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  1. Problems
  2. /A deep diver breathes a mixture of 80 % helium and 20 % oxygen by moles, at a total pressure of 5.0…
HardChemistryHigh School (11–12)Solutions › Solubility, Vapour Pressure and Raoult's LawNumericalAPBoard Exam

A deep diver breathes a mixture of 80 % helium and 20 % oxygen by moles, at a total pressure of 5.0…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A deep diver breathes a mixture of 80 %80\ \%80 % helium and 20 %20\ \%20 % oxygen by moles, at a total pressure of 5.0 bar5.0\ \text{bar}5.0 bar. At 293 K293\ \text{K}293 K the Henry's law constants in water are 144.97 kbar144.97\ \text{kbar}144.97 kbar for helium and 34.86 kbar34.86\ \text{kbar}34.86 kbar for oxygen. How many millimoles of each gas dissolve in 1.00 kg1.00\ \text{kg}1.00 kg of water? Use 55.56 mol55.56\ \text{mol}55.56 mol of water in 1 kg1\ \text{kg}1 kg.

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Work it out on paper, then type your number. The answer itself stays hidden.

mmol
mmol
Need a reminder? The key ideaShowHide

Henry's law: p=KH xp = K_H\, xp=KH​x, where xxx is the mole fraction of the gas in the solution. KHK_HKH​ has the unit of pressure, and a larger KHK_HKH​ means a smaller solubility. In a weak water solution, the moles of gas in 1 kg1\ \text{kg}1 kg of water are x×55.6 molx \times 55.6\ \text{mol}x×55.6 mol. Heating lowers the solubility of a gas.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. Partial pressures. Helium: 80100×5.0=4.0 bar\dfrac{80}{100} \times 5.0 = 4.0\ \text{bar}10080​×5.0=4.0 bar. Oxygen: 20100×5.0=1.0 bar\dfrac{20}{100} \times 5.0 = 1.0\ \text{bar}10020​×5.0=1.0 bar.

Step 2. Helium. x=4.0144 970=2.76×10−5x = \dfrac{4.0}{144\,970} = 2.76 \times 10^{-5}x=1449704.0​=2.76×10−5. In 1 kg1\ \text{kg}1 kg of water that is 2.76×10−5×55.56=1.53×10−3 mol=1.53 mmol2.76 \times 10^{-5} \times 55.56 = 1.53 \times 10^{-3}\ \text{mol} = 1.53\ \text{mmol}2.76×10−5×55.56=1.53×10−3 mol=1.53 mmol.

Step 3. Oxygen. x=1.034 860=2.87×10−5x = \dfrac{1.0}{34\,860} = 2.87 \times 10^{-5}x=348601.0​=2.87×10−5, which is 2.87×10−5×55.56=1.59 mmol2.87 \times 10^{-5} \times 55.56 = 1.59\ \text{mmol}2.87×10−5×55.56=1.59 mmol.

Answer: 1.53 mmol1.53\ \text{mmol}1.53 mmol of helium and 1.59 mmol1.59\ \text{mmol}1.59 mmol of oxygen.

What does it show? The helium has four times the partial pressure of the oxygen, yet about the same amount dissolves. Helium is much less soluble, so it adds little dissolved gas to the blood.

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  1. Gases in liquids and Henry's lawConcentration and Solubility

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