Question

The vapor pressure of water at blood temperature is 47 Torr. What is the partial pressure of dry air in our lungs when the total pressure is 760 Torr? (1Torr = 133.3 Pa )

Answer 1

Solution:

Find the partial pressure of dry air in our lungs.

By the formula PB=xB⋅PT

We have PT=760 Torr (total pressure)

• Find xB

By xB+xA=1⇒xB=1−xA

But PA=xA⋅PT⇒xA=PAPT

Since PA=47 Torr (pressure of water at blood temperature)

Then xA=47760=0.06

So \(x_B = 1 - 0.06 = 0.94

We get PB=0.94×760=714.4 Torr

We know 1 Torr =133.3 Pa

Therefore, \(P_B = 714.4 \times 133.3 = 95229.52 \mathrm{~Pa}

Therefore, the partial pressure of dry air in our lungs is 95229.52 Pa

• In other words:

From the formula Ptotal=PA+PB (Both are gases)

Therefore, PB=Ptotal−PA

= 760 Torr - 47 Torr

= 713 Torr

But 1 Torr =133.3 Pa

Therefore, \(P_B = 713 \times 133.3 = 95042.9 \mathrm{~Pa}

Therefore, the partial pressure of dry air in our lungs is 95229.52 Pa