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Old 12-16-2013
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Lungs Surfactant and Surface Area

So I'm having a bit of trouble piecing together the relationship between surfactant, surface area, and lung expansion.

According to Costanzo:
"During inflation of the lung (inspiration limb), surfactant, which is newly produced by type II alveolar cells, enters the liquid layer lining the alveoli and breaks up these intermolecular forces to reduce surface tension. In the initial part of the inspiration curve, at lowest lung volumes, the lung surface area is increasing faster than surfactant can be added to the liquid layer; thus, surfactant density is low, surface tension is high, compliance is low, and the curve is flat. As inflation proceeds, the surfactant density increases, which decreases surface tension, increases compliance, and increases the slope of the curve."

However, according to Wiki:
"As the alveoli increase in size, the surfactant becomes more spread out over the surface of the liquid. This increases surface tension effectively slowing the rate of expansion of the alveoli. This also helps all alveoli in the lungs expand at the same rate, as one that expands more quickly will experience a large rise in surface tension slowing its rate of expansion. It also means the rate of shrinking is more regular, as if one reduces in size more quickly the surface tension will reduce more, so other alveoli can contract more easily than it can. Surfactants reduce surface tension more readily when the alveoli is smaller because surfactants are more concentrated"

It seems that Costanzo is saying that as the aveolus expands, surfactant density increases (thus increasing compliance), but wiki says that as the aveolus expands the surfactant density decreases (surfactant more spread out). Also, if surfactant is added as the lung expands as Costanzo suggests, why does the surface tension increase with relative area???

THank you!
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Physiology-, Respiratory-

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