In the gas-exchange diagram (alveolus $p_{a}O_2 = 13.0\,\text{kPa}$; capillary A $= 11.7\,\text{kPa}$, B $= 11.4\,\text{kPa}$, C $= 5.7\,\text{kPa}$), explain how the information shows that A is the venous end of the pulmonary capillary.

What the diagram data is telling you

You use the partial pressure of oxygen ($p_{a}O_2$) to work out where blood has gained oxygen and where it is still deoxygenated.

Following the $p_{a}O_2$ along the capillary

• The alveolus has $p_{a}O_2 = 13.0\,\text{kPa}$.
• Blood in the capillary is lower than this at all three points, so oxygen will diffuse from alveolus to blood.
• The blood values increase from C to B to A:
• C: $5.7\,\text{kPa}$ (lowest, most deoxygenated)
• B: $11.4\,\text{kPa}$
• A: $11.7\,\text{kPa}$ (highest, most oxygenated)

Why that identifies A as the venous end

At the venous end of a pulmonary capillary, blood is about to drain into the pulmonary vein, so it should have a high $p_{a}O_2$ after gas exchange. Since A has the highest $p_{a}O_2$ and is closest to the alveolar value, A matches the end where oxygenated blood leaves the capillary into a vein.