Problem 65 - Chemistry: An Atoms-Focused Approach

Question

Temperature Effects on Bicycle Tires A bicycle racer inflates his tires to 7.1 atm on a warm autumn afternoon when temperatures reach \(27^{\circ} \mathrm{C} .\) By morning the temperature has dropped to \(5.0^{\circ} \mathrm{C} .\) What is the pressure in the tires if we assume that the volume of the tire does not change significantly?

Step-by-Step Solution

Answer

Answer: The pressure in the tires when the temperature drops to 5°C is approximately 6.53 atm.

Step 1: Convert the given temperatures to Kelvin

To convert the Celsius temperatures to Kelvin, we add 273.15 to each value. Initial temperature \(T_1 = 27^{\circ} \mathrm{C} + 273.15 = 300.15 \mathrm{K}\) Final temperature \(T_2 = 5^{\circ} \mathrm{C} + 273.15 = 278.15 \mathrm{K}\)

Step 2: Apply Gay-Lussac's Law

Now, we will use the simplified ideal gas law (Gay-Lussac's Law) and the known initial conditions to find the final pressure \(P_2\): \(\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}\) Given, \(P_1 = 7.1 \mathrm{atm}\), \(T_1 = 300.15 \mathrm{K}\), and \(T_2 = 278.15 \mathrm{K}\). We are looking for \(P_2\).

Step 3: Solve for P2

Rearrange the equation for \(P_2\) \(P_2 = \dfrac{P_1 \times T_2}{T_1}\) Now, substitute the known values and calculate the final pressure: \(P_2 = \dfrac{7.1 \mathrm{atm} \times 278.15 \mathrm{K}}{300.15 \mathrm{K}}\) \(P_2 = 6.53 \mathrm{atm}\) (rounded to two decimal places)

Step 4: State the final answer

The pressure in the tires when the temperature drops to \(5^{\circ} \mathrm{C}\) is approximately 6.53 atm.

Key Concepts

Ideal Gas Law

The ideal gas law is an important concept in chemistry and physics that describes the behavior of gases under various conditions. The full form of the ideal gas law is given by the equation:\[PV=nRT\] where:

• \(P\) is the pressure of the gas.
• \(V\) is the volume of the gas.
• \(n\) is the number of moles of the gas.
• \(R\) is the ideal gas constant.
• \(T\) is the temperature of the gas in Kelvin.

The ideal gas law simplifies to other forms under specific conditions. Gay-Lussac's Law is a simplified version employed when the volume of a gas is constant, specifically showing the relationship between pressure and temperature:\[\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}\]In this scenario, when volume and the amount of gas remain unchanged, the pressure of a gas is directly proportional to its absolute temperature. This allows us to calculate what happens to a gas's pressure with temperature changes, like in the scenario of an inflated bicycle tire on a cold morning.
With this understanding, you can see how the ideal gas law, and its simplified forms, help predict how gases behave in differing conditions.

Temperature Conversion

Converting temperatures from Celsius to Kelvin is a straightforward process vital for applying gas laws. Kelvin is the standard unit of temperature in these calculations because it starts from absolute zero, meaning there are no negative temperatures. The conversion is quite simple and follows this rule:\[T(K) = T(^{\circ}C) + 273.15\]In this example:

• For the initial temperature of \(27^{\circ}C\), the conversion is: \(27 + 273.15 = 300.15 \, K\).
• For the final temperature of \(5^{\circ}C\), the conversion is: \(5 + 273.15 = 278.15 \, K\).

Using Kelvin ensures our calculations are compatible with the ideal gas law equations, as it provides an accurate scale for scientific work.
Understanding temperature conversion to Kelvin is crucial for tackling problems involving gases since temperature significantly affects gas behavior.

Pressure Calculation

In problems concerning gases, calculating pressure changes is often necessary to understand what conditions will affect a contained gas. As per Gay-Lussac's law, when the volume remains constant, pressure varies with temperature. The formula is:\[P_2 = \dfrac{P_1 \times T_2}{T_1}\]Taking the given values in our tire problem, where the initial pressure \(P_1\) is 7.1 atm, initial temperature \(T_1\) is 300.15 K, and final temperature \(T_2\) is 278.15 K, let's calculate:\[P_2 = \dfrac{7.1 \, \text{atm} \times 278.15 \, \text{K}}{300.15 \, \text{K}}\]Performing this calculation yields approximately 6.53 atm for the final pressure in the bicycle tire after the temperature drops. This perfectly demonstrates pressure's dependence on temperature.
Mastering such calculations allows you to better assess how pressure fluctuations may occur in real-life scenarios, including everyday tools like bicycle tires.