Let's use the ideal gas law to find the volume of the tire. The ideal gas law is given by:
[tex]\[ PV = nRT \][/tex]
where:
- [tex]\( P \)[/tex] is the pressure (in Pascals),
- [tex]\( V \)[/tex] is the volume (in cubic meters),
- [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).
Given values:
- [tex]\( P = 220 \, \text{kPa} \)[/tex]
- [tex]\( T = 297 \, \text{K} \)[/tex]
- [tex]\( n = 4.8 \, \text{moles} \)[/tex]
- [tex]\( R = 8.314 \, \text{J/(mol}\cdot\text{K)} \)[/tex]
First, convert the pressure from kilopascals to pascals since 1 kPa = 1000 Pa:
[tex]\[ P = 220 \, \text{kPa} \times 1000 = 220000 \, \text{Pa} \][/tex]
Now, apply the ideal gas law to solve for [tex]\( V \)[/tex]:
[tex]\[ V = \frac{nRT}{P} = \frac{4.8 \, \text{moles} \times 8.314 \, \text{J/(mol}\cdot\text{K)} \times 297 \, \text{K}}{220000 \, \text{Pa}} \][/tex]
Calculate this step-by-step:
1. Calculate the numerator:
[tex]\[ 4.8 \times 8.314 \times 297 = 11860.4544 \, \text{J} \][/tex]
2. Divide by the pressure:
[tex]\[ \frac{11860.4544 \, \text{J}}{220000 \, \text{Pa}} = 0.05387472727272727 \, \text{m}^3 \][/tex]
Since 1 cubic meter (m[tex]\( ^3 \)[/tex]) equals 1000 liters, convert the volume to liters:
[tex]\[ 0.05387472727272727 \, \text{m}^3 \times 1000 = 53.87472727272727 \, \text{liters} \][/tex]
Rounded to two significant figures, the volume of the car tire is:
[tex]\[ 54 \, \text{liters} \][/tex]
Thus, the volume of the car tire is:
[tex]\[ \boxed{54} \, \text{liters} \][/tex]
