Question

A tire on a car is inflated to a gauge pressure of \(32 \mathrm{lb} / \mathrm{in}^{2}\) at a temperature of \(27^{\circ} \mathrm{C}\). After the car is driven for \(30 \mathrm{mi}\), the pressure has increased to \(34 \mathrm{lb} / \mathrm{in}^{2}\). What is the temperature of the air inside the tire at this point? a) \(40 .{ }^{\circ} \mathrm{C}\) b) \(23^{\circ} \mathrm{C}\) c) \(32^{\circ} \mathrm{C}\) d) \(54^{\circ} \mathrm{C}\)

Short answer

a) 40°C b) 42°C c) 46°C d) 48°C Answer: a) 40°C

Step-by-step solution

Convert units

Conversion factors: 1 atm = 14.7 lb/in² - 1 K = 273.15 °C Minimum allowed figure for working out this exercise : 4 significant. First, convert the given gauge pressures and temperature to standard units. Initial gauge pressure: \(32 {\rm lb}/{\rm in^2}\) Initial temperature: \(27^{\circ} \rm C\) Initial absolute pressure (\(P_1\)): \((32 {\rm lb}/{\rm in^2}) \times (\dfrac{1 {\rm atm}}{14.7 {\rm lb}/{\rm in^2}}) = 2.18\, {\rm atm}\) (approx) Initial temperature (\(T_1\)): \(27^{\circ}{\rm C} + 273.15 = 300.15 {\rm K}\) (approx) Final gauge pressure: \(34 \, {\rm lb}/{\rm in^2}\) Final absolute pressure (\(P_2\)): \((34 \,{\rm lb}/{\rm in^2}) \times (\dfrac{1 {\rm atm}}{{14.7 \, \rm lb}/{\rm in^2}}) = 2.31\, {\rm atm}\) (approx) We'll now use the Ideal Gas Law to find the final temperature.

Apply the Ideal Gas Law

The Ideal Gas Law is given by: \(PV = nRT\) Since \(nR\) is constant, we can write: \(P_1V_1/T_1 = P_2V_2/T_2\) Since the volume of the tire remains constant, we can write: \(P_1/T_1 = P_2/T_2\) Now, we can find the final temperature (\(T_2\)): \(T_2 = P_2T_1/P_1\)

Calculate the final temperature

Plug the values into the equation and solve for \(T_2\): \(T_2 = \frac{2.31 {\rm atm} \times 300.15\, {\rm K}}{2.18{\rm atm}}\) \(T_2 \approx 317.15 {\rm K}\) Now, let's convert this temperature back to °C. \(T_2 = 317.15 {\rm K} - 273.15 \approx 44^{\circ} \rm C\) The closest answer from the given options is (a) \(40^{\circ} \mathrm{C}\). So we choose this answer.