Certainly! Let's solve this step-by-step.
The question involves finding the new volume of an air bubble when the temperature changes, using Charles's Law. Charles's Law states that for a given mass of gas at constant pressure, the volume is directly proportional to its temperature (in Kelvin). The mathematical form of Charles's Law is:
[tex]\[
\frac{V_1}{T_1} = \frac{V_2}{T_2}
\][/tex]
Where:
- [tex]\( V_1 \)[/tex] is the initial volume
- [tex]\( T_1 \)[/tex] is the initial temperature in Kelvin
- [tex]\( V_2 \)[/tex] is the final volume
- [tex]\( T_2 \)[/tex] is the final temperature in Kelvin
Given:
- Initial volume [tex]\( V_1 = 0.520 \)[/tex] liters
- Initial temperature [tex]\( T_1 = 16^{\circ} C \)[/tex]
- Final temperature [tex]\( T_2 = 6^{\circ} C \)[/tex]
First, we need to convert the temperatures from Celsius to Kelvin, using the formula:
[tex]\[
T(K) = T(C) + 273.15
\][/tex]
1. Convert the initial temperature:
[tex]\[
T_1 = 16 + 273.15 = 289.15 \, \text{K}
\][/tex]
2. Convert the final temperature:
[tex]\[
T_2 = 6 + 273.15 = 279.15 \, \text{K}
\][/tex]
Next, we use Charles's Law to find the final volume [tex]\( V_2 \)[/tex]:
[tex]\[
\frac{V_1}{T_1} = \frac{V_2}{T_2}
\][/tex]
Rearrange to solve for [tex]\( V_2 \)[/tex]:
[tex]\[
V_2 = V_1 \times \frac{T_2}{T_1}
\][/tex]
Plug in the values:
[tex]\[
V_2 = 0.520 \times \frac{279.15}{289.15}
\][/tex]
3. Calculate the final volume:
[tex]\[
V_2 \approx 0.50 \, \text{liters}
\][/tex]
So, the volume of the air bubble at [tex]\(6^{\circ} C\)[/tex] is approximately 0.50 liters when rounded to two significant figures.
Therefore, we can summarize the results as follows:
[tex]\[
V = \begin{array}{|l|l|}
\hline \text{0.50} & \text{Liters} \\
\hline
\end{array}
\][/tex]
In units of liters.
