Sure! To solve for the volume of the air bubble at a different temperature while keeping the pressure and the number of moles constant, we can use 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 formula can be written as:
[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, and
- [tex]\( T_2 \)[/tex] is the final temperature in kelvin.
Given:
- Initial volume, [tex]\( V_1 = 0.520 \)[/tex] liters,
- Initial temperature, [tex]\( T_1 = 16 \)[/tex] degrees Celsius,
- Final temperature, [tex]\( T_2 = 420 \)[/tex] kelvin.
First, we need to convert the initial temperature from degrees Celsius to kelvin using the formula:
[tex]\[ T_{\text{kelvin}} = T_{\text{Celsius}} + 273.15 \][/tex]
So,
[tex]\[ T_1 = 16 + 273.15 = 289.15 \text{ K} \][/tex]
Now we can substitute the known values into Charles's Law formula and solve for [tex]\( V_2 \)[/tex]:
[tex]\[ \frac{0.520 \text{ L}}{289.15 \text{ K}} = \frac{V_2}{420 \text{ K}} \][/tex]
To find [tex]\( V_2 \)[/tex], we rearrange the equation:
[tex]\[ V_2 = 0.520 \text{ L} \times \frac{420 \text{ K}}{289.15 \text{ K}} \][/tex]
[tex]\[ V_2 \approx 0.76 \text{ L} \][/tex]
So, the volume of the air bubble at 420 K is approximately [tex]\( 0.76 \)[/tex] liters. This result is expressed to two significant figures and includes the appropriate units.
Final answer:
The volume of the air bubble at 420 K is [tex]\( 0.76 \)[/tex] L.
