500 mL of nitrogen at 0.936 bar pressure and 1000 mL oxygen at 0.80 bar pressure are put together in a 2 L flask. IF temperature is kept constant, calculate the final pressure of the mixture.

Explanation

To solve the problem of calculating the final pressure of a mixture of nitrogen and oxygen gases in a 2 L flask, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Volume of nitrogen (V_N) = 500 mL
- Pressure of nitrogen (P_N) = 0.936 bar
- Volume of oxygen (V_O) = 1000 mL
- Pressure of oxygen (P_O) = 0.80 bar
- Final volume of the mixture (V_f) = 2000 mL (since 2 L = 2000 mL)

2. **Convert the Volumes to Consistent Units:**
- Since we are working with pressures in bar, we will keep the volumes in mL for consistency.

3. **Calculate the Total Pressure Contribution from Each Gas:**
- The total pressure exerted by a gas can be calculated using the formula:
\[ P_{total} = \frac{P \times V}{V_f} \]
- For nitrogen:
\[ P_{N, total} = \frac{P_N \times V_N}{V_f} = \frac{0.936 \, \text{bar} \times 500 \, \text{mL}}{2000 \, \text{mL}} = \frac{468}{2000} = 0.234 \, \text{bar} \]
- For oxygen:
\[ P_{O, total} = \frac{P_O \times V_O}{V_f} = \frac{0.80 \, \text{bar} \times 1000 \, \text{mL}}{2000 \, \text{mL}} = \frac{800}{2000} = 0.40 \, \text{bar} \]

4. **Calculate the Final Pressure of the Mixture:**
- The final pressure (P_f) of the mixture is the sum of the partial pressures from both gases:
\[ P_f = P_{N, total} + P_{O, total} = 0.234 \, \text{bar} + 0.40 \, \text{bar} = 0.634 \, \text{bar} \]

### Final Answer:
The final pressure of the mixture in the 2 L flask is **0.634 bar**.