Respuesta :
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 [tex]O_2[/tex] 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 = [tex]n_1=\frac{51.8}{32 g/mol}=1.619 mol[/tex]
Moles of helium gas = [tex]n_2=\frac{33.1 g}{4 g/mol}=8.275 mol[/tex]
Total moles of gas = [tex]n_1+n_2=(1.619 +8.275 ) mole=9.894 mol[/tex]
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 )
[tex]P=\frac{nRT}{V}[/tex]
[tex]=\frac{9.894 mol\times 0.0821 atm L/mol K\times 294 K}{10.0 L}[/tex]
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
[tex]p_{1}=P\times \chi_{1}=P\times \frac{n_1}{n_1+n_2}[/tex]
[tex]p_1=23.88 atm\times \frac{1.619 mol}{9.894 mol}=3.91 atm[/tex]
Partial pressure of the helium gas
[tex]p_{2}=P\times \chi_{2}=P\times \frac{n_2}{n_1+n_2}[/tex]
[tex]p_2=23.88 atm\times \frac{8.275 mol}{9.894 mol}=19.97 atm\approx 20.0 atm[/tex]
