Partial Pressure in a Flask

Question

A rigid 9.45 L flask contains a mixture of 1.50 moles of O₂,

0.500 mole of N₂ and

sufficient Ar so that the partial pressure of Ar in the flask is 2.60 atm.

The temp is 80 degrees C.

1. total pressure in the flask?
2. Mole fraction of N₂ in the flask?
3. Density of mixture of gases in the flask?

Answer 1

A rigid 9.45 L flask contains a mixture of 1.50 moles of O₂,

0.500 mole of N₂ and

sufficient Ar so that the partial pressure of Ar in the flask is 2.60 atm.

The temp is 80 degrees C.

GIVEN

volume=9.45 l

moles of O2= 1.5 m O2

moles of N2= .5 moles of N2

pressure of Ar= 2.60 atm

temperature= 80 degree C 273 K +80 C =353 K

RELATIONSHIPS

PV=nRT

mole fraction = n of one type of gass / n+n+n+ of all gasses

1. total pressure in the flask?

First find the moles of Ar using PV=nRT 2.60 atm X 9.45L = n X .08206 L atm/moles K X 353K

solving for n = .848 moles

Next get the sum of all moles =2.848 moles

Lastly use PV=nRT and solve for total pressure=8.74 atm

2.Mole fraction of N₂ in the flask?

add all the moles of the 3 gasses together to get total moles of 2.848 moles

Then divide moles of N2 by total moles .5 moles N2/2,848 moles =.177 mole fraction

3.Density of mixture of gases in the flask?

Density= mass / volume

find the mass of each gas

1.5 moles of O2 X 32 g/mole=48 g O2

.5 moles N2 X 28 g/mole =14 g of N2

.848 moles of Ar X 40 g/mole = 34 g of Ar

total gas mass =96 g

density= 96g/9.45L = 10.16 g/L

Answer 2

There are several different approaches to this problem. Here is just one....

Let us first find the pressure in the absence of Ar

PV = nRT .. ideal gas law

(P)(9.45 L) = (2.00 mols)(0.0821 Latm/Kmol)(353K)

P = 6.13 atm = pressure provided by 1.50 mols O₂ + 0.500 mols N₂

Now add in the pressure from Ar to get the total pressure of the system...

Total pressure = 6.13 atm + 2.60 atm = 8.73 atm

Moles Ar present:

PV = nRT

n = PV/RT = (2.60)(9.45) / (0.0821)(353)

n = 0.848 moles Ar

Total moles present:

mols O₂ + mols N₂ + mols Ar = 1.50 + 0.500 + 0.848 = 2.85 moles

Mole fraction of N₂ = moles N₂ / total moles

0.500 mols / 2.85 mols = 0.175 = mole fraction of N₂

Density of gas = mass / volume

mass = g O₂ + g N₂ + g Ar

g O₂ = 1.50 mols O₂ x 32 g / mol = 48 g O₂

g N₂ = 0.500 mols N₂ x 29 g / mol = 14 g N₂

g Ar = 0.848 mols Ar x 40 g / mol = 34 g Ar

total mass = 96 g

Volume = 9.45 L

Density = 96 g / 9.45 L

Density = 10.2 g / L