Many of you may have noticed the phenomenon that basketballs get flat if the weather is cold. If a basketball was inflated to a gauge pressure of 60,000 Pa when the temperature outside was 20∘C, what is the gauge pressure inside the basketball in Pa when the temperature is 10∘C?
Details and assumptions
The outside air maintains a constant pressure of 1 atm=101,325 Pa as the temperature changes.
The basketball always has a circumference of 0.75 m.
There's a reason gauge is in bold face. If you don't know what gauge pressure is, you should look it up before attempting the problem.
At 60,000 Pa gauge pressure, the absolute pressure inside the ball is
101,325 + 60,000 = 161,325 Pa
Dropping the absolute temperature from 293 to 283 K decreases the absolute pressure in the ball by a factor 283/293, resulting in 15,582 Pa. Subtract ambient pressure from that for the new gauge pressure. The gauge pressure becomes 54,494 Pa.
To solve this problem, we need to apply the ideal gas law, which states that the pressure of a gas is directly proportional to its temperature and inversely proportional to its volume when the number of moles of gas and the gas constant are constant.
1. First, let's determine the initial temperature and pressure of the basketball. We are given that the basketball was inflated to a gauge pressure of 60,000 Pa when the temperature outside was 20∘C. Since we are given the gauge pressure, we need to convert it to absolute pressure by adding the atmospheric pressure. The atmospheric pressure is given as 1 atm = 101,325 Pa, so the initial absolute pressure is 60,000 Pa + 101,325 Pa = 161,325 Pa. The initial temperature is 20∘C, which needs to be converted to Kelvin by adding 273.15. So the initial temperature in Kelvin is (20+273.15) K = 293.15 K.
2. Next, we need to determine the final temperature. It is given that the temperature outside is now 10∘C. We convert this to Kelvin by adding 273.15. So the final temperature in Kelvin is (10+273.15) K = 283.15 K.
3. Now, let's use the ideal gas law equation:
(P1/T1) = (P2/T2)
Where P1 and T1 are the initial pressure and temperature, and P2 and T2 are the final pressure and temperature.
4. Plug in the values:
(P1/T1) = (P2/T2)
(161,325 Pa / 293.15 K) = (P2 / 283.15 K)
5. Solve for P2:
P2 = (161,325 Pa / 293.15 K) * 283.15 K
P2 = 155,795.93 Pa
Therefore, the gauge pressure inside the basketball when the temperature is 10∘C is approximately 155,795.93 Pa.