If an alveolus with an initial volume of 3 ml of air with a total pressure of 810 mmhg decreases in volume to 1.7 ml, what would the new pressure be and in which direction would air flow? assume you are at sea level.

This question is answered by using Boyle's law which states that P1V1=P2V2. Therefore 3ml x 810mmHg = 1.7ml * P2. P2 in this case is 1429mmHg. Air flow would be out of the body into the environment, where atmospheric pressure is approximately 760 mmHg. Higher pressure flows into lower pressure areas.

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johanrusli johanrusli · Answer 2 · 2020-01-09T08:39:48+01:00

The new pressure would be about 1430 mmHg.

Air would flow out of the alveolus.

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Further explanation

The Ideal Gas Law and Work formula that needs to be recalled is:

[tex]\boxed {PV = nRT}[/tex]

[tex]\boxed { W = P \Delta V }[/tex]

where:

W = Work ( J )

P = Pressure ( Pa )

V = Volume ( m³ )

n = number of moles ( moles )

R = Gas Constant ( 8.314 J/mol K )

T = Absolute Temperature ( K )

Let us now tackle the problem !

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Given:

initial volume = V₁ = 3 ml

initial pressure = P₁ = 810 mmHg

final volume = V₂ = 1.7 ml

Asked:

final pressure = P₂ = ?

Solution:

We will use Boyle's Law to solve this problem as follows:

[tex]P_1V_1 = P_2V_2[/tex]

[tex]810 \times 3 = P_2 \times 1.7[/tex]

[tex]P_2 = 2430 \div 1.7[/tex]

[tex]P_2 \approx 1430 \texttt{ mmHg}[/tex]

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At sea level , the atmospheric pressure (Po) is about 760 mmHg.

Because P₂ > Po , air would flow out of the alveolus.

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Answer details

Grade: High School

Subject: Physics

Chapter: Pressure