Question

On a hot day, the pressure in an automobile tire increases. Assuming that the air in a tire at \(59^{\circ} \mathrm{F}\) increases in pressure from \(28.0 \mathrm{lbs} / \mathrm{in}^{2}\) to \(30.0 \mathrm{lbs} / \mathrm{in}^{2},(\mathrm{a})\) what is the temperature of the air in the tire, assuming no change in volume? (b) what will the pressure be if the temperature rises to \(106^{\circ} \mathrm{F}\) ?

Short answer

(a) The temperature of the air in the tire after pressure increases to 30.0 lbs/in² is approximately \(309\,K\). (b) The pressure when the temperature rises to 106°F will be approximately \(32.2\,lbs/in^2\).

Step-by-step solution

Convert given temperatures to the Kelvin scale

First, we should convert the given temperatures in Fahrenheit (59°F and 106°F) to Kelvin by using the following formula: \(K = \frac{5}{9}(°F - 32) + 273\). Applying the formula, we will find the initial and higher temperatures in Kelvin. \(T_1 = \frac{5}{9}(59 - 32) + 273 = 288 \, K\) \(T_2' = \frac{5}{9}(106 - 32) + 273 = 313 \, K\) where \(T_1\) is the initial temperature, \(T_2\) is the final temperature, and \(T_2'\) is the 106°F in Kelvin.

Calculate the final temperature inside the tire

Now, we can apply the formula relating the initial and final pressures and temperatures: \(P_1 / T_1 = P_2 / T_2\). Rearrange the equation to find \(T_2\): \(T_2 = T_1\times \frac{P_2}{P_1}\) We have \(P_1 = 28.0\,lbs/in^2\), \(P_2 = 30.0\,lbs/in^2\), and \(T_1 = 288\, K\). Now, calculate: \(T_2 =288\,K\times \frac{30.0\,lbs/in^2}{28.0\,lbs/in^2} \approx 309 \, K\)

Calculate the pressure when the temperature rises to 106°F

Now that we have both initial and higher temperatures in Kelvin, we can calculate the pressure when the temperature rises to 106°F. We will use the same formula as before, but now we will plug in \(T_2'\) as the final temperature: \(P_1'/ T_1 = P_2'/ T_2'\) Rearrange the equation to find \(P_2'\): \(P_2' = P_1'\times \frac{T_2'}{T_1}\) We have \(P_1' = 28.0\,lbs/in^2\), \(T_1 = 288\,K\), and \(T_2' = 313\,K\). Calculate: \(P_2' = 28.0\,lbs/in^2\times \frac{313\,K}{288\,K} \approx 32.2\,lbs/in^2\) #Answer# (a) The temperature of the air in the tire after pressure increases to 30.0 lbs/in² is approximately \(309\,K\). (b) The pressure when the temperature rises to 106°F will be approximately \(32.2\,lbs/in^2\).

Key concepts

Temperature-Pressure Relationship

Understanding the temperature-pressure relationship is essential when studying gas behavior under changing conditions.

According to this relationship, when a gas is kept at a constant volume, its pressure is directly proportional to its temperature. This means if the temperature of a gas increases, its pressure also increases when the volume doesn't change. Conversely, if the temperature decreases, the pressure does too, maintaining a proportional relationship as long as the volume of the gas doesn't change.

This relationship plays a crucial role when analyzing problems like the one presented where the temperature inside an automobile tire on a hot day causes changes in the pressure. Remember not to confuse this with the ideal gas law which combines pressure, volume, temperature, and the number of gas moles in one formula.

Kelvin Temperature Scale

The Kelvin temperature scale is fundamental in the realm of thermodynamics and is a must-know for understanding gas laws. Unlike Celsius or Fahrenheit, Kelvin is an absolute temperature scale, starting at absolute zero, where theoretically, particles cease their motion.

One key aspect is that Kelvin does not have degrees, we simply refer to Kelvin units. For gas law calculations, temperatures must be in Kelvin because they represent the absolute temperature directly proportional to the average kinetic energy of the gas particles.

The conversion formula from Fahrenheit to Kelvin, which is used in the automobile tire problem, reflects this need for an absolute scale. By converting Fahrenheit to Celsius and then to Kelvin, we ensure our temperature is absolute and ready for accurate gas law applications.

Gay-Lussac's Law

Gay-Lussac's Law describes the pressure-temperature relationship for a fixed mass of gas at a constant volume. The law is stated as: the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. Thus, if the temperature increases, the pressure will also increase, provided the volume remains unchanged.

Mathematically, Gay-Lussac's law can be written as: \(P_1 / T_1 = P_2 / T_2\), where \(P_1\) and \(P_2\) are the initial and final pressures, \(T_1\) and \(T_2\) are the initial and final temperatures respectively.

In our exercise related to the automotive tire, by using Gay-Lussac's law, we were able to calculate the temperature associated with the pressure increase (part a) and the pressure at a new higher temperature (part b). These calculations illustrate the law's practical application in real-world scenarios.