Answer: 0.0016 kg
Explanation: To solve this problem, we need to use the ideal gas law, which states that the pressure of a gas is directly proportional to its temperature and the number of moles of gas present. The ideal gas law is given by the following equation: PV = nRT
where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of gas present, R is the ideal gas constant, and T is the temperature of the gas.
In this case, we are given the pressure of the gas inside the tire (760 kPa and 550 kPa), the temperature of the gas (20 °C), and the diameter and circumference of the tire (2.2 cm and 2.0 m). We need to use this information to calculate the mass of air that was lost from the tire.
First, we need to find the volume of the gas inside the tire. Since the circumference of the tire is 2.0 m and the diameter is 2.2 cm, we can use the formula for the circumference of a circle to calculate the radius of the tire:
C = 2 * pi * r
2.0 m = 2 * pi * r
r = 0.3 m
Next, we can use the formula for the volume of a circle to calculate the volume of the gas inside the tire:
V = pi * r^2 * h
V = pi * (0.3 m)^2 * 2.2 cm
V = 0.0183 m^3
Now that we have the volume of the gas inside the tire, we can use the ideal gas law to calculate the number of moles of air inside the tire:
PV = nRT
n = PV / RT
n = (760 kPa * 0.0183 m^3) / (8.314 J/molK * 293 K)
n = 0.0059 mol
Next, we can use the ideal gas law to calculate the number of moles of air lost from the tire when the pressure decreased from 760 kPa to 550 kPa:
n = PV / RT
n = (550 kPa * 0.0183 m^3) / (8.314 J/molK * 293 K)
n = 0.0043 mol
Finally, we can use the molar mass of air to calculate the mass of air that was lost from the tire:
m = n * M
m = (0.0059 mol - 0.0043 mol) * 28.97 g/mol
m = 0.0016 kg
Therefore, the mass of air lost from the tire when the pressure decreased from 760 kPa to 550 kPa is 0.0016 kg.