Before beginning a long trip on a hot day, a driver inflates an automobile tire to a gauge pressure of 2.20 atm at 300 K. At the end of the trip, the gauge pressure has increased to 2.70 atm.

Is there a question here?

To solve this problem, we can utilize the ideal gas law equation, which relates the pressure, volume, and temperature of a gas:

PV = nRT

Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature

In this case, we are given the initial and final gauge pressures of the tire (P1 and P2) and the initial temperature (T1). We are asked to find the final temperature (T2).

Step-by-step process to solve the problem:

1. Convert the given pressures from gauge pressure to absolute pressure. Absolute pressure is the sum of gauge pressure and atmospheric pressure. We assume atmospheric pressure to be 1 atm.

P1_abs = P1 + 1 atm
P2_abs = P2 + 1 atm

In this case, P1_abs = 2.20 atm + 1 atm = 3.20 atm
P2_abs = 2.70 atm + 1 atm = 3.70 atm

2. Convert the absolute pressures to the SI unit of pressure, which is pascal (Pa). 1 atm = 101325 Pa.

P1_abs_SI = P1_abs × 101325 Pa
P2_abs_SI = P2_abs × 101325 Pa

In this case, P1_abs_SI = 3.20 atm × 101325 Pa/atm ≈ 323760 Pa
P2_abs_SI = 3.70 atm × 101325 Pa/atm ≈ 375925 Pa

3. Convert the initial temperature from Kelvin (K) to Celsius (°C), if necessary.

4. Rearrange the ideal gas law equation to solve for the final temperature (T2):

T2 = (P2_abs_SI × V) / (P1_abs_SI × R)

Note: Since we are considering the same volume before and after the trip, we can cancel out the V term in the equation.

5. Substitute the known values into the equation:

T2 = (375925 Pa) / (323760 Pa) × 300 K

6. Calculate the final temperature (T2).

T2 ≈ 349.4 K

Therefore, at the end of the trip, the temperature of the tire is approximately 349.4 K.