To find the gauge pressure of the air in the tire when the temperature and volume change, we can use the Ideal Gas Law, which states that the pressure, volume, and temperature of a gas are related. The formula we will use is derived from the combined gas law, which is expressed as:
Understanding the Variables
In this scenario, we have the following variables:
- P1: Initial gauge pressure = 24.2 lb/in²
- V1: Initial volume = 988 in³
- T1: Initial temperature in Kelvin = -2.60ºC + 273.15 = 270.55 K
- P2: Final gauge pressure (what we want to find)
- V2: Final volume = 1020 in³
- T2: Final temperature in Kelvin = 25.6ºC + 273.15 = 298.75 K
Applying the Ideal Gas Law
The combined gas law can be expressed as:
(P1 × V1) / T1 = (P2 × V2) / T2
We can rearrange this equation to solve for P2:
P2 = (P1 × V1 × T2) / (V2 × T1)
Plugging in the Values
Now, let's substitute the known values into the equation:
P2 = (24.2 lb/in² × 988 in³ × 298.75 K) / (1020 in³ × 270.55 K)
Calculating Step-by-Step
First, calculate the numerator:
Numerator = 24.2 × 988 × 298.75
Calculating this gives:
Numerator = 24.2 × 988 = 23957.6
23957.6 × 298.75 ≈ 7150000.2
Next, calculate the denominator:
Denominator = 1020 × 270.55
Calculating this gives:
Denominator = 1020 × 270.55 ≈ 275000.1
Final Calculation
Now, we can find P2:
P2 = 7150000.2 / 275000.1 ≈ 26.0 lb/in²
Understanding Gauge Pressure
Since we are looking for gauge pressure, we need to subtract atmospheric pressure from the absolute pressure. The atmospheric pressure is given as 14.7 lb/in². Therefore, the gauge pressure P2 is:
Gauge Pressure = P2 - patm
Gauge Pressure = 26.0 lb/in² - 14.7 lb/in² = 11.3 lb/in²
Conclusion
The gauge pressure of the air in the tire when the temperature rises to 25.6ºC and the volume increases to 1020 in³ is approximately 11.3 lb/in². This calculation illustrates how temperature and volume changes affect the pressure of a gas, demonstrating the principles of the Ideal Gas Law in a practical context.
