Question

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?

Answer 1

Answer:

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:

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