Tire pressure problem?

Question

Using your tire gauge you read a pressure of 30.0 psig. It's about 10.0 degrees celsius outside. You drive to class. When you arrive you check the pressure again and it now reads 35.0 psig. By how many degrees celsius did the drive to class increase the temperature of the air in your tire?

Answer 1

UPDATED 4/7/2007

Temperatures must be changed from degrees Celsius to Kelvin.

Celsius degrees + 273.25 = Kelvin

10.0 deg. Celsius + 273.15 . = 283.15 Kelvin = T1

PRESSURE:

Gauge pressure (psig) must be changed to absolute pressure (psia) which is required by the Combined Gas Equation used below. (The Combined Gas Equation combines both Boyle’s and Charles’ Law)

NOTE: One must use absolute pressure and NOT gauge pressure for a correct solution.

gauge pressure + one atmosphere = absolute pressure

30 psig + 14.7 = 44.7 psia. = P1

35 psig +14.7 = 49.7 psia = P2

NOTE 1: psig = pounds per square inch guage.
psia = pounds per square inch absolute
one atmosphere = 14.7 psi


NOTE 1: One does not have to convert pressure in pounds per square inch absolute (psia) to Pascals (Pa) since the two pressures (P1) and (P2) will be divided one by the other and the units will cancel. The same answer will result if pressures are converted to SI, however, one must be careful to use absolute pressure and NOT gauge pressure in whatever units one chooses to use.

Now substitute these values into the Combined Gas Equation below:

(P1)(V1)/T1 = (P2)(V2)/T2

(44.7 psia)(V1)/283.15 K = (49.7 psia)(V2)/T2

The volume of the tire is constant so V1 = V2, so when one solves this equation for T2 the two volumes cancel out.

T2 = (P2)(V2)(T1)/(P1)(V1) = (P2)(T1)/(P1)

T2 = (49.7)(283.15)/(44.7)

T2 = 314.82 Kelvin

T2 – T1 = Change in temperature.
314.82K - 283.15K = 31.7K or 31.7 degrees Celsius

The temperature of the air in the tire from the drive to class has increased by 31.7 degrees Celsius (or 31.7 Kelvin). Remember this is a change in temperature and not an actual temperature, so the change in temperature is the same in either Celsius or Kelvin.

Answer 2

The pressure is proportional to the absolute temperature, which has increased from 283 K to some higher value x. This gives x/283 = 35/30, where x will be in Kelvin; subtract 273 from x to get Celsius, and subtract 10 to get the temperature increase.

Answer 3

P1/T1 = P2/T2

30.0psig/283K = 35.0psig/T2

T2 = (35.0)(283)/(30.0) = 330K

330-283 = 47degC difference (increase)