Questions on Partial Pressures

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DATA

EQUILIBRIUM VAPOUR PRESSURE FOR WATER.

 Temp / K P / mmHg P / kPa 294 18.7 2.49 298 23.8 3.17 300 26.8 3.57

1

(a) A mixture of hydrogen (1.01 g) and chlorine (17.73 g) in a container at 300 K has a total gas pressure of 98.8 kPa. What is the partial pressure of hydrogen in the mixture?

(b) The mixture of gases is then transferred into another container which has a volume two-thirds that of the original container, the temperature remaining unchanged. How many molecules are present under these new conditions? (The increased pressure does not cause any chemical reaction).

(c) The gas mixture in the new container is exploded by continuously passing a spark, and the temperature is subsequently returned to 300 K. What is the total gas pressure now? 2

A container of volume 10.00 litre is evacuated and held at a constant temperature. Into it is injected at the same temperature

• 2.00 litre of oxygen gas at an original pressure of 202.6 kPa and
• 3.00 litre of neon gas at an original pressure of 303.9 kPa.

What pressure is produced inside the container? 3

A sample of nitrogen gas is bubbled through water at 298 K and the volume collected is 250 mL. The total pressure of the gas, which is saturated with water vapour, is found to be 98.8 kPa at 298 K. What amount (in mole) of nitrogen is in the sample? 4

A bulb is found to weigh 5.76 gram more when filled with air at 298 K and 1.00 atmosphere pressure than when it is evacuated. What is the volume of the bulb? (Consider air to be 80% nitrogen and 20% oxygen by volume). 5

(a) A mixture of hydrogen (4.04 g) and oxygen (32.0 g) in a container at 300 K has a total gas pressure of 0.906 atmosphere. What is the partial pressure of hydrogen in the mixture?

(b) The gas mixture in the container is exploded by continuously passing a spark. The temperature is subsequently returned to 300 K and droplets of liquid are observed on the inside walls of the container. What is the pressure inside the container now? 6

Nitrogen gas is collected over water in a measuring cylinder, the level of water being the same inside and outside the cylinder. The volume of nitrogen collected was 120 mL, the temperature was read as 21 oC. Barometric pressure was observed to be 750.1 mmHg. Calculate the volume of dry nitrogen obtained at 298 K and 101.3 kPa. 7

The volume of hydrogen collected over water following a reaction is 96.5 mL at 300 K and 103.0 kPa. Calculate the volume of dry hydrogen at 298 K and 101.3 kPa. 8

Nitrogen gas is produced by heating ammonium nitrite. The reaction equation is

NH4NO2 2H2O + N2

If the products are collected over water at 298 K and 101.3 kPa, the volume observed is 1.043 L. What mass of dry nitrogen was obtained? 9

Carbon dioxide (1.100 g) was introduced into a 1.00 L flask which contained some pure oxygen gas. The flask was warmed to 373 K and the pressure was then found to be 608 mmHg. If CO2 and O2 were the only gases present, what was the mass of oxygen in the flask? 10

At 338 K, pure PCl5 gas is present in a flask at a pressure of 26.7 kPa. At 473 K this is completely dissociated into PCl3 gas and Cl2 gas. Calculate the pressure in the flask at 473 K. Preamble

When two or more gases are introduced into the same container, each gas individually expands to uniformly occupy that container. Thus each gas in the mixture has the same volume but, depending on how many moles of each is present, exerts a different pressure, called its partial pressure. Dalton's Law of Partial Pressures states that in a mixture of gases, the total pressure is the sum of the individual pressures of each gas present in the mixture - i.e. the sum of the partial pressures. When a gas is collected over a liquid such as water, the gas is mixed with vapour from the liquid (e.g. water) and so is a mixture of the gas + water vapour. Then the total pressure = partial pressure of the gas + partial pressure of water vapour. The partial pressure of water vapour depends on the temperature alone and is tabulated as the Equilibrium Vapour Pressure (equilibrium vapour pressure) of water. See the data given for equilibrium vapour pressure values for water at the beginning of the questions section.

1

Note preamble at start of answer section (above).

(a) Moles of H2 = mass/molar mass = 1.02/2.02 = 0.500 mol
Moles of Cl2 = mass/molar mass = 17.73/70.90 = 0.250 mol
Total moles = 0.500 + 0.250 = 0.750 mol.
Total pressure = 98.8 kPa.

Partial pressure of each gas is proportional to its mole fraction in the mixture.

Therefore partial pressure of H2 = (0.500/0.750) x 98.8 = 65.9 kPa.

(b) The number of molecules does not change, only the volume (reduced)
and therefore the partial pressure of each gas (increased).
Total moles of gas = 0.750 mol

Therefore number of molecules = moles x NA
= 0.750 x 6.022 x 1023 = 4.52 x 1023 molecules.

(c) The reaction equation is
H2(g) + Cl2(g)
2HCl(g)
1 mol   1 mol        2 mol

From the stoichiometry, 2 moles of reactants produce 2 moles of product,
so the total moles of gas remains unchanged and also the temperature is unchanged.
Thus Boyle's Law can be used to calculate the new pressure.

P1V1 = P2V2
As V2 = (2/3)/V1,
98.8 x V1 = P2 x (2/3)V1
P2 = 148 kPa. 2

Each gas is undergoing an expansion when introduced into the 10.00 L container, so each will exert a partial pressure which is less than its original pressure. As temperature and number of moles of each gas in unchanged, Boyle's Law can be used to calculate the new (partial) pressure of each gas in the mixture.

P1V1 = P2V2

Oxygen:
202.6 x 2.00 = P(O2) x 10.00
P(O2) = 40.5 kPa

Neon:
303.9 x 3.00 = P(Ne) x 10.00
P(Ne) = 91.2 kPa

Total pressure = sum of partial pressures
= 40.5 + 91.2 = 131.7 kPa 3

Gases collected over water are saturated with water vapour (see preamble).
Therefore Ptotal = Pgas + Pwater

The partial pressure of water in the mixture, Pwater, is the equilibrium vapour pressure of water at the temperature specified.

At 298 K, from the data at the beginning of the questions section,
Pwater = 3.17 kPa.

Therefore, P(N2) = 98.8 - 3.17 = 95.6 kPa

Using the Ideal Gas Equation, the number of moles of N2 can be calculated.
PV = nRT
n = PV/(RT) = (95.6 x 0.250)/(8.314 x 298)

[Note that V must be in L and P in kPa for R to be 8.314 J K-1 mol-1.]
n = 9.65 x 10-3 mol. 4

Air is 20% oxygen and 80% nitrogen by volume, and therefore 20% oxygen and 80% nitrogen by moles.

[Gases expand to uniformly fill their container, so the number of moles of any gas in a given volume is directly proportional to the volume, provided the temperature is unchanged.]

Let total number of moles present = n.
= moles(O2) + moles(N2)
Moles(O2) = 0.20 x total moles = 0.20 x n mol.
Moles(N2) = 0.80 x total moles = 0.80 x n mol.

Total mass = mass(O2) + mass(N2)
Therefore 5.76 = 0.20 x n x 32.00 + 0.80 x n x 28.02
= 6.400 n + 22.42 n = 28.8 n
n = 0.199 mol.

Use the ideal gas equation to calculate V:
PV = nRT
V = nRT/P = (0.199 x 8.314 x 298)/101.3
= 4.89 L

[Note that V will be in L if P is in kPa and R = 8.314 J K-1 mol-1.] 5

See preamble (above).

(a) Moles of H2 = 4.04/2.02 = 2.00 mol
Moles of O2 = 32.0/32.00 = 1.00 mol.

Therefore total moles = 3.00 mol.

The mole fraction of H2 in the mixture = 2.00/3.00 = 0.667

The partial pressure of hydrogen = mole fraction of hydrogen x total pressure
= 0.667 x 0.906 = 0.604 atmosphere or
0.604 x 101.3 kPa = 61.2 kPa.

(b) The reaction equation is 2H2(g) + O2(g) 2H2O(l)

From the stoichiometry of this reaction, 2 mol of H2 reacts exactly with 1 mol of O2,
so there is complete reaction of the two gases in the mixture with no hydrogen or oxygen remaining.

The product of the reaction is water which is a liquid at 300 K.
It is in equilibrium with its vapour and exerts just its equilibrium vapour pressure in the container.

From the data table, equilibrium vapour pressure of water at 300 K = 3.57 kPa
or 3.57/101.3 atm = 0.0352 atm. 6

Gases collected over water are saturated with water vapour (see preamble).
Therefore Ptotal = Pgas + Pwater

The partial pressure of water in the mixture, Pwater, is the equilibrium vapour pressure of water at the temperature specified.

At 294 K, from the data at the beginning of the questions section,
Pwater = 2.49 kPa.

The pressure inside the container is the same as that outside = 750.1 mmHg
= (750.1/760.0) x 101.3 kPa = 100.0 kPa.
100.0 = Pnitrogen + 2.49

Therefore, Pnitrogen = 100.0 - 2.49 = 97.5 kPa.
The combined Boyle's and Charles's Laws can then be used to calculate
the volume of dry nitrogen at 298 K and 101.3 kPa.

P1V1/T1 = P2V2/T2
97.5 x 120/294 = 101.3 x V2/298
V2 = 97.5 x 120 x 298/(101.3 x 294)
= 117 mL. 7

Using the same reasoning as in the previous question,
Ptotal = Phydrogen + Pwater

The equilibrium vapour pressure for water at 300 K = 3.57 kPa

Therefore 103.0 = Phydrogen + 3.57
Phydrogen = 99.4 kPa.

The combined Boyle's and Charles's Laws can then be used to calculate
the volume of dry hydrogen at 298 K and 101.3 kPa.

P1V1/T1 = P2V2/T2
99.4 x 96.5/300 = 101.3 x V2/298
V2 = 99.4 x 96.5 x 298/(101.3 x 300)
= 94.1 mL. 8

Using the same reasoning as in the previous question,
Ptotal = Pnitrogen + Pwater

The equilibrirum vapour pressure for water at 298 K = 3.17 kPa

Therefore 101.3 = Pnitrogen + 3.17
Pnitrogen = 98.1 kPa.

Using the Ideal Gas Equation, the number of moles of nitrogen can be calculated.
PV = nRT
or
n = PV/(RT)

n = (98.1 x 1.043)/(8.314 x 298)
= 0.04130 mol.

Therefore mass of nitrogen = moles x molar mass
= 0.04130 x 28.02 = 1.16 g.

[Note that the value of R used requires pressure to be expressed in kPa and volume to be in L.] 9

Moles of CO2 = mass/molar mass
= 1.100/44.01 = 0.02499 mol.

The partial pressure of carbon dioxide can then be calculated from the Ideal Gas Equation.
V = 1.00 L
T = 373 K

Pcarbon dioxide = nRT/V
= 0.02499 x 8.314 x 373/1.00
= 77.5 kPa

The total pressure Ptotal = Poxygen + Pcarbon dioxide

Ptotal = 608 mmHg = (608/760) x 101.3 kPa
= 81.0 kPa

Therefore 81.0 = Poxygen + 77.5
Poxygen = 3.5 kPa

From the ideal gas equation, the moles of O2 can be deduced.

n = PV/(RT) = 3.5 x 1.00/(8.314 x 373)
= 1.13 x 10-3 mol.

Mass of oxygen = moles x molar mass
= 1.13 x 10-3 x 32.00 g
= 0.0361 g. 10

No volume is specified, so take a convenient value, say V = 1.00 L.

At 338 K, moles of PCl5 gas = PV/(RT)
= (26.7 x 1.00)/(8.314 x 338)
= 9.50 x 10-3 mol.

The equation for the reaction that occurs at 473 K is

PCl5(g) PCl3(g) + Cl2(g)

which shows that for each mole of reactant, two moles in total of products are formed.

Therefore at 473 K, moles of gas present = 2 x 9.50 x 10-3
= 0.0190 mol.

Then, using the Ideal Gas Equation, the new pressure can be deduced.
P = nRT/V
= 0.0190 x 8.314 x 473/1.00
= 74.7 kPa.

[An alternative method would be to take 1.00 mole of the reactant and calculate the volume it would occupy at the specified T and P. Then at the higher temperature, the volume would be unchanged but the number of moles is doubled to 2.00 moles. Thence the new P could be calculated.]  