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A 10.0 L balloon contains helium gas at a pressure of 660 mmHg . What is the final pressure, in millimeters of mercury, of the helium gas at each of the following volumes, if there is no change in temperature and amount of gas?

1 Answer

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Answer:

(a)= 264mmHg

(b)= 2000mmHg

(c)474.82mmHg

(d)= 511.63mmHg

Step-by-step explanation:

the question deals with boyles law, which states that the volume of a given mass of gas at constant temperature is inversely proportional to its pressure

V ∝ 1/P

P₁V₁ = P₂V₂

making V₂ as the subject of formular

P₂ = P₁V₁/ V₂

with a volume of 25.0L

P₂ = 660×10 / 25

= 264mmHg

with a volume of 3.30 L

P₂ = 660 × 10 / 3.30

= 2000mmHg

with a volume of 13900 mL

= 13.9L

P₂ =660× 10 / 13.9

474.82mmHg

with a volume of 12900 mL

P₂ =660×10 / 12.9

= 511.63mmHg

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