Question

A tire is inflated to a gauge pressure of 28.0 psi at \(71^{\circ} \mathrm{F}\). Gauge pressure is the pressure above atmospheric pressure, which is 14.7 psi. After several hours of driving, the air in the tire has a temperature of \(115^{\circ} \mathrm{F}\). What is the gauge pressure of the air in the tire? What is the actual pressure of the air in the tire? Assume that the tire volume changes are negligible.

Short answer

Short answer: After increasing the temperature from 71°F to 115°F, the final gauge pressure of the air in the tire is 30.203 psi, and the actual pressure is 44.903 psi.

Step-by-step solution

1. Convert Temperatures to Kelvin:

First, we need to convert the given temperatures from Fahrenheit to Kelvin using the following formula: \(T_{K} = (T_{F} - 32)\times\frac{5}{9} + 273.15\). The initial temperature \(T_{1}\) is \(71^{\circ} \mathrm{F}\) and the final temperature \(T_{2}\) is \(115^{\circ} \mathrm{F}\).

2. Calculate \(T_{1}\) and \(T_{2}\) in Kelvin:

Use the conversion formula to find the initial and final temperature in Kelvin: \(T_{1K} = (71 - 32) \times \frac{5}{9} + 273.15 = 294.261\,\mathrm{K}\) \(T_{2K} = (115 - 32) \times \frac{5}{9} + 273.15 = 318.150\,\mathrm{K}\)

3. Use the combined gas law formula:

To relate the initial and final conditions of pressure and temperature, the combined gas law, \(\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\), can be used. Here, \(P_{1}\) is the initial gauge pressure, \(T_{1}\) is the initial temperature in Kelvin, \(P_{2}\) is the final gauge pressure, and \(T_{2}\) is the final temperature in Kelvin.

4. Calculate the final gauge pressure \(P_{2}\):

Using the combined gas law formula, we can now find \(P_{2}\) as follows: \(P_{2} = P_{1} \times \frac{T_{2}}{T_{1}}\) From the problem, we know that \(P_{1} = 28.0\,\text{psi}\). Therefore, \(P_{2} = 28.0 \times \frac{318.150}{294.261} = 30.203\,\text{psi}\) The final gauge pressure is \(30.203\,\text{psi}\).

5. Calculate the actual pressure in the tire:

Now, we need to find the actual pressure inside the tire. We know from the problem that atmospheric pressure is \(14.7\,\text{psi}\). Actual pressure is the sum of gauge pressure and atmospheric pressure. Thus, Actual pressure = Final gauge pressure + Atmospheric pressure Actual pressure = \(30.203\,\text{psi} + 14.7\,\text{psi} = 44.903\,\text{psi}\) The actual pressure of the air in the tire is \(44.903\,\text{psi}\).

Key concepts

Gas Laws

Gas laws are fundamental scientific principles describing the behavior of gases under various conditions of pressure, temperature, and volume. Understanding these laws helps in predicting how gases will react when subjected to changes in conditions, which is crucial in fields such as chemistry, physics, and engineering.

One of the essential gas laws is the Combined Gas Law, which is a combination of three simpler laws: Boyle's Law (pressure-volume relationship), Charles's Law (temperature-volume relationship), and Gay-Lussac's Law (pressure-temperature relationship). The Combined Gas Law formula is expressed as \( \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2} \), where \( P \) stands for pressure, \( V \) for volume, and \( T \) for temperature, with the subscripts 1 and 2 referring to the initial and final states of the gas, respectively.

Using this law, one can predict the final state of a gas when any two of the three properties (pressure, volume, temperature) change, as long as the amount of gas remains constant. In the tire pressure example used in the exercise, we could assume constant volume to simplify the calculation to involve only pressure and temperature changes.

Pressure-Temperature Relationship

The pressure-temperature relationship is a key part of understanding gas behavior and is represented in Gay-Lussac's Law, a component of the Combined Gas Law. This relationship states that for a given mass of gas at constant volume, the pressure of the gas is directly proportional to its temperature measured in Kelvin.

This relationship can be expressed as \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \) when volume is held constant. When the temperature of a gas increases, its pressure increases, if volume can't expand, and vice versa. When dealing with problems like the tire pressure scenario, you're typically dealing with a closed system where the volume of the tire doesn't change significantly, making the pressure-temperature relationship directly applicable.

Kelvin Temperature Scale

The Kelvin temperature scale is an absolute thermodynamic temperature scale used extensively in the physical sciences. Unlike Celsius or Fahrenheit, it starts at absolute zero, the theoretical point where all kinetic energy in atoms ceases — meaning there is no heat and the particles are at rest.

The formula to convert Fahrenheit to Kelvin, which is vital for gas law calculations, is given by \( T_K = (T_F - 32) \times \frac{5}{9} + 273.15 \). In the context of the tire pressure exercise, temperatures must be converted to the Kelvin scale because gas laws require temperature to be in Kelvin to ensure proportional relationships between temperature and pressure.

Gauge Pressure

Gauge pressure is the pressure of a system above atmospheric pressure. It's the additional pressure in a system that is often read by instruments like tire gauges. The gauge pressure does not account for atmospheric pressure, which is why it tends to be lower than the actual pressure (absolute pressure) inside the system.

In many practical situations, it's gauge pressure that's of interest rather than absolute pressure. For example, when inflating tires, the recommended pressures are gauge pressures. To obtain the actual pressure, you must add the atmospheric pressure (average atmospheric pressure at sea level is \(14.7 \text{psi}\)) to the gauge pressure, which gives a complete picture of the force exerted by the gas within a container, such as a tire.