To solve this problem, we'll use the Ideal Gas Law, which is given by the equation:
[tex]\[ PV = nRT \][/tex]
Where:
- [tex]\( P \)[/tex] is the pressure.
- [tex]\( V \)[/tex] is the volume.
- [tex]\( n \)[/tex] is the amount of substance (in moles).
- [tex]\( R \)[/tex] is the ideal gas constant.
- [tex]\( T \)[/tex] is the temperature in Kelvin.
### Step-by-Step Solution:
1. Convert the Temperature from Celsius to Kelvin:
The given temperature is [tex]\(35^{\circ}C\)[/tex]. To convert this to Kelvin:
[tex]\[ T = 35 + 273.15 = 308.15 \, K \][/tex]
2. Using the Ideal Gas Law to Find Pressure in atm:
- Given:
- [tex]\( n = 0.250 \)[/tex] moles
- [tex]\( V = 6.23 \)[/tex] liters
- [tex]\( R = 0.0821 \, \frac{L \cdot atm}{mol \cdot K} \)[/tex]
- [tex]\( T = 308.15 \)[/tex] K
- Plugging these values into the Ideal Gas Law equation:
[tex]\[ P \cdot 6.23 = 0.250 \cdot 0.0821 \cdot 308.15 \][/tex]
- Solving for [tex]\( P \)[/tex]:
[tex]\[ P = \frac{0.250 \cdot 0.0821 \cdot 308.15}{6.23} \][/tex]
[tex]\[ P \approx 1.015 \, atm \][/tex]
3. Convert the Pressure from atm to kPa:
- Given that [tex]\(1 \, atm = 101.3 \, kPa\)[/tex]:
[tex]\[ P = 1.015 \, atm \times 101.3 \, \frac{kPa}{atm} \][/tex]
[tex]\[ P \approx 102.841 \, kPa \][/tex]
Thus, the absolute pressure of the air in the balloon is approximately [tex]\(102.841 \, kPa\)[/tex].
Rounded to three significant figures, the absolute pressure is:
[tex]\[ \boxed{103 \, kPa} \][/tex]
