A gas mixture with a total pressure of 795 mmHg contains each of the following gases at the indicated partial pressures: 137 mmHg
CO2, 241 mmHg Ar, and 184 mmg 02. The mixture also contains helium gas.
What mass of helium gas is present in a 12.9-L sample of this mixture at 279 K ?
To find the mass of helium gas present in the mixture, we first need to find the moles of helium using the ideal gas law:
PV = nRT
where:
P = total pressure = 795 mmHg = 795/760 atm = 1.047 atm
V = volume = 12.9 L
n = moles (unknown)
R = gas constant = 0.0821 L.atm/mol.K
T = temperature = 279 K
First, we need to find the moles of each gas present in the mixture using their partial pressures:
For CO2:
nCO2 = (137 mmHg / 760 mmHg/atm) * 12.9 L / (0.0821 L.atm/mol.K * 279 K) = 0.805 moles
For Ar:
nAr = (241 mmHg / 760 mmHg/atm) * 12.9 L / (0.0821 L.atm/mol.K * 279 K) = 1.42 moles
For O2:
nO2 = (184 mmHg / 760 mmHg/atm) * 12.9 L / (0.0821 L.atm/mol.K * 279 K) = 1.09 moles
Now, we can find the total moles of gas present in the mixture:
ntotal = nCO2 + nAr + nO2 + nHe (helium)
1.047 atm * 12.9 L = (0.805 + 1.42 + 1.09 + nHe) * 0.0821 L.atm/mol.K * 279 K
nHe = (1.047 * 12.9) / (0.0821 * 279) - (0.805 + 1.42 + 1.09)
nHe = 0.063 moles
Finally, we can convert moles of helium gas to grams using the molar mass of helium:
Molar mass of helium (He) = 4.0026 g/mol
mass = nHe * molar mass
mass = 0.063 moles * 4.0026 g/mol
mass ≈ 0.252 grams
Therefore, the mass of helium gas present in the 12.9-L sample of the mixture is approximately 0.252 grams.