Question

an automobile tire has a maximum rating of 38.0 psi (gauge pressure). the tire is inflated (while cold) to a volume of 11.8 l and a gauge pressure of 36.0 psi at a temperature of 12.0 °c. driving on a hot day, the tire warms to 65.0 °c and its volume expands to 12.2 l. does the pressure in the tire exceed its maximum rating? (note: the gauge pressure is the difference between the total pressure and atmospheric pressure. in this case, assume that atmospheric pressure is 14.7 psi.)

Explanation:

Step1: Convert initial and final temperatures to Kelvin

Initial temperature $T_1 = 12.0 + 273.15=285.15\ K$. Final temperature $T_2=65.0 + 273.15 = 338.15\ K$.

Step2: Calculate initial total pressure

Initial gauge - pressure $P_{g1}=36.0\ psi$, atmospheric pressure $P_{atm}=14.7\ psi$. Initial total pressure $P_1=P_{g1}+P_{atm}=36.0 + 14.7=50.7\ psi$.

Step3: Use the combined - gas law $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

We know $V_1 = 11.8\ L$, $V_2 = 12.2\ L$, $P_1 = 50.7\ psi$, $T_1 = 285.15\ K$, $T_2 = 338.15\ K$. Rearranging for $P_2$, we get $P_2=\frac{P_1V_1T_2}{V_2T_1}$.
Substitute the values: $P_2=\frac{50.7\times11.8\times338.15}{12.2\times285.15}$.
$P_2=\frac{50.7\times11.8\times338.15}{12.2\times285.15}\approx57.5\ psi$.

Step4: Calculate final gauge - pressure

Final gauge - pressure $P_{g2}=P_2 - P_{atm}=57.5-14.7 = 42.8\ psi$.

Answer:

Yes, the pressure in the tire exceeds its maximum rating since $42.8\ psi>38.0\ psi$.