To find the resulting pressure when the volume of the container is decreased, we can use Boyle's Law, which states that for a given amount of gas at constant temperature, the pressure of the gas is inversely proportional to its volume. Mathematically, Boyle's Law is expressed as:
[tex]\[ P_1 \cdot V_1 = P_2 \cdot V_2 \][/tex]
where:
- [tex]\( P_1 \)[/tex] is the initial pressure,
- [tex]\( V_1 \)[/tex] is the initial volume,
- [tex]\( P_2 \)[/tex] is the final pressure, and
- [tex]\( V_2 \)[/tex] is the final volume.
Given the values:
- [tex]\( V_1 = 2.0 \, L \)[/tex]
- [tex]\( P_1 = 1.5 \, atm \)[/tex]
- [tex]\( V_2 = 1.0 \, L \)[/tex]
We need to find the final pressure [tex]\( P_2 \)[/tex]. Let's rearrange the formula to solve for [tex]\( P_2 \)[/tex]:
[tex]\[ P_2 = \frac{P_1 \cdot V_1}{V_2} \][/tex]
Now, substituting the known values into the equation:
[tex]\[ P_2 = \frac{1.5 \, \text{atm} \cdot 2.0 \, \text{L}}{1.0 \, \text{L}} \][/tex]
Perform the multiplication and division:
[tex]\[ P_2 = \frac{3.0 \, \text{atm} \cdot \text{L}}{1.0 \, \text{L}} \][/tex]
[tex]\[ P_2 = 3.0 \, \text{atm} \][/tex]
Therefore, the resulting pressure when the volume of the container is decreased to [tex]\( 1.0 \, L \)[/tex] is [tex]\( 3.0 \, \text{atm} \)[/tex], using the correct number of significant figures.
