An automobile tire has a volume of 1.62 x 10-2 m3 and contains air at a gauge pressure (pressure above atmospheric pressure) of 167 kPa when the temperature is 0.00°C. What is the gauge pressure of the air in the tires when its temperature rises to 24.7°C and its volume increases to 1.69 x 10-2 m3? Assume atmospheric pressure is 1.01 x 105 Pa.

1.79 x 10^5 Pa

Explanation:

V1 = 1.62 x 10^-2 m^3

Absolute pressure is the sum of gauge pressure and the atmospheric pressure.

P1 = 167 kPa + P0

Where, P0 is the atmospheric pressure.

P1 = 1.67 x 10^5 + 1.01 x 10^5 = 2.68 x 10^5 Pa

T1 = 0 degree C = 273 K

V2 = 1.69 x 10^-2 m^3

T2 = 24.7 degree C = 24.7 + 273 = 297.7 K

Let the new absolute pressure is P2.

Use the relation given below

P1 V1 / T1 = P2 V2 / T2

(2.68 x 10^5 x 1.62 x 10^-2) / 273 = P2 x 1.69 x 10^-2 / 297.7

15.9 x 297.7 / (1.69 x 10^-2) = P2

P2 = 2.80 x 10^5 Pa

New gauge pressure = P2 - P0 = 2.80 x 10^5 - 1.01 x 10^5 = 1.79 x 10^5 Pa