Sure! Let's solve this step-by-step.
### A. A gas has a volume of 4 mL at a temperature of 2 K. What would the temperature be if the volume increased to 10 mL?
#### Formula:
We will use Charles's Law, which states:
[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
- [tex]\( V_2 \)[/tex] is the final volume
- [tex]\( T_2 \)[/tex] is the final temperature
We need to solve for [tex]\( T_2 \)[/tex]. Rearrange the formula to isolate [tex]\( T_2 \)[/tex]:
[tex]\[ T_2 = \frac{V_2 \cdot T_1}{V_1} \][/tex]
#### Input numbers:
Now, we substitute the given values into the formula:
- [tex]\( V_1 = 4 \)[/tex] mL
- [tex]\( T_1 = 2 \)[/tex] K
- [tex]\( V_2 = 10 \)[/tex] mL
So,
[tex]\[ T_2 = \frac{10 \, \text{mL} \cdot 2 \, \text{K}}{4 \, \text{mL}} \][/tex]
#### Answer:
Now, perform the math:
[tex]\[ T_2 = \frac{10 \times 2}{4} \][/tex]
[tex]\[ T_2 = \frac{20}{4} \][/tex]
[tex]\[ T_2 = 5 \][/tex]
So, the temperature would be 5 K if the volume increased to 10 mL.
