Question

1. the front tire on a station wagon holds a pressure of 39 atm on a cold october day (5.0 °c). after a 300 km journey the tire has a temperature of 65.°c. assuming the tire has not changed in volume the new pressure will likely be

0.0211 atm
32.1 atm
507 atm
47.4 atm

Explanation:

Step1: Convert temperatures to Kelvin

$T_1 = 5 + 273 = 278$ K, $T_2=65 + 273 = 338$ K

Step2: Apply Gay - Lussac's Law ($\frac{P_1}{T_1}=\frac{P_2}{T_2}$)

$P_2=\frac{P_1\times T_2}{T_1}$
Substitute $P_1 = 39$ atm, $T_1 = 278$ K and $T_2 = 338$ K into the formula:
$P_2=\frac{39\times338}{278}$
$P_2=\frac{13182}{278}\approx47.4$ atm

Answer:

47.4 atm