To solve this problem, we can use the ideal gas law, which is given by:
[tex]\[ PV = nRT \][/tex]
Here:
- [tex]\( P \)[/tex] is the absolute pressure in kilopascals (kPa),
- [tex]\( V \)[/tex] is the volume in liters (L),
- [tex]\( n \)[/tex] is the number of moles of gas,
- [tex]\( R \)[/tex] is the ideal gas constant,
- [tex]\( T \)[/tex] is the temperature in Kelvin (K).
First, let's convert the given temperature from Celsius to Kelvin:
[tex]\[ T(K) = T(°C) + 273.15 \][/tex]
[tex]\[ T(K) = 32°C + 273.15 = 305.15 K \][/tex]
Next, we need to convert the gauge pressure to absolute pressure. Gauge pressure is the pressure relative to atmospheric pressure:
[tex]\[ \text{Absolute pressure} = \text{Gauge pressure} + \text{Atmospheric pressure} \][/tex]
[tex]\[ P = 61 \text{ kPa} + 101.3 \text{ kPa} = 162.3 \text{ kPa} \][/tex]
We are given:
- [tex]\( V = 5.2 \text{ L} \)[/tex],
- [tex]\( R = 8.314 \frac{\text{L} \cdot \text{kPa}}{\text{mol} \cdot \text{K}} \)[/tex],
- [tex]\( T = 305.15 \text{ K} \)[/tex],
- [tex]\( P = 162.3 \text{ kPa} \)[/tex].
Now, we can rearrange the ideal gas law equation to solve for the number of moles [tex]\( n \)[/tex]:
[tex]\[ n = \frac{PV}{RT} \][/tex]
Substitute the known values:
[tex]\[ n = \frac{(162.3 \text{ kPa}) \times (5.2 \text{ L})}{(8.314 \frac{\text{L} \cdot \text{kPa}}{\text{mol} \cdot \text{K}}) \times (305.15 \text{ K})} \][/tex]
Calculating this, we get:
[tex]\[ n \approx 0.33 \text{ mol} \][/tex]
Thus, the number of moles of air in the soccer ball is approximately [tex]\( 0.33 \text{ mol} \)[/tex].
So, the correct answer is:
[tex]\[ \boxed{B. \ 0.33 \text{ mol}} \][/tex]
