Question

A bicycle tire is filled to a pressure of \(110 .\) psi at a temperature of \(30.0^{\circ} \mathrm{C}\). At what temperature will the air pressure in the tire decrease to 105 psi? Assume that the volume of the tire remains constant.

Short answer

The final temperature is approximately 16.02°C.

Step-by-step solution

Identify the given values and the unknown

Given: Initial pressure, \(P_1 = 110 \text{ psi}\), Initial temperature, \(T_1 = 30.0^\text{°C} = 303.15 \text{ K}\) (converted to Kelvin). Final pressure, \(P_2 = 105 \text{ psi}\). To find: Final temperature, \(T_2 \).

Use the relationship between pressure and temperature

Since the volume of the tire remains constant, we use Gay-Lussac's law, which is \[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]

Rearrange the formula to solve for T2

From the formula, solving for \(T_2 \) gives \[ T_2 = \frac{P_2 \times T_1}{P_1} \]

Substitute the known values

Substitute \(P_1 = 110 \text{ psi}\), \(T_1 = 303.15 \text{ K}\) and \(P_2 = 105 \text{ psi}\) into the equation: \[ T_2 = \frac{105 \text{ psi} \times 303.15 \text{ K}}{110 \text{ psi}} \]

Calculate the final temperature

Perform the calculation: \[ T_2 = \frac{105 \times 303.15}{110} = 289.17 \text{ K}\]

Convert the temperature back to Celsius

To convert back to Celsius: \(T_2 = 289.17 \text{ K} - 273.15 = 16.02^\text{°C}\)

Key concepts

Pressure-Temperature Relationship

When studying gases and their behaviors, it's crucial to understand the relationship between pressure and temperature. This relationship is clarified by Gay-Lussac's law. According to this law, for a given mass of an ideal gas, the pressure exerted on the sides of its container is directly proportional to its temperature, provided the volume remains constant.
In mathematical terms, Gay-Lussac's law is expressed as: \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \). This formula shows that when temperature increases, so does pressure, and vice-versa.
In our problem, the volume of the bicycle tire remains constant. Hence, we can apply Gay-Lussac's law seamlessly to find the necessary changes in temperature when pressure changes.

Ideal Gas Law

The ideal gas law is another fundamental concept in understanding gas behavior. It combines several gas laws including Boyle's, Charles's, and Gay-Lussac's laws into one comprehensive equation: \( PV = nRT \). Here, \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles of gas, \( R \) is the gas constant, and \( T \) is temperature in Kelvin.
While the ideal gas law is useful for a wide range of problems, in this particular exercise, we can simplify our approach utilizing only Gay-Lussac’s law because the volume stays constant. This makes \( n \) and \( R \) constants, hence they cancel out of the equation.

Temperature Conversion

Temperature is a key variable when working with gas laws. However, it is often given in Celsius, which isn't suitable for these calculations. The Kelvin scale, which starts at absolute zero, is the preferred unit in gas law equations. To convert Celsius to Kelvin, use the formula: \( T(K) = T(°C) + 273.15 \).
For example, in the given problem: \( 30.0^\text{°C} = 303.15 \) K. After solving the equation to find the final temperature, converting back to Celsius: \( T_2 = 289.17 \text{ K} - 273.15 = 16.02^\text{°C} \).
It's vital to ensure you always convert your temperatures to the same units before starting your calculations to prevent any errors.