In a car tire there is air at a pressure of 5.9 * 10 ^ 5 Pa at a temperature of 20 C.
In a car tire there is air at a pressure of 5.9 * 10 ^ 5 Pa at a temperature of 20 C. When the car is moving, the temperature rises to 35 C. How much does the air pressure inside the tire increase?
By the conditions of the problem, we have:
P1 – initial air pressure in the tire, 5.9 * 10 ^ 5 Pa;
T1 is the initial temperature of the air in the tire, 20 degrees C or 293 degrees K;
T2 is the temperature of the air in the tire while the vehicle is in motion, 35 degrees C or 308 degrees K;
It is required to find P2 – the air pressure in the tire while driving in the car.
Since, according to the conditions of the problem, the volume of air does not change (isochoric process), then P2 can be found by the formula:
P1 / T1 = P2 / T2 or
P2 = P1 * T2 / T1 = 5.9 * 10 ^ 5 * 308/293 = 6.2 * 10 ^ 5 Pa.
P2 – P1 = 6.2 * 10 ^ 5 – 5.9 * 10 ^ 5 = 0.3 * 10 ^ 5 Pa.
Answer: the pressure inside the tire is 6.2 * 10 ^ 5 Pa, that is, it increases by 0.3 * 10 ^ 5 Pa.
