Gas Exchange DSL
Gas Exchange basics
If we analyse air, we find it contains two main gases, around 21% O 2 and 79% N2,
these are approximations and there are tiny fractions of other gases present
including CO2, but this is so small it can be rounded for our purposes to 0%.
If barometric pressure (PB) is 100kPa (a fair assumption, when close to sea level),
each of the two main gases makes an individual contribution to this total pressure,
the pressure exerted by each is its partial pressure.
So PB = 100kPa = PO2 + PN2
PO2 = 21% of 100kPa, so PO2 = 21kPa
PN2 = 79% of 100kPa, so PN2 = 79kPa
Working in kilopascals is useful because unless you climb a mountain (and therefore
reduce PB) the partial pressures of gases in air are more or less equal to their
contribution as a percentage.
If we did climb a mountain and PB was reduced, then the partial pressures would
change, but the percentage composition would remain the same.
E.g. Climb a mountain and PB is now 75kPa
PO2 = 21% of 75kPa = 15.8kPa
PN2 = 79% of 75kPa = 59.3kPa
We often need to deal with gases dissolved in blood and transported in the body (i.e.
in solution). When in solution, gases don’t exert a pressure, so we solve this and
measure them through the following-
1. Boil water in a beaker, driving off any gas already in the water
2. Cover beaker so closed to atmosphere and allow to cool
3. Open beaker to atmosphere
When the surface of the water is exposed to the atmosphere, gases from the
atmosphere start to dissolve in the water. They do this down their concentration
gradient, so for O2, the gradient is 21kPa in air to 0 initially.
Given enough time, the gases will equilibrate so partial pressure of O2 in water is
equal to the atmosphere (i.e. 21kPa). So, the important point is when we say water
has a partial pressure of O2 equal to 21kPa, what we are really saying is that
whatever is dissolved in the water is equivalent to allowing it to equilibrate with
21kPa of gaseous O2.
Hypothetical Lung Scenario
Imagine a model lung with a volume of 1 litre, with no gas exchange (e.g. a balloon
with 1l of air in it). If we fill this lung with air, what is the total volume of O 2?
Gas Exchange basics
If we analyse air, we find it contains two main gases, around 21% O 2 and 79% N2,
these are approximations and there are tiny fractions of other gases present
including CO2, but this is so small it can be rounded for our purposes to 0%.
If barometric pressure (PB) is 100kPa (a fair assumption, when close to sea level),
each of the two main gases makes an individual contribution to this total pressure,
the pressure exerted by each is its partial pressure.
So PB = 100kPa = PO2 + PN2
PO2 = 21% of 100kPa, so PO2 = 21kPa
PN2 = 79% of 100kPa, so PN2 = 79kPa
Working in kilopascals is useful because unless you climb a mountain (and therefore
reduce PB) the partial pressures of gases in air are more or less equal to their
contribution as a percentage.
If we did climb a mountain and PB was reduced, then the partial pressures would
change, but the percentage composition would remain the same.
E.g. Climb a mountain and PB is now 75kPa
PO2 = 21% of 75kPa = 15.8kPa
PN2 = 79% of 75kPa = 59.3kPa
We often need to deal with gases dissolved in blood and transported in the body (i.e.
in solution). When in solution, gases don’t exert a pressure, so we solve this and
measure them through the following-
1. Boil water in a beaker, driving off any gas already in the water
2. Cover beaker so closed to atmosphere and allow to cool
3. Open beaker to atmosphere
When the surface of the water is exposed to the atmosphere, gases from the
atmosphere start to dissolve in the water. They do this down their concentration
gradient, so for O2, the gradient is 21kPa in air to 0 initially.
Given enough time, the gases will equilibrate so partial pressure of O2 in water is
equal to the atmosphere (i.e. 21kPa). So, the important point is when we say water
has a partial pressure of O2 equal to 21kPa, what we are really saying is that
whatever is dissolved in the water is equivalent to allowing it to equilibrate with
21kPa of gaseous O2.
Hypothetical Lung Scenario
Imagine a model lung with a volume of 1 litre, with no gas exchange (e.g. a balloon
with 1l of air in it). If we fill this lung with air, what is the total volume of O 2?
