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Gas sample contained in a cylinder equipped with a moveable piston occupied 300.0 mL at a pressure of 2.00atm. What would be the final pressure if the volume were increased to 500.00 mL at a constant temperature?

Answer :

Answer:

The final pressure of the gas sample at constant temperature is 1.20 atm.

Explanation:

Boyle's Law : Volume occupied by fixed amount of gas at a given temperature is inversely proportional to pressure .(moles, temperature are constant )

[tex]P\alpha \frac{1}{V}[/tex]

[tex]P = \frac{k}{V}[/tex]

on transposing ,

PV = constant

[tex]P_{1}V_{1} = P_{1}V_{1}[/tex]............(1)

Initial pressure = P1 = 2.00 atm

Final pressure = P2 = ?

Initial volume = V1 = 300.0 mL

Final volume = V2 = 500.0 mL

Putting values of P1 ,V1, V2 in equation (1) and solving for P2

[tex]P_{1}V_{1} = P_{1}V_{1}[/tex]

[tex]2\times 300 = P_{2}\times 500 [/tex]

[tex]P_{2} = \frac{2\times 300 }{500}[/tex]

[tex]P_{2} = 1.2[/tex]

P2 = 1.20 atm  at constant temperature

Taking into account the Boyle's law, the final pressure would be 1.2 atm.

Boyle's law

Boyle's law is one of the gas laws that relates the volume and pressure of a certain quantity of gas kept at constant temperature.

This law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant. That is, if the pressure increases, the volume decreases; while if the pressure decreases, the volume increases.

Mathematically, Boyle's law states that the product of pressure and volume is constant:

P×V= k

Studying two different states, an initial state 1 and a final state 2, it is satisfied:

P1× V1= P2×V2

Final pressure in this case

In this case, you know:

  • P1= 2 atm
  • V1= 300 mL
  • P2= ?
  • V2= 500 mL

Replacing in Boyle's law:

2 atm× 300 mL= P2×500 mL

Solving:

P2= (2 atm× 300 mL) ÷500 mL

P2= 1.2 atm

Finally, if the volume were increased to 500.00 mL at a constant temperature, the final pressure would be 1.2 atm.

Learn more about Boyle's law:

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