Question

A motorist inflates the tires of her car to a pressure of 180 kPa on a day when the temperature is -8.0° C. When she arrives at her destination, the tire pressure has increased to 245 kPa. What is the temperature of the tires if we assume that the tires do not expand?

Answer 1

Apply Gay-Lussac's law:

P/T = const.

P = pressure, T = temperature, the quotient of P/T must stay constant.

Initial P and T values:

P = 180kPa, T = -8.0°C = 265.15K

Final P and T values:

P = 245kPa, T = ?

Set the initial and final P/T values equal to each other and solve for the final T:

180/265.15 = 245/T

T = 361K