{or} \, 4.55 \, \text{MPa} \]