Question

An automobile tire has a volume of \(988 \mathrm{in} .^{3}\) and contains air at a gauge pressure of \(24.2 \mathrm{lb} / \mathrm{in}^{2}\) where the temperature is \(-2.60^{\circ} \mathrm{C}\). Find the gauge pressure of the air in the tire when its temperature rises to \(25.6^{\circ} \mathrm{C}\) and its volume increases to \(1020 \mathrm{in} .{ }^{3}\). (Hint: It is not necessary to convert from British to SI units. Why? Use \(p_{\text {atm }}=14.7 \mathrm{lb} / \mathrm{in}^{2} .\) )

Short answer

The gauge pressure of the air in the tire when its temperature rises to \(25.6^{\circ}C\) and its volume increases to \(1020 \, in^3 \) is \(26.5 \, lb/in^2\).

Step-by-step solution

Identify Given Variables and Unknown

We are given the initial volume (\(V_1 = 988 \, in^3\)), and pressure (\(P_1 = 24.2 \, lb/in^2\)) and the initial temperature (\(T_1 = -2.60^{\circ}C = 270.55\,K\)). We are also given the final volume (\(V_2 = 1020 \, in^3\)) and final temperature (\(T_2 = 25.6^{\circ}C = 298.75\, K\)). We need to find the final pressure (\(P_2\)).

Apply the Ideal Gas Law

According to the ideal gas law, \(P_1V_1/T_1 = P_2V_2/T_2\). We can rearrange the formula to solve for \(P_2\): \(P_2 = P_1V_1T_2/T_1V_2\).

Solve for \(P_2\)

Substituting the given values into the formula: \(P_2 = (24.2\, lb/in^2)(988 \,in^3)(298.75 \, K)/ (270.55 \, K)(1020 \, in^3) = 26.5 \, lb/in^2\).

Key concepts

Pressure-Temperature Relationship

The pressure-temperature relationship is a fundamental concept in understanding gases. This principle is rooted in Gay-Lussac's Law which states that for a constant volume, the pressure of a gas is directly proportional to its temperature.

To visualize this, imagine a closed container filled with gas. When the temperature inside this container increases, the gas molecules start moving faster. This increased speed causes more frequent and forceful collisions with the walls of the container, thus increasing the pressure.

• Formula: The relationship can be described with the formula: \( P_1 / T_1 = P_2 / T_2 \), where \( P \) represents pressure and \( T \) temperature.
• Direct Proportionality: If the temperature of the gas goes up, the pressure goes up too, provided the volume remains constant.

In the context of the automobile tire, as the air within warms from \(-2.6^{\circ}C\) to \(25.6^{\circ}C\), pressure increases assuming the volume stays the same. However, changes in volume are also considered, requiring ideal gas law adjustments.

Volume-Temperature Relationship

The volume-temperature relationship is encapsulated by Charles's Law. This states that when the pressure of a gas is held constant, its volume is directly proportional to its absolute temperature.

Consider a balloon inflated with air. As the temperature rises, the air molecules inside the balloon move faster and expand, increasing the volume of the balloon.

• Mathematical Representation: This can be represented as \( V_1 / T_1 = V_2 / T_2 \), where \( V \) is the volume and \( T \) is the temperature.
• Direct Proportionality: As the temperature increases, the volume increases similarly, if the pressure is constant.

In our tire example, since both temperature and volume increase, Charles’s Law helps explain how the gas expands to fill the increased volume, responding dynamically to changes in physical conditions.

Gauge Pressure

Gauge pressure is an important measurement in practical applications of pressure. It is the pressure of a system above the atmospheric pressure. This is distinct from absolute pressure which includes atmospheric pressure.

When a tire gauge is used to check a tire's pressure, it reflects the gauge pressure, which excludes the atmospheric pressure already acting on the tire.

• Formulation: Gauge Pressure = Absolute Pressure - Atmospheric Pressure
• Practical Use: It's used because it effectively measures pressure as experienced in most applications, such as the pressure in car tires.

Returning to our tire problem, initially, the air has a gauge pressure of \(24.2 \, lb/in^2\). When the temperature rises, utilizing the ideal gas law and adjusting for changes in both temperature and volume, the new gauge pressure is calculated to be \(26.5 \, lb/in^2\). This reflects how increased energy or heat can cause physical systems like a tire to further pressurize.