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?

Question

Nero

Physics

15 April, 06:22

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?

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Answers (1)

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Camilla · Answer 1

(a) = 264mmHg

(b) = 2000mmHg

(c) 474.82mmHg

(d) = 511.63mmHg

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