Sure, let's solve the given problem step-by-step using Charles's Law. Charles's Law states that for a fixed amount of gas at a constant pressure, the volume of the gas is directly proportional to its absolute temperature. The formula is:
[tex]\[
\frac{V_1}{T_1} = \frac{V_2}{T_2}
\][/tex]
Where:
- [tex]\( V_1 \)[/tex] is the initial volume of the gas.
- [tex]\( T_1 \)[/tex] is the initial absolute temperature in Kelvin.
- [tex]\( V_2 \)[/tex] is the final volume of the gas.
- [tex]\( T_2 \)[/tex] is the final absolute temperature in Kelvin.
Given:
- [tex]\( V_1 = 1.0 \, \text{L} \)[/tex]
- [tex]\( T_1 = 22^{\circ} \text{C} \)[/tex]
- [tex]\( T_2 = 52^{\circ} \text{C} \)[/tex]
We need to find [tex]\( V_2 \)[/tex], the final volume of the tire.
First, we need to convert the temperatures from Celsius to Kelvin. The conversion formula is:
[tex]\[
T (\text{K}) = T (\text{C}) + 273.15
\][/tex]
So, the initial temperature [tex]\( T_1 \)[/tex] in Kelvin is:
[tex]\[
T_1 = 22 + 273.15 = 295.15 \, \text{K}
\][/tex]
And the final temperature [tex]\( T_2 \)[/tex] in Kelvin is:
[tex]\[
T_2 = 52 + 273.15 = 325.15 \, \text{K}
\][/tex]
Now, apply Charles's Law to find the final volume [tex]\( V_2 \)[/tex]:
[tex]\[
\frac{V_1}{T_1} = \frac{V_2}{T_2}
\][/tex]
Rearranging the equation to solve for [tex]\( V_2 \)[/tex]:
[tex]\[
V_2 = V_1 \times \frac{T_2}{T_1}
\][/tex]
Substitute the known values:
[tex]\[
V_2 = 1.0 \, \text{L} \times \frac{325.15 \, \text{K}}{295.15 \, \text{K}}
\][/tex]
Calculating the value:
[tex]\[
V_2 \approx 1.0 \times 1.1016432322547858 = 1.1016432322547858 \, \text{L}
\][/tex]
Considering the significant figures, we notice that all the initial values [tex]\(1.0 \, \text{L}\)[/tex], [tex]\(22^{\circ} \text{C}\)[/tex], and [tex]\(52^{\circ} \text{C}\)[/tex] are given to two significant figures. Therefore, our final result should be rounded to two significant figures as well.
Thus, the final volume [tex]\( V_2 \)[/tex] of the tire, rounded to two significant figures, is:
[tex]\[
V_2 \approx 1.1 \, \text{L}
\][/tex]
So, the resulting volume of the tire is [tex]\( \boxed{1.1} \, \text{L} \)[/tex] when the temperature is increased to [tex]\( 52^{\circ} \text{C} \)[/tex].
