Question

An automobile tire has a volume of 1.62 x 10^{-2}
m^{3} and contains air at a gauge pressure (pressure above
atmospheric pressure) of 164 kPa when the temperature is 0.00°C.
What is the gauge pressure of the air in the tires when its
temperature rises to 21.8°C and its volume increases to 1.68 x
10^{-2} m^{3}? Assume atmospheric pressure is 1.01
x 10^{5} Pa.

Answer #1

Atmospheric pressure = P_{atm} = 1.01 x 10^{5}
Pa

Initial gauge pressure of the air in the tire = P_{1g} =
164 kPa = 164 x 10^{3} Pa = 1.64 x 10^{5} Pa

Initial absolute pressure of the air in the tire =
P_{1}

P_{1} = P_{1g} + P_{atm}

P_{1} = 1.64x10^{5} + 1.01x10^{5}

P_{1} = 2.65 x 10^{5} Pa

Initial volume of the tire = V_{1} = 1.62 x
10^{-2} m^{3}

Initial temperature of the air in the tire = T_{1} = 0
^{o}C = 0 + 273 K = 273 K

Final absolute pressure of the air in the tire =
P_{2}

Final volume of the tire = V_{2} = 1.68 x
10^{-2} m^{3}

Final temperature of the air in the tire = T_{2} = 21.8
^{o}C = 21.8 + 273 K = 294.8 K

P_{2} = 2.76 x 10^{5} Pa

Final gauge pressure of the air in the tires =
P_{2g}

P_{2g} = P_{2} - P_{atm}

P_{2g} = 2.76x10^{5} - 1.01x10^{5}

P_{2g} = 1.75 x 10^{5} Pa

**Final gauge pressure of the air in the tires = 1.75 x
10 ^{5} Pa**

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