- Convert initial and final temperatures from Celsius to Kelvin by adding 273.15: \(T_1 = 295.15 \text{ K}\) and \(T_2 = 325.15 \text{ K}\).
- Use Charles' Law \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) to solve for \(V_2 = V_1 \cdot \frac{T_2}{T_1}\).
- Calculate the final volume: \(V_2 = 1.0 \cdot \frac{325.15}{295.15} \approx 1.10164 \text{ L}\).
- Round the result to two significant figures: \(\boxed{1.1}\) \(\text{L}\).
Explanation
- Analysis of the Problem and Data
The problem asks for the final volume of a bicycle tire when the temperature increases from \(22^{\circ} \text{C}\) to \(52^{\circ} \text{C}\), given an initial volume of \(1.0 \text{ L}\). We use Charles' Law, which states that the ratio of volume to temperature is constant for an ideal gas at constant pressure, expressed as \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\). Note that in all gas law calculations, temperature must be converted from Celsius to the Kelvin scale by adding \(273.15\). The input values of \(1.0 \text{ L}\), \(22^{\circ} \text{C}\), and \(52^{\circ} \text{C}\) have two significant figures, so our final result should also be reported with two significant figures.
- Calculating the Final Volume
First, convert the temperatures from Celsius to Kelvin:
\[T_1 = 22^{\circ} \text{C} + 273.15 = 295.15 \text{ K}\]
\[T_2 = 52^{\circ} \text{C} + 273.15 = 325.15 \text{ K}\]
Using Charles' Law, we rearrange the formula to solve for the final volume \(V_2\):
\[V_2 = V_1 \times \frac{T_2}{T_1}\]
Substituting the given values:
\[V_2 = 1.0 \text{ L} \times \frac{325.15 \text{ K}}{295.15 \text{ K}}\]
Performing the calculation:
\[V_2 \approx 1.0 \times 1.10164 = 1.10164 \text{ L}\]
- Final Answer and Significant Figures
Given the initial value \(1.0 \text{ L}\) and the temperature values, we maintain two significant figures in our final calculation. Rounding \(1.10164 \text{ L}\) to two significant figures gives us \(1.1 \text{ L}\). Thus, the final volume of the tire is \(\boxed{1.1}\) \(\text{L}\).
Examples
Understanding Charles' Law is crucial in many real-world scenarios. For example, it explains why tires might appear under-inflated in winter or over-inflated in summer. As the temperature of the air inside the tire changes, the volume and pressure of the trapped air adjust, which engineers must account for when designing tires for different climates and driving conditions. By using the relationship \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\), we can predict these changes accurately and ensure the safety and longevity of our vehicle components.