Chemistry: Deep sea divers use a mixture of helium and oxygen to breathe. assume that a diver is going to a depth of 150 feet where the total pressure is 4.42 atm. the partial pressure of oxygen at this depth is to be maintained at 0.20 atm, the same as at sea level. what must be the percent by volume of oxygen in the gas mixture?

The lighter components are able to rise higher in the column before they are cooled to their condensing temperature, allowing them to be removed at slightly higher levels.I hope this helps...

Deep sea divers use a mixture of helium and oxygen

Ответ:

meganwintergirl

03.01.2022

Chemistry

4.525% is the percentage by volume of oxygen in the gas mixture.

Explanation:

Total pressure of the mixture = p = 4.42 atm

Partial pressure of the oxygen = p_1=0.20 atm

Partial pressure of the helium = p_2

$p_1=p\times \chi_1$ (Dalton law of partial pressure)

$0.20 atm=4.42 atm\times \chi_1$

$\chi_1=\frac{0.20 atm}{4.42 atm}=0.04525$

$\chi_2=1-\chi_1=1-0.04525=0.95475$

chi_1+chi_2=1

n_1=0.04525 mol,n_2=0.95475 mol

According Avogadro law:

$Moles\propto Volume$ (At temperature and pressure)

Volume occupied by oxygen gas = V_1

Total moles of gases = n = 1 mol

Total Volume of the gases = V

$\frac{n_1}{V_1}=\frac{n}{V}$

$\frac{V_1}{V}=\frac{n_1}{n}=\frac{0.04525 mol}{1 mol}$

Percent by volume of oxygen in the gas mixture:

$\frac{V_1}{V}\times 100=\frac{0.04525 mol}{1 mol}\times 100=4.525\%$

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