Question

5. A steel-belted radial automobile tire is inflated to a gauge pressure of 2.00 × 10 5 Pa when the temperature is 62 °F. Later in the day, the temperature rises to 101 °F. Assuming the volume of the tire remains constant, what is the gauge pressure at the elevated temperature? [Hint: Remember that the ideal gas law uses absolute pressure.]

Answer 1

Answer:

From ideal gas law PV=nRT here Vin, and Rare constant so PIT E CONSTANT OR P1/T1=P2/T2 now first we need to conver Fahrenheit in kelvin 62°F = 289.82K and 101 ºf = 311.48 K formula (f-32)x5/9+273.15 =K and absolute pressure at 68°F = (2+1.013)x 10^5 Pa so 3.013x10^5 289.82=P2/311.48 so P2= 3.238x10^5 Pa or gauge pressure at 101 ºf is P= (3.238 - 1.013)x1045 = 2.225 x10^5 Pa (Answer)