400 mL of `N_(2)` gas at 700 mm and 300 mL of `H_(2)` gas at 800 mm were introduced into a vessel of 2 litres at the same temperature. Calculate the final pressure of the gas mixture.

Answer

To solve the problem of finding the final pressure of the gas mixture, we can use the ideal gas law and the principle of partial pressures. Here's a step-by-step solution:

### Step 1: Identify the initial conditions of the gases
- For Nitrogen (N₂):
- Volume (V₁) = 400 mL = 0.4 L
- Pressure (P₁) = 700 mmHg

- For Hydrogen (H₂):
- Volume (V₂) = 300 mL = 0.3 L
- Pressure (P₂) = 800 mmHg

### Step 2: Convert the volumes to liters
- N₂: 400 mL = 0.4 L
- H₂: 300 mL = 0.3 L

### Step 3: Calculate the number of moles of each gas using the ideal gas law
Using the formula \( n = \frac{PV}{RT} \), we can express the number of moles for each gas. However, since R and T are constant and the same for both gases, we can ignore them for the final calculation of pressure.

### Step 4: Calculate the total moles of each gas
- For N₂:
\[ n₁ = \frac{P₁ \times V₁}{RT} = \frac{700 \, \text{mmHg} \times 0.4 \, \text{L}}{RT} \]

- For H₂:
\[ n₂ = \frac{P₂ \times V₂}{RT} = \frac{800 \, \text{mmHg} \times 0.3 \, \text{L}}{RT} \]

### Step 5: Calculate the total pressure in the final volume
The total pressure in the final volume (2 L) can be calculated using the formula for partial pressures:
\[ P_{\text{final}} = \frac{P₁ \times V₁ + P₂ \times V₂}{V_{\text{final}}} \]
Substituting the values:
\[ P_{\text{final}} = \frac{(700 \, \text{mmHg} \times 0.4 \, \text{L}) + (800 \, \text{mmHg} \times 0.3 \, \text{L})}{2 \, \text{L}} \]

### Step 6: Perform the calculations
Calculating the numerator:
\[ = (700 \times 0.4) + (800 \times 0.3) = 280 + 240 = 520 \, \text{mmHg} \]

Now, divide by the final volume:
\[ P_{\text{final}} = \frac{520 \, \text{mmHg}}{2 \, \text{L}} = 260 \, \text{mmHg} \]

### Final Answer
The final pressure of the gas mixture is **260 mmHg**.

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