1.00 mol of `N_(2)` and 3.00 mol of `H_(2)` are present in a container of volume `10.0 dm^(3)` at 298 K. What is the total pressure of the mixture?

Explanation

To find the total pressure of the mixture of nitrogen (N₂) and hydrogen (H₂) in a container, we can use the Ideal Gas Law, which is given by the equation:

\[ PV = nRT \]

Where:
- \( P \) = pressure (in bar)
- \( V \) = volume (in dm³)
- \( n \) = number of moles
- \( R \) = ideal gas constant (0.0831 dm³·bar/(K·mol))
- \( T \) = temperature (in K)

### Step 1: Calculate the Partial Pressure of Nitrogen (N₂)

1. **Identify the number of moles of nitrogen (N₂)**:
- Given: \( n_{N_2} = 1.00 \, \text{mol} \)

2. **Use the Ideal Gas Law to find the partial pressure of nitrogen**:
\[ P_{N_2} = \frac{n_{N_2}RT}{V} \]
Substituting the values:
- \( n_{N_2} = 1.00 \, \text{mol} \)
- \( R = 0.0831 \, \text{dm³·bar/(K·mol)} \)
- \( T = 298 \, \text{K} \)
- \( V = 10.0 \, \text{dm³} \)

\[ P_{N_2} = \frac{(1.00 \, \text{mol})(0.0831 \, \text{dm³·bar/(K·mol)})(298 \, \text{K})}{10.0 \, \text{dm³}} \]

\[ P_{N_2} = \frac{24.78}{10.0} = 2.478 \, \text{bar} \approx 2.48 \, \text{bar} \]

### Step 2: Calculate the Partial Pressure of Hydrogen (H₂)

1. **Identify the number of moles of hydrogen (H₂)**:
- Given: \( n_{H_2} = 3.00 \, \text{mol} \)

2. **Use the Ideal Gas Law to find the partial pressure of hydrogen**:
\[ P_{H_2} = \frac{n_{H_2}RT}{V} \]
Substituting the values:
- \( n_{H_2} = 3.00 \, \text{mol} \)
- \( R = 0.0831 \, \text{dm³·bar/(K·mol)} \)
- \( T = 298 \, \text{K} \)
- \( V = 10.0 \, \text{dm³} \)

\[ P_{H_2} = \frac{(3.00 \, \text{mol})(0.0831 \, \text{dm³·bar/(K·mol)})(298 \, \text{K})}{10.0 \, \text{dm³}} \]

\[ P_{H_2} = \frac{74.34}{10.0} = 7.434 \, \text{bar} \approx 7.43 \, \text{bar} \]

### Step 3: Calculate the Total Pressure of the Mixture

1. **Add the partial pressures**:
\[ P_{total} = P_{N_2} + P_{H_2} \]
\[ P_{total} = 2.48 \, \text{bar} + 7.43 \, \text{bar} = 9.91 \, \text{bar} \]

### Final Answer:
The total pressure of the mixture is approximately **9.91 bar**.

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