A motorist inflates the tires of her car to a gauge 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 (a) the tires do not expand, or (b) that the tires expand so the volume of the enclosed air increases by 7 percent?

You want the final temperature of a tire after the pressure has increased to 245 kPa when it was inflated to 180 kPa at -8.0°C, (a) if the volume is unchanged, and (b) if the volume increases by 7%.

Temperature

The absolute temperature of a constant amount of air is proportional to the pressure and inversely proportional to the volume. The absolute temperature is 273.15°C more than the Celsius temperature.

(a) Constant volume

The absolute temperature starts at -8.0+273.15 = 265.15K. The final temperature will be ...

(245 kPa)/(180 kPa) · 265.15K = 360.899K

The Celsius temperature of the tire will be ...

360.899K -273.15K ≈ 87.7°C

The temperature at the destination is about 87.7°C.

(b) Increased volume

The final absolute temperature will be inversely proportional to the volume:

360.899K/(1 +7%) ≈ 337.288K

The Celsius temperature of the larger tire will be ...

337.288K -273.15K ≈ 64.1°C

The temperature of the expanded tire at the destination is about 64.1°C.