You fill your tires with air on a cold morning (-5degreesC) to 220-kPa gauge pressure, then drive into a 32degreesC desert. (a) Assuming the volume of air in the tires remains constant, what's the new gauge pressure? (b) what would be the gauge pressure if the volume of air had expanded by 3%?
To answer these questions, we can use the ideal gas law and assume that the air in the tires behaves as an ideal gas. The ideal gas law can be expressed as:
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, R is the ideal gas constant, and T is the temperature in Kelvin.
Given:
Initial temperature (T1) = -5 degrees Celsius
Final temperature (T2) = 32 degrees Celsius
Initial pressure (P1) = 220 kPa (gauge pressure)
Note: When using the ideal gas law, temperatures must be converted to Kelvin, so we can convert the given temperatures to Kelvin by adding 273.15 to each.
(a) Assuming the volume of air in the tires remains constant:
To solve for the new gauge pressure (P2), we can rearrange the ideal gas law equation:
P1/T1 = P2/T2
Let's plug in the values:
P1 = 220 kPa - atmospheric pressure
T1 = -5 degrees Celsius + 273.15 = 268.15 K
T2 = 32 degrees Celsius + 273.15 = 305.15 K
P1/T1 = P2/T2
Solving for P2:
P2 = (P1 * T2) / T1
P2 = (220 kPa * 305.15 K) / 268.15 K
P2 ≈ 250.90 kPa
Therefore, the new gauge pressure when driving into a 32 degrees Celsius desert (assuming constant volume) would be approximately 250.90 kPa.
(b) If the volume of air had expanded by 3%:
In this case, we need to consider the change in volume of the gas. Let's assume the initial volume is V1 and the final volume is V2.
Since the volume increases by 3%, the final volume (V2) can be calculated as:
V2 = V1 + (0.03 * V1)
Now, let's use the combined gas law, which combines the ideal gas law equation with the assumption of constant number of moles and rearranges it to calculate the final pressure:
(P1 * V1) / T1 = (P2 * V2) / T2
P1/V1 = P2/V2
Substituting the values:
P1 = 220 kPa - atmospheric pressure
V1 = initial volume
T1 = -5 degrees Celsius + 273.15 = 268.15 K
T2 = 32 degrees Celsius + 273.15 = 305.15 K
V2 = V1 + (0.03 * V1)
Solving for P2:
(P1 * V1) / T1 = (P2 * (V1 + 0.03 * V1)) / T2
Simplifying:
P2 = (P1 * V1 * T2) / ((V1 + 0.03 * V1) * T1)
P2 = (P1 * T2) / (V1 + 0.03 * V1) * T1
P2 ≈ (220 kPa * 305.15 K) / (1.03 * V1 * 268.15 K)
Therefore, the gauge pressure if the volume of air had expanded by 3% would be approximately (220 kPa * 305.15 K) / (1.03 * V1 * 268.15 K) kPa.