Question

Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is 21 ∘C∘C. Express the pressures in atmospheres to three significant digits separated by commas.

Answer 1

The question is incomplete, complete question is ;

A deep-sea diver uses a gas cylinder with a volume of 10.0 L and a content of 51.8 g of $O_2$ and 33.1 g of He. Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is 21°C.Express the pressures in atmospheres to three significant digits separated by commas.

Answer:

Partial pressure of the oxygen gas is 3.91 atm.

Partial pressure of the helium gas is 20.0 atm

Total pressure of the gases is 24.0 atm

Explanation:

Moles of oxygen gas = $n_1=\frac{51.8}{32 g/mol}=1.619 mol$

Moles of helium gas = $n_2=\frac{33.1 g}{4 g/mol}=8.275 mol$

Total moles of gas = $n_1+n_2=(1.619 +8.275 ) mole=9.894 mol$

Volume of the cylinder = V = 10.0 L

Total pressure in the cylinder = P = ?

Temperature of the gas in cylinder = T = 21°C = 21 + 273 K = 294 K

PV = nRT ( ideal gas equation )

$P=\frac{nRT}{V}$

$=\frac{9.894 mol\times 0.0821 atm L/mol K\times 294 K}{10.0 L}$

P = 23.88 atm ≈ 23.9

Partial pressure of the individual gas will be determined by the help of Dalton's law:

partial pressure = Total pressure × mole fraction of gas

Partial pressure of the oxygen gas

$p_{1}=P\times \chi_{1}=P\times \frac{n_1}{n_1+n_2}$

$p_1=23.88 atm\times \frac{1.619 mol}{9.894 mol}=3.91 atm$

Partial pressure of the helium gas

$p_{2}=P\times \chi_{2}=P\times \frac{n_2}{n_1+n_2}$

$p_2=23.88 atm\times \frac{8.275 mol}{9.894 mol}=19.97 atm\approx 20.0 atm$