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 mmHg
O2
. 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 partial pressure of helium gas.
The total pressure of the gas mixture is 795 mmHg, and the partial pressures of CO2, Ar, and O2 are given as 137 mmHg, 241 mmHg, and 184 mmHg respectively. Therefore, the partial pressure of helium gas can be calculated as:
Partial pressure of helium gas = Total pressure - partial pressures of CO2, Ar, and O2
Partial pressure of helium gas = 795 mmHg - 137 mmHg - 241 mmHg - 184 mmHg
Partial pressure of helium gas = 233 mmHg
Next, we need to convert the volume of the gas mixture to liters as it is given in mmHg. 1 L = 760 mmHg, so:
Volume of gas mixture in liters = 12.9 L
Now, we can use the ideal gas law to find the number of moles of helium gas present:
PV = nRT
(233 mmHg) (12.9 L) = n (0.0821 L.atm/mol.K) (279 K)
n = (233 x 12.9) / (0.0821 x 279)
n = 458.086
Finally, we can find the mass of helium gas present using the formula:
Mass = number of moles x molar mass
Molar mass of helium = 4 g/mol (approx)
Mass of helium gas present = 458.086 moles x 4 g/mol
Mass of helium gas present = 1832.344 g or 1.832 kg
Therefore, the mass of helium gas present in the 12.9 L sample of this mixture at 279 K is approximately 1.832 kg.