Question

1. a mountain climber has partial pressures of o₂ and co₂ in her lungs of 4.7 kpa and 1.0 kpa respectively. the atmospheric pressure on the mountain is 33.7 kpa. assume the lungs are saturated with water vapour at a temperature of 310 k (the vapour pressure of water at this temperature is 6.3 kpa).

a) assuming the lungs only contain n₂, o₂, co₂ and h₂o gases, what is the partial pressure of n₂ in bar?
b) assuming the volume of the lungs is 6.00 l, how many molecules of o₂ are in her lungs?

Explanation:

Step1: Calculate partial pressure of \(N_2\)

According to Dalton's law of partial pressures \(P_{total}=P_{O_2}+P_{CO_2}+P_{H_2O}+P_{N_2}\).
We know \(P_{total} = 33.7\ kPa\), \(P_{O_2}=4.7\ kPa\), \(P_{CO_2}=1.0\ kPa\), \(P_{H_2O}=6.3\ kPa\).
So \(P_{N_2}=P_{total}-(P_{O_2}+P_{CO_2}+P_{H_2O})\)
\(P_{N_2}=33.7-(4.7 + 1.0+6.3)\)
\(P_{N_2}=33.7 - 12\)
\(P_{N_2}=21.7\ kPa\)
Since \(1\ bar = 100\ kPa\), then \(P_{N_2}=\frac{21.7}{100}=0.217\ bar\)

Step2: Calculate moles of \(O_2\) using ideal gas law

The ideal gas law is \(PV = nRT\).
We have \(P = 4.7\ kPa=4700\ Pa\), \(V = 6.00\ L = 6.00\times10^{- 3}\ m^3\), \(R = 8.314\ J/(mol\cdot K)\), \(T = 310\ K\)
From \(n=\frac{PV}{RT}\), \(n=\frac{4700\times6.00\times10^{-3}}{8.314\times310}\)
\(n=\frac{28.2}{2577.34}\approx0.01094\ mol\)

Step3: Calculate number of \(O_2\) molecules

Using \(N = nN_A\), where \(N_A=6.022\times10^{23}\ mol^{-1}\)
\(N=0.01094\times6.022\times10^{23}\)
\(N = 6.6\times10^{21}\)

Answer:

a) \(0.217\ bar\)
b) \(6.6\times10^{21}\)