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 h elium gas at each of the following volumes, if there is no change in temperature and amount of gas?
1 answer:
Answer:
(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
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