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BOYLE'S LAW (SHOW FORMULA AND COMPLETE SOLUTION).

A sample of He gas was confined to a container at constant temperature. The volume of the gas is measured at 5.00 mL when the pressure applied is 6.50 atm. Determine the volume of He gas if the pressure is reduce to 2.00 atm.​

Sagot :

Given:

[tex]P_{1} = \text{6.50 atm}[/tex]

[tex]V_{1} = \text{5.00 mL}[/tex]

[tex]P_{2} = \text{2.00 atm}[/tex]

Required:

[tex]V_{2}[/tex]

Strategy:

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]

Solution:

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{5.00 mL} \times \dfrac{\text{6.50 atm}}{\text{2.00 atm}}[/tex]

Therefore, the final volume is

[tex]\boxed{V_{2} = \text{16.25 mL}}[/tex]

Answer:

[tex]V_{2} = \text{16.25 mL}[/tex]

[tex]\\[/tex]

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