To solve this problem, we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature when pressure and the number of moles are kept constant. Mathematically, this is expressed 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
- [tex]\( V_2 \)[/tex] is the final volume
- [tex]\( T_2 \)[/tex] is the final temperature
Given the problem:
- [tex]\( V_1 = 30 \)[/tex] liters
- [tex]\( T_1 = 100 \)[/tex] Kelvin
- [tex]\( T_2 = 200 \)[/tex] Kelvin
We need to find [tex]\( V_2 \)[/tex]. Rearrange Charles's Law to solve for [tex]\( V_2 \)[/tex]:
[tex]\[ V_2 = V_1 \cdot \frac{T_2}{T_1} \][/tex]
Substitute the given values into the equation:
[tex]\[ V_2 = 30 \cdot \frac{200}{100} \][/tex]
[tex]\[ V_2 = 30 \cdot 2 \][/tex]
[tex]\[ V_2 = 60 \][/tex]
So, the volume of the gas at 200 degrees K is [tex]\( 60 \)[/tex].
