A car tire contains air at a pressure of 24 lb/in^2 at 0 degrees C. If the volume of the tire remains unchanged, what will the pressure be when the temperature has increased to 32 degrees C?
Charles' law, P1/T1=P2/T2, or P/T=constant, as long as volume is constant. T must be in °K.
24/(273.15+0)=P/(273.15+32)
Solve for P
P=24/(273.15)*(305.15)
=26.8 lb/in²
It's Gay-Lussac's Law, not Charles' law (T vs V).
To solve this problem, you can use the ideal gas law, which states that the pressure of a gas is directly proportional to its temperature.
First, let's convert the pressure from pounds per square inch (lb/in^2) to a more widely used unit, such as Pascals (Pa).
1 lb/in^2 is approximately equal to 6894.76 Pascals. Therefore, the initial pressure is:
24 lb/in^2 * 6894.76 Pa/lb/in^2 ≈ 165,474.24 Pa
Next, let's convert the temperatures from Celsius to Kelvin. The Kelvin temperature scale is commonly used in gas laws. To convert from Celsius to Kelvin, you simply have to add 273.15 to the Celsius temperature.
Initial temperature (in Kelvin): 0 degrees C + 273.15 = 273.15 K
Final temperature (in Kelvin): 32 degrees C + 273.15 = 305.15 K
Now that we have the initial and final temperatures in Kelvin, we can solve for the final pressure using the ideal gas law equation:
P1 / T1 = P2 / T2
Where:
P1 = Initial pressure
P2 = Final pressure (what we want to find)
T1 = Initial temperature
T2 = Final temperature
Plugging in the values we have:
165,474.24 Pa / 273.15 K = P2 / 305.15 K
Now we can solve for P2:
P2 = (165,474.24 Pa / 273.15 K) * 305.15 K
Calculating the expression on the right side:
P2 ≈ 197,684.24 Pa
So, when the temperature has increased to 32 degrees C, the pressure in the car tire will be approximately 197,684.24 Pascals.