A sample of oxygen gas occupies 431 mL at standard temperature and pressure. Calculate the volume when the temperature is `35^(@)C` and pressure is 1.05 bar.

Explanation

To solve the problem of finding the volume of oxygen gas under new conditions of temperature and pressure, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas. The formula we will use is:

\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

### Step-by-Step Solution:

1. **Identify the known values:**
- Initial volume (V1) = 431 mL
- Initial temperature (T1) = 0°C = 273 K (convert to Kelvin by adding 273)
- Initial pressure (P1) = 1 bar
- Final temperature (T2) = 35°C = 308 K (convert to Kelvin)
- Final pressure (P2) = 1.05 bar

2. **Rearrange the formula to solve for V2:**
\[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]

3. **Substitute the known values into the equation:**
\[ V_2 = \frac{(1 \, \text{bar}) \times (431 \, \text{mL}) \times (308 \, \text{K})}{(1.05 \, \text{bar}) \times (273 \, \text{K})} \]

4. **Calculate the numerator:**
\[ \text{Numerator} = 1 \times 431 \times 308 = 132,188 \, \text{mL K} \]

5. **Calculate the denominator:**
\[ \text{Denominator} = 1.05 \times 273 = 286.65 \, \text{bar K} \]

6. **Now, divide the numerator by the denominator to find V2:**
\[ V_2 = \frac{132,188 \, \text{mL K}}{286.65 \, \text{bar K}} \approx 460.5 \, \text{mL} \]

7. **Final result:**
The volume of the oxygen gas at 35°C and 1.05 bar is approximately **460.5 mL**.