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A balloon is inflated to a volume of 7.0 L on a day when the atmospheric pressure is 765 mmHg. The next day, as a storm front arrives, the atmospheric pressure drops to 733 mmHg . Assuming the temperature remains constant, what is the new volume of the balloon, in liters?

Respuesta :

skyluke89

Answer:

7.3 L

Explanation:

We can solve the problem by using Boyle's law, which states that for a gas at constant temperature, the product between pressure and volume of the gas remains the same:

[tex]pV=const.[/tex]

where

p is the pressure

V is the volume

The equation can also be rewritten as

[tex]p_1 V_1 = p_2 V_2[/tex]

where

[tex]p_1 = 765 mm Hg[/tex] is the initial pressure

[tex]V_1 = 7.0 L[/tex] is the initial volume

[tex]p_2 = 733 mm Hg[/tex] is the final pressure

[tex]V_2[/tex] is the final volume

Solving for V2, we find the new volume of the balloon:

[tex]V_2=\frac{p_1 V_1}{p_2}=\frac{(765)(7.0)}{733}=7.3 L[/tex]

nuhulawal20

At constant temperature, if the atmospheric pressure drops to the given values, the new volume of the balloon is 7.3L.

Given the data in the question;

  • Initial volume of balloon; [tex]V_1 = 7.0L[/tex]
  • Initial pressure; [tex]P_1 = 765mmHg = 1.00658atm[/tex]
  • Final pressure; [tex]P_2 = 733mmHg = 0.964474atm[/tex]
  • Final volume; [tex]V_2 = ?[/tex]

Boyle's law

Boyle's law states that the volume V of any given quantity of gas is inversely proportional to its pressure P as long as temperature remains constant.

Boyle's law is expressed as;

[tex]P_1V_1 = P_2V_2[/tex]

To determine the final volume, We substitute our given values into the expression above.

[tex]P_1V_1 = P_2V_2\\\\V_2 = \frac{P_1V_1}{P_2} \\\\V_2 = \frac{1.00658atm\ *\ 7.0L}{0.964474atm}\\\\V_2 = \frac{7.04606Latm}{0.964474atm}\\ \\V_2 = 7.3L[/tex]

Therefore, at constant temperature, if the atmospheric pressure drops to the given values, the new volume of the balloon is 7.3L.

Learn more about Boyle's law: brainly.com/question/1437490

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