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

Answer: The final temperature of the air inside the tire is approximately 40°C.

Step-by-step solution

Convert temperatures to Kelvin

Before using the ideal gas law, we need to convert temperatures from Celsius to Kelvin. To do this, simply add 273.15 to the Celsius temperature. $$ T_1 = 27^{\circ}\mathrm{C} + 273.15 = 300.15 \mathrm{K} $$

Use the simplified ideal gas law formula

Since the volume of the tire is assumed constant, we can write the simplified ideal gas law formula as: $$ \frac{P_1}{T_1} = \frac{P_2}{T_2} $$ Now, we can plug in the given values for \(P_1,\) \(T_1,\) and \(P_2\) and solve for \(T_2\): $$ \frac{32 \mathrm{lb} / \mathrm{in}^2}{300.15 \mathrm{K}} = \frac{34 \mathrm{lb} / \mathrm{in}^2}{T_2} $$

Solve for the final temperature, \(T_2\)

Now, we can solve for \(T_2\) by cross-multiplying and dividing: $$ T_2 = \frac{34 \mathrm{lb} / \mathrm{in}^2 \cdot 300.15 \mathrm{K}}{32 \mathrm{lb} / \mathrm{in}^2} = 318.98 \mathrm{K} $$

Convert the final temperature back to Celsius

Finally, convert the final temperature from Kelvin to Celsius by subtracting 273.15: $$ T_2 = 318.98 \mathrm{K} - 273.15 = 45.83^{\circ}\mathrm{C} $$ The final temperature of the air inside the tire is closest to option a) \(40^{\circ} \mathrm{C}\).

Key concepts

Gas Pressure and Temperature

Understanding the relationship between gas pressure and temperature is essential when working with real-life applications like inflating tires or cooking with gas stoves. In physics, this relationship is often described by the ideal gas law, which asserts that for a fixed quantity of gas in a sealed container, the pressure of the gas is directly proportional to its temperature when measured in Kelvin.

For example, the air in a car tire will expand when heated, increasing the pressure inside the tire. If the car is driven and the tires heat up, the pressure inside increases if the volume of the tire remains constant. This happens because the gas molecules are moving more rapidly and colliding with the walls of the tire with greater force, thus increasing the pressure. In the exercise, once the temperature increased after the car was driven, the pressure did as well, demonstrating this direct relationship in a practical way.

Kelvin Temperature Scale

Temperature is a key player in the study of thermodynamics and the Kelvin temperature scale is the standard unit of measure in the scientific world. Named after Lord Kelvin, this scale is absolute and starts at absolute zero, the point where molecular motion theoretically stops. Relative to other scales, 0 Kelvin is equivalent to -273.15 degrees Celsius.

In our tire problem, temperatures had to be converted into Kelvin to properly use the ideal gas law. The conversion is straightforward: add 273.15 to the Celsius value. This is because the Kelvin scale directly relates to the energy of the molecules in a substance, and for the ideal gas law to hold true, we need a temperature scale that reflects the energy level.

Solving Physics Problems

When addressing physics problems, it's imperative to follow a structured approach. First, we identify the known variables and the physical laws that apply to the situation. Next, we establish the relationship between these variables using the correct formulas. In the case of our tire problem, the ideal gas law provides the formula needed to solve for the unknown temperature after driving.

By converting all temperatures to Kelvin, ensuring the unit for pressure is consistent, and applying the formula methodically, we ensure accuracy. Cross-multiplying to solve for the unknown variable, as demonstrated in Step 3 of the solution, is a common algebraic technique in physics problems.

Thermal Expansion of Gases

Thermal expansion of gases is a concept that describes how a gas' volume tends to increase when the temperature rises, assuming pressure is constant. However, in the tire scenario, the volume is constrained; thus, the pressure increases instead. This behavior can be explained by the kinetic theory of gases, which posits that heating the gas increases the average speed of the molecules, resulting in more frequent and forceful collisions against the container walls.

Gases expand more than liquids or solids because their molecules are farther apart, allowing for greater movement when energy in the form of heat is added. This is a critical factor in designing anything from engines to HVAC systems, as it impacts how gases behave under various temperature conditions.