[tex]P_{1} = \text{1.00 atm}[/tex]
[tex]V_{1} = \text{3.60 L}[/tex]
[tex]P_{2} = \text{2.50 atm}[/tex]
[tex]V_{2}[/tex]
This is a gas law problem. What gas law should we use?
Since the given quantities are pressure and volume, we will use Boyle's law. According to this gas law, the pressure of a gas is inversely proportional to its volume keeping the amount of gas and its temperature constant.
The formula used for Boyle's law is
[tex]\boxed{P_{1}V_{1} = P_{2}V_{2}}[/tex]
where:
[tex]P_{1} = \text{initial pressure}[/tex]
[tex]V_{1} = \text{initial volume}[/tex]
[tex]P_{2} = \text{final pressure}[/tex]
[tex]V_{2} = \text{final volume}[/tex]
Starting with the formula of Boyle's law
[tex]P_{1}V_{1} = P_{2}V_{2}[/tex]
Dividing both sides of the equation by P₂ to solve for V₂
[tex]V_{2} = V_{1} \times \dfrac{P_{1}}{P_{2}}[/tex]
Substituting the given values and solving for V₂
[tex]V_{2} = \text{3.60 L} \times \dfrac{\text{1.00 atm}}{\text{2.50 atm}}[/tex]
Therefore, the final volume is
[tex]\boxed{V_{2} = \text{1.44 L}}[/tex]
[tex]V_{2} = \text{1.44 L}[/tex]
[tex]\\[/tex]
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