Questions on Partial Pressures
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Q: 1 2 3
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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 twothirds 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 ^{o}C. 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
NH_{4}NO_{2} → 2H_{2}O
+ N_{2}
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 CO_{2} and O_{2} were the only gases present,
what was the mass of oxygen in the flask?


10 
At 338 K, pure PCl_{5} gas is present
in a flask at a pressure of 26.7 kPa. At 473 K this is completely
dissociated into PCl_{3} gas and Cl_{2} gas. Calculate
the pressure in the flask at 473 K.


Partial Pressures
(Answers)


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 H_{2} = mass/molar mass = 1.02/2.02 = 0.500 mol
Moles of Cl_{2} = 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 H_{2} = (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 N_{A}
= 0.750 x 6.022 x 10^{23} = 4.52 x 10^{23} molecules.
(c)
The reaction equation is
H_{2}(g) + Cl_{2}(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.
P_{1}V_{1}
= P_{2}V_{2}
As V_{2} = (2/3)/V_{1},
98.8 x V_{1} = P_{2} x (2/3)V_{1}
P_{2} = 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.
P_{1}V_{1}
= P_{2}V_{2}
Oxygen:
202.6 x 2.00 = P(O_{2}) x 10.00
P(O_{2}) = 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 P_{total} = P_{gas} + P_{water}
The partial pressure of water in the mixture,
P_{water}, is the equilibrium vapour pressure of water at the temperature
specified.
At 298 K, from the data at the beginning of
the questions section,
P_{water} = 3.17 kPa.
Therefore, P(N_{2}) = 98.8  3.17 =
95.6 kPa
Using the Ideal Gas Equation, the number of
moles of N_{2} 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(O_{2}) + moles(N_{2})
Moles(O_{2}) = 0.20 x total moles = 0.20 x n mol.
Moles(N_{2}) = 0.80 x total moles = 0.80 x n mol.
Total mass = mass(O_{2})
+ mass(N_{2})
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 H_{2} = 4.04/2.02 = 2.00 mol
Moles of O_{2} = 32.0/32.00 = 1.00 mol.
Therefore total moles
= 3.00 mol.
The mole fraction of
H_{2} 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 2H_{2}(g) + O_{2}(g) →
2H_{2}O(l)
From the stoichiometry
of this reaction, 2 mol of H_{2} reacts exactly with 1 mol
of O_{2},
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 P_{total} = P_{gas} + P_{water}
The partial pressure of water in the mixture, P_{water},
is the equilibrium vapour pressure of water at the temperature specified.
At 294 K, from the data at the beginning of the questions section,
P_{water} = 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 = P_{nitrogen} + 2.49
Therefore, P_{nitrogen} = 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.
P_{1}V_{1}/T_{1} = P_{2}V_{2}/T_{2}
97.5 x 120/294 = 101.3 x V_{2}/298
V2 = 97.5 x 120 x 298/(101.3 x 294)
= 117 mL.


7 
Using the same reasoning
as in the previous question,
P_{total} = P_{hydrogen} + P_{water}
The equilibrium
vapour pressure for water at 300 K = 3.57 kPa
Therefore 103.0 = P_{hydrogen}
+ 3.57
P_{hydrogen} = 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.
P_{1}V_{1}/T_{1}
= P_{2}V_{2}/T_{2}
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,
P_{total} = P_{nitrogen} + P_{water}
The equilibrirum vapour
pressure for water at 298 K = 3.17 kPa
Therefore 101.3 = P_{nitrogen}
+ 3.17
P_{nitrogen} = 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 CO_{2}
= 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
P_{carbon dioxide}
= nRT/V
= 0.02499 x 8.314 x 373/1.00
= 77.5 kPa
The total pressure P_{total}
= P_{oxygen} + P_{carbon dioxide}
P_{total} = 608
mmHg = (608/760) x 101.3 kPa
= 81.0 kPa
Therefore 81.0 = Poxygen
+ 77.5
P_{oxygen} = 3.5 kPa
From the ideal gas equation,
the moles of O_{2} 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 PCl_{5}
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
PCl_{5}(g) →
PCl_{3}(g) + Cl_{2}(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.]

