A basketball is inflated in a garage at 25 degrees Celsius to a gauge pressure of 8.0 psi. Gauge pressure is the pressure above atmospheric pressure, which is 14.7 psi. The ball is used on the driveway at a temperature of -7 degrees celsius and feels flat. What is the actual pressure of the air in the ball? What is the gauge pressure?
I'm stuck as to how to start this problem. Any help would be appreciated.
(P1V1)/T1=(P2V2)/T2
Don't forget to change Celsius to Kelvin.
To solve this problem, we can use the Ideal Gas Law formula: P1V1/T1 = P2V2/T2.
Let's break down the information given:
- P1 is the initial gauge pressure in the garage, which is 8.0 psi. We need to convert this to absolute pressure by adding the atmospheric pressure of 14.7 psi.
- V1 is the initial volume of the basketball, which we do not have.
- T1 is the initial temperature in the garage, which is 25 degrees Celsius. Remember to convert Celsius to Kelvin by adding 273.15.
- P2 is the desired pressure of the air inside the basketball.
- V2 is the final volume of the basketball, which we do not have.
- T2 is the final temperature on the driveway, which is -7 degrees Celsius. Convert Celsius to Kelvin by adding 273.15.
Now we can set up the equation:
(P1 + Patm)V1 / T1 = P2V2 / T2
Since we are interested in the actual pressure (P2), we can rearrange the equation to solve for P2:
P2 = [(P1 + Patm)V1 * T2] / [T1 * V2]
Keep in mind that we also need the final volume of the basketball (V2). If it remains the same during the temperature change, we can cancel it out when solving for P2. Otherwise, we would need the information about the final volume to solve accurately.
To find the actual pressure of the air in the ball, use the above formula with the given values and any missing values you may have.