An automobile tire contains air at a pressure of 200. kPa and 25°C. The tire and its air heat up during traveling. If the pressure increases to 250. kPa but the volume remains the same, what is the temperature of the air inside the tire?

(p1/t1) = (p2/t2)

t1 and t2 must be in kelvin.

35°C

To solve this problem, we can use the ideal gas law, which states that the pressure, volume, and temperature of a gas are related. The equation for the ideal gas law is:

PV = nRT

Where:
P = Pressure
V = Volume
n = Number of moles of gas
R = Ideal gas constant
T = Temperature in Kelvin

Given:
Initial pressure (P1) = 200 kPa
Final pressure (P2) = 250 kPa
Initial temperature (T1) = 25°C

First, we need to convert the temperature from Celsius to Kelvin. The Kelvin temperature scale is obtained by adding 273.15 to the Celsius temperature.

T1 = 25°C + 273.15 = 298.15 K

Since the volume remains the same, we can cancel it out in our calculation:

P1/T1 = P2/T2

Now, we can rearrange the equation to solve for T2 (the final temperature):

T2 = (P2 * T1) / P1

Plugging in the values:

T2 = (250 kPa * 298.15 K) / 200 kPa

Calculating:

T2 = 372.68 K

Converting the temperature back to Celsius:

T2 = 372.68 K - 273.15 = 99.53°C

Therefore, the temperature of the air inside the tire is approximately 99.53°C.