AP GAS LAWS
A. MEASURING GAS PRESSURE
Definition: PRESSURE is force per unit area.
That is: Pressure =
force
area
The unit of pressure is a derived unit called the "pascal".
1 Pa = 1 N/m2 , where "N" (the "Newton") is the SI unit of force.

Gas pressure is normally measured in kilopascals (kPa), atmospheres (atm) or millimetres of mercury
(torr).
ONE STANDARD ATMOSPHERE = 1 atm = 101.325 kPa = 760 torr
The following three types of pressure–measuring devices are of interest.
a) The Barometer
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Whe n th e do wn wa rd pu sh o f Hg e qual s
the pus h of the Hg in to the tube by the
air, th e Hg l evel d oes not move.
vacu um
mercury pu shes down
PAI R
A p ressu re of 1 a tm can supp ort a
co lumn of mercury which is 760 mm
hig h.
air press ure p ushe s
mercury ba ck up
The refore:
1 a tm = 760 mm Hg
1 torr = 1 mm Hg
mercury
Uses: A barometer CANNOT be used to measure the pressure of a gas. It can ONLY be used to
measure atmospheric pressure.
b) The Closed–End Manometer
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sto pco ck (gas inl et val ve )
vacu um
sam ple bulb
gas
vacu um
put gas into bu lb
PGAS
mercury
The closed end manometer gives a DIRECT READING of gas pressure in the bulb, as shown below.
PMEAS = PGAS – PVAC = PGAS – 0 = PGAS
c) The Open–End Manometer
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Hebden : Chemistry AP
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op en e nd
74 5 mm Hg
vacuum
70 0 mm Hg
atmosp heric
pre ssure
ad d g as i nto bulb
74 5 mm Hg
PMEAS
= 745 – 70 0
= 45 m m Hg
ad d g as i nto bulb
74 5 mm Hg
74 5 mm Hg
14 65 mm Hg
74 5 mm Hg
ad d g as i nto bulb
PMEAS = 720 mm Hg
PMEAS = 0
As seen above, the open end manometer is read relative to the atmospheric pressure.
PMEAS = | PATM – PGAS |
EXERCISES:
1. A closed–end manometer has a mercury level of 25.4 cm on the bulb side and 58.7 cm on the
arm side. If the atmospheric pressure is 745 mm Hg, what is the pressure of the gas in the bulb?
2. A open–end manometer has a mercury level of 65.2 cm on the bulb side and 37.4 cm on the arm
side. If the atmospheric pressure is 738 mm Hg, what is the pressure of the gas in the bulb?
3. A open–end manometer has a mercury level of 12.9 cm on the bulb side and 81.4 cm on the arm
side. If the atmospheric pressure is 764 mm Hg, what is the pressure in the bulb?
4. Express a pressure of 723 mm Hg in
(a) torr
(b) kilopascals
5. Express a pressure of 55.0 kPa in
(a) torr
(b) atmospheres
6. Express a pressure of 0.0250 atm in
(a) torr
(b) kilopascals
(c) atmospheres
7. What is the maximum height that a vacuum pump attached to a tall glass column immersed in
water can cause the water to rise? Assume that the atmospheric pressure is 1 atm, water has a
density of 1.00 g/mL and that mercury has a density of 13.6 g/mL.
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B. AVOGADRO'S HYPOTHESIS (A Review of Chemistry 11 Material)
In 1808, the French chemist J. L. Gay–Lussac experimentally showed that
Gases combine completely in volume ratios that are whole numbers.
This statement is known as “Gay–Lussac's Law”. For example, he found that
1 L H2 + 1 L Cl2
2 L HCl .
Based on Gay–Lussac's Law, in 1811 the Italian chemist Amedeo Avogadro stated –
Avogadro's Principle or Avogadro's Hypothesis:
Equal volumes of gases, measured at the same temperature and pressure,
contain the same numbers of moles (or molecules).
This implies that the VOLUMES occupied by ANY TWO GASES at the same conditions MUST BE THE
SAME, and suggests that a STANDARD MOLAR VOLUME exists (22.4 L).
Since equal volumes must contain equal numbers of moles, then if one gas sample is twice as heavy as
another, this occurs because molecules of the first gas are twice as heavy as the other gas molecules.
Avogadro's Hypothesis helped find how many atoms were actually present in a molecule, allowing chemists to
calculate the atomic masses of elements more accurately.
Experimentally, the following data is found.
1.00 L of NH3(g) reacts with exactly 1.00 L of HCl(g).
1.00 L of NH3(g) has a mass of 0.759 g
1.00 L of HCl(g) has a mass of 1.63 g .
Therefore
and:
mass of HCl(g)
1.63 g
=
= 2.15
mass of NH 3 (g)
0.759 g
mass of HCl(g) = 2.15 x mass of NH3(g).
According
to Avogadro's

 Hypothesis, 1 L of NH3(g) and 1 L of HCl(g) contain the same number of particles, so
that if:
1 L of HCl(g) is 2.15 times heavier than 1 L of NH3(g), and
1 L of HCl(g) has the same number of molecules as 1 L of NH3(g),
then each particle of HCl must be 2.15 times heavier than each particle of NH3 .
Therefore you can find the mass of one gas particle relative to the mass of another gas particle by comparing
the masses of equal gas volumes.
EXAMPLE: 1 L of hydrogen gas at 0oC and 101.3 kPa has a mass of 0.0893 g and 1 L of oxygen gas under
the same conditions has a mass of 1.43 g. Assigning a mass of 2.00 to a hydrogen particle,
what is the mass of an oxygen particle relative to hydrogen?
Setting up the ratio
mass of oxygen gas
mass of oxygen particle
=
mass of hydrogen gas
mass of hydrogen particle
1.43 g
M
=
, which solves to give: M = 32.0
0.0893 g
2.00


Therefore a particle of oxygen has a mass of 32.0.
then


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EXAMPLE: 1.85 L of helium have a mass of 0.400 g. Under the same conditions, 5.68 L of nitrogen have a
mass of 8.60 g. Assigning a mass of 4.00 to helium, what is the mass of a nitrogen particle
relative to a helium particle?
The gases cannot be directly compared since they have different volumes. An additional
calculation is required: finding the amount of gas which exists in an arbitrarily selected volume,
say 1 L; that is, calculating the density of the gas.
helium density =
0.400 g
g
= 0.216
1.85 L
L
and
nitrogen density =
8.60 g
g
= 1.51
5.68 L
L
mass of nitrogen gas
mass of nitrogen particle
=
mass of helium gas
mass of helium particle



which gives M = mass of nitrogen molecule = 28.0
Now equal (1 L) volumes are compared:
1.51g  M
=
,
0.216 g
4.00

EXERCISES:

8. At 25oC and 101.3
 kPa pressure, 1.00 L of hydrogen chloride gas has a mass of 1.49 g, and 1.00 L of
oxygen gas has a mass of 1.31 g. What is the mass of a hydrogen chloride molecule relative to an
oxygen molecule having a mass of 32 units?
9. At 25oC and 101.3 kPa pressure, 3.50 L of sulphur dioxide have a mass of 9.14 g, and 3.50 L of
nitrous oxide have a mass of 4.29 g. What is the mass of a nitrous oxide molecule relative to a mass
of 64 units for a molecule of sulphur dioxide?
10. Under the same conditions, 7.00 L of fluorine have a mass of 11.9 g and 7.00 L of chlorine have a
mass of 22.2 g. If fluorine has a mass of 38 units, what is the mass of chlorine?
11. Under the same conditions, 8.00 L of hydrogen bromide have a mass of 28.9 g and 14.0 L of xenon
gas have a mass of 81.9 g. If a xenon atom has a mass of 131 units, what is the mass of a hydrogen
bromide molecule?
12. Under the same conditions, 15.0 L of sulphur trioxide have a mass of 53.6 g and 4.00 L of carbon
monoxide have a mass of 5.00 g. If a carbon monoxide molecule has a mass of 28.0 units, what is
the mass of a sulphur trioxide molecule?
13. At 0oC and 101.3 kPa pressure, 10.0 L of hydrogen have a mass of 0.89 g and 10.0 L of oxygen have
a mass of 14.3 g. At 25oC and 202.6 kPa pressure, 1.00 L of argon has a mass of 1.64 g and 1.00 L
of hydrogen has a mass of 0.082 g. What is the mass of an argon atom relative to an oxygen
molecule having a mass of 32.0 units?
14. At 0oC and 101.3 kPa pressure, 3.00 L of helium gas have a mass of 0.536 g and 3.00 L of krypton
have a mass of 11.2 g. At 25oC and 152.0 kPa pressure, 2.00 L of krypton have a mass of 10.3 g and
2.00 L of neon have a mass of 2.47 g. What is the mass of a neon atom relative to a mass of 4.00
units for a helium atom?
C. BOYLE'S LAW
In 1660, Robert Boyle found that the volume of a gas decreased when the pressure increased. That is
1
cons tant
or
V=
or
P•V = constant
V
P
P
If either the pressure or volume is changed, then

that is:
EXAMPLE:
P1•V1 = constant = P2•V2

where “1”= before
and “2” = after
P1•V1 = P2•V2
A gas at 0.500 atm pressure was expanded to 3.50 L and 0.0850 atm. What was the previous
volume of the gas?
AP : GAS LAWS
5
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Let
P1 = previous pressure , P2 = new pressure
V1 = previous volume , V2 = new volume
P1•V1 = P2•V2
so that
V1 =
0.0850 atm x 3.50 L
P2 V2
=
= 0.595 L
0.500 atm
P1
EXERCISES:
15. A cylinder contains a movable pistonand has
a pressure of 825 torr in a volume of 0.575 L. Calculate
the pressure if the piston compresses the gas in the cylinder to a volume of 0.275 L.
16. A gas is compressed in a cylinder to a pressure of 1.50 x 105 kPa and a volume of 25.0 mL. If the
gas is allowed to expand such its pressure is 1.00 atm, what volume does the gas occupy?
D. DALTON'S LAW OF PARTIAL PRESSURES
Definition: The PARTIAL PRESSURE of a component X of a gas mixture is the pressure which X would
exert if it OCCUPIED THE WHOLE VOLUME ALONE.
The following law was put forward by the “father of Chemistry”, John Dalton, an English school teacher.
DALTON'S LAW: The total pressure of a gas mixture is equal to the sum of the
partial pressures of its components.
In other words, the pressure in a gaseous system is directly proportional to the number of moles of gas
molecules present. Whether or not the gas is pure or a mixture is irrelevant.
EXAMPLE: A mixture of gases contains 0.250 mol Ar(g), 0.130 mol of He(g) and 0.310 mol of Ne(g). If the
total pressure in the container is 740 mm Hg, what is the partial pressure of each gas?
PGAS
PTOT
=
moles of gas
total moles
First, find nTOT
 Now
Similarly

Let
n = # of moles
nTOT = 0.250 + 0.130 + 0.310 = 0.690 mol
PAr =
0.250 mol x 740 mm Hg
nAr PTOT
=
= 268 mm Hg
0.690 mol
n TOT
PHe = 139 mm Hg
and
PNe = 333 mm Hg
(Check: 268 + 139 + 333 = 740 mm Hg)


EXERCISES:
(Do enough of these to become an expert)
17. A glass bulb contains a mixture of 0.0150 mol of He(g) and 0.0350 mol of Ar(g). If the pressure in the
bulb is 60.0 kPa, what is the partial pressure of each gas in the mixture?
18. A bulb contains 0.0160 mol of CO(g), 0.005 00 mol of N2(g) and 0.0240 mol of SO2(g). If the
pressure in the bulb is 117 kPa, what is the partial pressure of each gas in the mixture?
19. A bulb contains a gas mixture which is 15.0% O2 , 30.0% SO2 and 55.0% SO3 . If the total pressure
in the bulb is 160.0 kPa, what is the partial pressure of each gas?
20. A steel tank contains 60.0 g of He(g), 200.0 g of O2(g) and 50.0 g of N2(g). What is the partial
pressure of each gas if the total pressure in the tank is 137 kPa?
21. A balloon contains 9.65 x 1024 molecules of H2(g), 8.50 x 1024 molecules of Ne(g) and 1.72 x 1025
molecules of He(g). The total pressure in the balloon is 127 kPa. What is the partial pressure of each
of the gases?
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22. A gas bulb contains 10.0 g of O2(g), 15.0 g of N2(g) and 20.0 g of Ar(g). If the partial pressure of
O2(g) is 33.0 kPa, what is the total pressure in the tank?
23. A storage tank contains 200.0 g of C2H2(g), 250.0 g of C2H4(g) and 400.0 g of C2H6(g). If the partial
pressure of the C2H2 is 122 kPa, what is the total pressure in the tank?
24. A steel cylinder contains 4.00 x 1022 molecules of CO2(g), 3.65 x 1022 molecules of CH4(g) and
2.90 x 1022 molecules of NO2(g). If the combined pressure of the CH4 and CO2 is 152.0 kPa, what is
the partial pressure of each of the gases in the cylinder?
25. A glass bulb contains 0.0210 mol of O2(g), 0.0165 mol of N2(g) and 0.0314 mol of CO2(g). When the
CO2 is removed by a reaction with NaOH: NaOH(s) + CO2(g)
NaHCO3(s), the pressure in the
system drops by 43.0 kPa. What is the partial pressure of each of O2 and N2?
E. VAPOUR PRESSURE
1. A Molecular Theory of Vapour Pressure
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Experiments lead to the following observations.
• Vapour pressure increases with temperature.
• As a liquid is added to a sealed, evacuated system, the gas pressure slowly increases and then
levels off. The final pressure exerted by the evaporating liquid is not affected by the amount of
liquid present as long as there is SOME liquid present.
Definition: The VAPOUR PRESSURE of a liquid is the MAXIMUM gas pressure that can be created by
the evaporation of the liquid into the gas phase at a specific temperature.
At a given temperature two separate processes occur in a liquid.
a) Molecules in the liquid are knocked out of the liquid by energetic neighbouring molecules, leaving a
“hole” or “cavity” in the smooth surface of the liquid.
b) Molecules in the gas phase dart about in random directions. Sometimes the gas molecules head in
the direction of the liquid and either
i) rebound off a molecule at the liquid surface, or
ii) enter a cavity and become part of the liquid again.
gas pha se
liq uid phas e
When a liquid is first introduced into a vessel, there are no molecules from the liquid in the gas phase.
However, there is a small (but non–zero) probability that a given molecule at the surface of the liquid will
be knocked into the gas phase, and therefore as time passes the number of molecules in the gas phase
increases and the pressure increases.
Once in the gas phase, there is a very small probability that a given molecule will be “aimed” in precisely
the correct direction so as to allow it to re–enter the liquid phase. However, as the number of gas phase
molecules continues to rise, there is an increasing total probability that some molecules will re–enter the
liquid. Eventually, gas molecules are re–entering the liquid as often as liquid molecules enter the gas
phase. At this point the gas pressure ceases to rise and stays at some constant value known as the
VAPOUR PRESSURE.
AP : GAS LAWS
7
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Note that some liquid must be present in order to know that the gas phase has become SATURATED
with gas molecules from the liquid. (A phase is said to be SATURATED with a particular substance when
the phase can contain no more of the substance.) When no further changes occur in the pressure, the
system is at EQUILIBRIUM at that temperature.
Since evaporation occurs when molecules in the liquid gain enough energy to escape into the gas phase,
the greater the temperature of the liquid the greater the kinetic energy of the molecules in the liquid. The
greater the kinetic energy, the greater the evaporation rate and the greater the vapour pressure. This
theoretical argument explains the observation that
The higher the temperature, the greater the vapour pressure.
Freeze–Drying
When the gas molecules above a liquid are continuously removed by a vacuum pump, no molecules
can re–enter the liquid. Therefore, liquid molecules continually go into the gas phase. But, only
molecules with a relatively high energy can evaporate, leaving behind a collection of liquid molecules
having an increasingly–lower average kinetic energy. As higher–energy molecules are pumped off, the
remaining liquid cools. In this way a solid in solution can be “freeze–dried” by pumping off the vapour
from the liquid and cooling the residue. The cooling effect can be quite dramatic, even to the point of
freezing the remaining liquid after some of the solvent has been removed.
The Effect of Compressing the Vapour above a Liquid
When the pressure is increased by compressing the gas phase above a liquid, no change in vapour
pressure occurs. Because the gas molecules are more “concentrated” when the volume decreases, gas
molecules are more likely to re–enter the liquid.
co mpres s to hal f volu me
Since molecules are still leaving the liquid at the original rate (determined by the liquid temperature) while
excess vapour molecules are re–entering the liquid at a faster rate (proportional to the concentration of
molecules in the gas phase), the gas pressure quickly goes back to its former value.
sud denl y decrease the volum e
gas pha se m olecule s go back into the liq uid
fas ter than liqu id m olecule s go into ga s ph ase
Press ure
sta rting
pres sure
equ ilib rium is
re-es tabl ishe d
Tim e
The VP of a pure liquid is only affected by the temperature and is not affected by the pressure applied to
the gas phase above the liquid (except on a momentary basis). In practice, the volume can be decreased
to the point where the piston touches the surface of the liquid. In this case, any additional pressure
applied will drive all the molecules into the liquid phase and reduce the vapour pressure to zero,
provided that a pressure greater than the vapour pressure is applied to the liquid.
2. Boiling Temperature
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The total gas pressure in the space immediately above the surface of an open vessel of water is due to a
mixture of air and water vapour. Therefore
PTOTAL = PWATER + PAIR
where
PWATER = the vapour pressure of water
PAIR
= the partial pressure of air.
But this total pressure equals the pressure of the atmosphere pushing down on the pan surface since the
pan is open to the air.
PATM = PTOTAL = PWATER + PAIR
When water is heated, PWATER increases. But PATM cannot be changed in an open system, so that if
PWATER is increasing just above the surface of the liquid then PAIR must be decreasing. That is, the water
vapour starts to displace the air.
Eventually, a maximum water temperature is achieved when
PWATER = PATM
which implies that PAIR = 0 just above the liquid; that is, the air molecules above the liquid have been
displaced by water molecules. But, if PWATER = PATM then molecules all through the liquid have enough
KE to overcome the downward pressure from the atmosphere (which forces them to stay in the liquid) and
go directly into the gas phase. As a result, gas “bubbles” appear all through the liquid and the liquid
BOILS.
When PWATER < PATM , some highly energetic molecules below the liquid
surface have sufficient KE to separate from their neighbours and form a
microscopic gas bubble (see figure at right) but the greater pressure of
the atmosphere crushes the bubble out of existence and thus no bubbles
are seen to form.
When PWATER = PATM , the pressure generated within the bubble (which equals PWATER) matches the
pressure of the atmosphere bearing down on the liquid and the gas bubble remains in existence. Note
that if a bubble forms at the bottom of the liquid, the weight of the water above the liquid causes a small
additional increase in the pressure bearing down on the bubble, so that a slightly greater temperature is
required to maintain sufficient vapour pressure within the bubble and prevent it from being compressed
out of existence.
As the low–density bubble slowly rises up through the higher–density liquid, the pressure on the outside
of the bubble decreases (less water is pushing down on it) and the bubble expands against the lowered
external pressure. This explains why BUBBLES EXPAND AS THEY RISE UP THROUGH A LIQUID.
Definitions: The BOILING TEMPERATURE of a liquid is the temperature at which the VP of the liquid
equals the pressure of the gases pushing down on the surface of the liquid.
The NORMAL BOILING TEMPERATURE is the temperature at which the VP of a liquid
equals 101.3 kPa (1 atm).
Therefore, when all the air is removed from a sealed flask containing water, the only gas above the liquid
is water vapour and
PTOTAL = PWATER .
The VP of the water is always equal to the pressure of the “atmosphere” pushing down on the liquid
BECAUSE THE ONLY GAS IN THE “ATMOSPHERE” IS WATER VAPOUR, and the water boils at room
temperature (or any temperature at which the flask exists!)
Similarly, on a mountain top the atmospheric pressure is less than 101.3 kPa, say 70 kPa for example.
Therefore, a lower temperature (90.0oC) is required to attain a water vapour pressure of 70 kPa than to
AP : GAS LAWS
9
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attain 101.3 kPa and water boils at a lower temperature on top of a mountain.
EXERCISES:
26. If you were to stay on top of a very tall mountain long enough to cook a meal that included a boiled
egg, what difficulties might you encounter with the cooking of the egg?
27. Why is the concept of normal boiling temperature necessary, especially with respect to reporting the
boiling temperatures of newly–discovered compounds?
28. Draw a graph to represent the pressure–versus–time behaviour that exists when a sample of gas
trapped over a sample of water is compressed to half of the original gas volume.
29. A sample of gas occupies a volume of 500.0 mL and is trapped in a piston and cylinder that contains
a few millilitres of water at 50oC. The total pressure inside the cylinder is 115.0 kPa. The vapour
pressure of the water is 12.3 kPa. What is the total pressure in the cylinder if the gas volume is
decreased to 250.0 mL?
30. A sample of gas occupies a volume of 1.55 L and is trapped in a piston and cylinder that contains a
few millilitres of water. If the water is at 25oC and has a vapour pressure of 3.17 kPa, and the
pressure due to the gas is 95.0 kPa, what is the total pressure in the cylinder if the gas volume is
expanded to 2.75 L?
31. Tennis ball manufacturers make different balls for people to use at low altitudes versus high altitudes.
Why might this be?
32. What should be true of the boiling temperature of water at the bottom of a deep mine?
F. CHARLES' LAW
In 1787 the French physicist Jacques Charles, who was the first person to design and fly in a balloon, found
that an increase in temperature causes an increase in the volume of a gas.
Kelvin (1848) noted that Charles' Law implies a lowest temperature exists, and that a gas thermometer can
establish an ABSOLUTE TEMPERATURE SCALE: this is now known as the Kelvin scale.
Plotting volume–vs–temp(oC) for many gases gives the following graph.
Vol ume
of 1 mo l
of gas
0
Tem perature (oC)
Experimentally, it is found that the line extrapolates to zero volume at –273oC.
0 K = –273oC
Definition:
and T(K) = T(oC) + 273
Charles' Law is expressed by the relationship
V
 T(K) or V = constant•T(K).
V1
T
= 1
V2
T2
so that

or
V1
V
= 2
T1
T2
EXAMPLE: A gas has a volume of 250.0 mL at –80oC. What volume does the gas occupy at 100oC?




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Hebden : Chemistry AP
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Let V1 = old volume , T1 = old temperature = –80 + 273 = 193 K
V2 = new volume , T2 = new temperature = 100 + 273 = 373 K
V2 =
250.0 mL x 373 K
V1  T2
=
= 483 mL
193 K
T1
G. THE IDEAL
GASLAW

Up to this point the following three laws have been investigated.
P•V = constant
(Boyle's Law)
V = constant •T
(Charles' Law)
P = constant •n
(Dalton's Law)
All these relationships (and more) can be summarized by the following master equation.
PV = nRT
where R is a proportionality constant called the GAS CONSTANT or IDEAL GAS CONSTANT.
The value of R can be found using the known values for P, V, n and T at STANDARD STATE.
R=
101.3 kPa x 22.4 L
PV
kPa L
=
= 8.31
1mol x 273 K
nT
mol K
EXERCISE:
 Calculate
 the pressure is expressed in: (a) torr.

33.
the value of R when
(b) atmospheres.
GAS LAW CALCULATIONS
There are two main types of gas law calculations.
1. A Single Set of Conditions: The Conditions Under Which a Gas Exists
——————————————————————————————————————————————
A SINGLE set of conditions is given, describing the conditions to which a gas is subjected. There is no
sense of time passing.
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EXAMPLE: How many moles of O2(g) are contained in a 500.0 mL bulb containing O2 at 0.200 atm and
60oC?
101.3 kPa
Assemble a “shopping list”
P = 0.200 atm x
= 20.3 kPa
1atm
V = 500.0 mL = 0.5000 L
n=?
R = 8.31 kPa•L/mol•K
T = 60 +273 = 333 k
Note that all the units used must agree with the units of R.
Now
n=
PV
20.0 kPa x 0.5000 L
=
= 0.00367 mol
8.31kPaL / mol K x 333 K
RT
EXAMPLE: If 16.0 g of an unknown gas occupies a volume of 2.50 L at 450 kPa and 100oC, what is the
molar mass of the gas?


P = 450 kPa
V = 2.50 L
n=?
R = 8.31 kPa•L/mol•K
T = 100 + 273 = 373 K
PV
450.0 kPa x 2.50 L
n=
=
= 0.363 mol
8.31kPaL / mol K x 373 K
RT
16.0 g
mass
But molar mass =
=
= 44.1 g/mol
0.363 mol
moles

 equation is used to solve the above problem.
NOTE: Sometimes the following
m

PV =
RT
MW
where
and
m = mass of gas
MW = molar mass of gas
However, there is little advantage to memorizing a special equation for one type of problem.
EXERCISES:

34. If 2.00 mol of O2(g) occupy 200.0 L at 0oC, what is the pressure of the O2?
35. How many moles of N2(g) are in a 3.50 L cylinder at 35oC and 202.6 kPa?
36. A glass bulb with a volume of 2.75 L contains 0.220 g of He(g) at a pressure of 20.3 kPa. What is
the temperature of the bulb?
37. When 0.560 g of N2(g) is cooled to –40oC, the pressure of the gas is 66.7 kPa. What is the
volume of the container holding the gas?
38. A 0.500 g sample of a gas has a pressure of 23.3 kPa at 25oC in a 0.750 L bulb. What is the
molar mass of the gas?
39. What mass of gaseous ethane, C2H6 , at 60.8 kPa pressure and –25oC contains the same
number of molecules as 9.20 g of NO2(g) at 101.3 kPa pressure and 0oC?
40. The best vacuum that can be attained presently on earth is about 1.0 x 10–14 kPa at 0oC. Under
these conditions, how many molecules are contained in a volume of 1.00 mL?
41. Calculate the mass of one molecule of a gas if 20.4 g occupy 11.4 L at 507 kPa and 273oC.
42. When 1.75 g of an unknown gas is put into a 1.25 L glass bulb at 39oC, the pressure in the bulb
is 86.4 kPa. What is the molar mass of the gas?
43. What is the pressure of a sample of helium gas having a density of 7.80 g/L at 273oC?
44. A 1.25 g sample of oxygen gas at STP is found to contain the same number of molecules as
12
Hebden : Chemistry AP
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0.664 g of an unknown gas at 50.0 kPa and –75oC. What is the molar mass of the unknown?
45. Calculate the density of a sample of nitrogen gas at –15oC and 100.0 kPa.
2. Two Sets of Conditions: Before and After
——————————————————————————————————————————————
Two sets of conditions are given. One set describes an initial set of conditions
P1V1 = n1RT1
(R cannot change).
A second set of values describes a final or changed set of conditions
P2V2 = n2RT2 .
P1V1
n RT
Comparing the two sets of values in a ratio gives
= 1 1
P2 V2
n2RT2
P1V1
nT
and canceling the constant, R, gives the final form
= 1 1
n2 T2

 P2 V2
EXAMPLE: A steel cylinder contains 2.00 L of N2(g) at 20oC and 90.0 kPa. When a piston is pushed

into the cylinder, the volume iscompressed
to 0.800 L and the pressure rises to 260.0 kPa.
What is the final temperature?
P1 = 90.0 kPa
V1 = 2.00 L
P2 = 260.0 kPa
V2 = 0.800 L
same
n1 =
T1 = 20 + 273 = 293 K
Rearranging
P1V1
nT
= 1 1
P2 V2
n2 T2
gives
T2 =


n2 =
T2 = ?
260.0 kPa x 0.800 L x 293 K
P2 V2 T1
=
= 339 K (66oC)
90.0 kPa x 2.00 L
P1V1
EXAMPLE: A 600.0 mL container at 80oC contains a gas at 45.0 kPa. A piston changes the volume to
400.0 mL, 
the temperature
is raised to 160oC and the mass of gas within the cylinder is

doubled. What is the final pressure?
P1 = 45.0 kPa
V1 = 600.0 mL
n1 = n
T1 = 80 + 273 = 353 K
Rearranging
and
P1V1
nT
= 1 1
P2 V2
n2 T2
P2 =



gives
P2 = ?
V2 = 400.0 mL
n2 = 2 n
T2 = 160 + 273 = 433 K
P2 =
n2 T2P1V1
V2n1T1
2n x 433 K x 45.0 kPa x 600.0 mL
= 166 kPa
n x 353 Kx 400.0 mL
AP : GAS LAWS
13
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A final comment on Avogadro's Hypothesis
According to Avogadro's Hypothesis, “Equal volumes of gases at the same temperature and pressure
contain the same numbers of moles”. That is, when V1 = V2 , T1 = T2 and P1 = P2 , then n1 = n2 .
P1V1
nT
n
This means
= 1 1
reduces to
1= 1
and therefore
n1 = n 2
P2 V2
n2 T2
n2
so if V1 = V2 , T1 = T2 and P1 = P2 , then n1 = n2 automatically, and therefore “Avogadro's Hypothesis” is
just a special case derivable from the Ideal Gas Law.



EXERCISES:
46. A gas at 1621 kPa pressure occupies 11.0 L at 36oC. What is its new volume if the temperature
is changed to 16oC and the pressure to 3647 kPa?
47. A gas occupying 2.30 L at –100oC is allowed to warm up to 25oC, What is the new gas volume, if
pressure is constant?
48. The reaction K(s) + H2O(l)
KOH(aq) + H2(g) produces 45.3 mL of H2(g) at 23oC and
97.8 kPa. What mass of K(s) is used in the reaction?
49. The fuel supply for a rocket engine on a satellite is contained in a steel sphere. The sphere has a
volume of 10.0 L and delivers 1400 L of gas at 25oC and 101.3 kPa. What pressure must the
sphere be able to withstand if the normal operating temperature of the sphere is –10oC? What
mass of O2(g) can be delivered from the sphere?
50. The average breath that a 16 year old male takes when resting is about 300 mL in volume at
20oC and 100 kPa. His respiratory rate is about 20 breaths per minute. What volume of air at
STP does he breathe in one minute?
51. The density of oxygen gas is 1.31 g/L at 25oC and 101.3 kPa pressure. What is the volume
occupied by 1.00 mol of oxygen under these conditions?
52. A steel cylinder holds 1.00 mol of a gaseous hydrocarbon and 4.00 mol of oxygen. The internal
temperature and pressure of the container is 25oC and 202.6 kPa. A spark ignites and
completely reacts the gases. If the final pressure is 709.1 kPa and the products of combustion
are 3.00 mol of CO2(g) and 4.00 mol of H2O(g), what is the temperature after the ignition occurs?
53. A rocket probe to the surface of Mars collected a sample of gas from a surface crack in the soil.
When compressed to 101.3 kPa pressure (only a wisp of gas was collected) and warmed to 0oC,
a 5.01 mL sample of the compressed gas had a mass of 6.27 mg. The scientists in charge of the
Martian Probe project hoped that the gas was CO2 , in support of an important theory of the
Martian atmosphere. They were also aware of the fact that the following reaction was used to
propel the descent rocket of the landing module: N2H4 + O2
2 H2O + N2 .
Was the gas CO2 , or did rocket exhaust contaminate the experiment?
54. Calculate the density of NH3(g) at STP.
55. A cylinder contains 3.00 mol of N2(g) at a temperature of 100oC and a pressure of 405.2 kPa in a
volume of 23.0 L. If the cylinder is heated to 400oC and the volume expands to 50.0 L by moving
out a piston, what mass of N2 must be removed to lower the pressure to 101.3 kPa?
56. Calculate the volume of oxygen gas at 150oC and 95.0 kPa required to react completely with
15.0 g of NH3 gas at 250oC and 115 kPa if the reaction is 4 NH3 + 7 O2
4 NO2 + 6 H2O.
57. A cylinder having a volume of 883.0 L contains 50.0 g of He(g) at 33.3 kPa and 10oC. If the total
volume is compressed to 500.0 L, the temperature is kept constant and another 120.0 g of helium
are added, what is the final pressure?
58. What volume of HCl(g) at 50oC and 120.0 kPa can be formed by reacting 15.0 L of H2(g) at
150oC and 85.0 kPa with 12.0 L of Cl2(g) at 100oC and 125.0 kPa?
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Hebden : Chemistry AP
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59. A glassblower uses an oxygen tank having a volume of 15.0 L. The tank remains at a
temperature of 21oC at all times. If the tank pressure goes from 190.5 atm to 187.2 atm, what
mass of oxygen is delivered from the tank?
60. In order to make 125 kg of (NH4)2SO4, what volume of 9.00 M H2SO4 and what volume of NH3
gas at 25.0 atm and 5oC is required for the reaction?
61. A 700.0 mL bulb contains 0.00100 mol of N2(g) at 3.60 kPa. If the temperature is unchanged,
how many moles of N2 have to be added to bring the pressure up to 20.00 kPa?
62. A sample bulb contains 0.712 g of H2(g) at a certain temperature and pressure. Under the same
temperature and pressure conditions, the bulb can hold 13.0 g of an unknown gas. What is the
molar mass of the unknown gas?
63. A 2.00 L storage tank contains F2(g) at 126 kPa and 30oC. A 1.00 L reaction vessel contains
Xe(g) at 50.4 kPa and 30oC. The F2(g) is pumped into the reaction vessel containing the Xe(g)
and the reaction mixture is heated for a few hours, allowing the reaction
Xe(g) + 2 F2(g)
XeF4(s)
to occur. After all the Xe(g) is used up, the reaction vessel is cooled to –10oC. Assuming all the
XeF4 remains in the solid phase, what pressure of F2(g) remains in the reaction vessel?
H. GRAHAM'S LAW OF EFFUSION
Definition: EFFUSION RATE is the rate at which a gas passes through a tiny hole into a vacuum.
Care: Do not confuse EFFUSION with DIFFUSION, which is the mixing of fluids.
The kinetic energy of an effusing particle is given by
KE =
1
2
m•v2
where m = mass of particle
and
v = effusion rate.
Since temperature is just a measurement of the average KE of molecules, if gases A and B are at the same
temperature they must have equal kinetic energies.

KEA = KEB
and
mA•vA2 = mB•vB2
Note: This implies light particles have a large velocity (a high effusion rate) and heavy particles have a small
velocity (a small effusion rate). Therefore, both light and heavy particles strike with the same impact
(KE).
Now
vA mA = vB mB
mB
vA

=
mA
vB



and rearranging this gives
Note: In this equation and the two below, the left side has gas A on top BUT
on the right side gas A is on the bottom. Therefore, this is an
INVERSE RELATIONSHIP, reflecting the fact that THE HEAVIER A
MOLECULE, THE SMALLER ITS AVERAGE VELOCITY.
AP : GAS LAWS
15
————————————————————————————————————————————————–
Recall that: d = m / V . If V = the molar volume, then m = the molar mass and the density of a gas is
proportional to its molar mass. These relationships lead to two forms of Graham's Law of Effusion (below).
(effusion rate) A
=
(effusion rate)B
where

dB
dA
MWB
MWA
(effusion rate) A
=
(effusion rate)B
or
dA and dB are the densities of gases A and B,
 the molar masses
MW
 A and MW B are
 of gases A and B,
effusion rates are usually given in metres/second or moles/second
EXAMPLE: If the average velocity of H2 molecules is 1700 m/s at 0oC, what is the average velocity of He
atoms at 0oC?
Let vB = v(H2) , vA = v(He)
mB = m(H2) , mA = m(He)
vA = vB
mB
= 1700 m/s x
mA
2.0 g
= 1200 m/s
4.0 g
EXAMPLE: If O2 effuses through a pinhole at a rate of 2.30 x 10–4 mol/s and an unknown gas effuses
–4
through 
the same pinhole under
 identical conditions at a rate of 1.96 x 10 mol/s, what is the
molar mass of the unknown?
(effusion rate) A
=
(effusion rate)B
Solving:

EXERCISES:
MWB
MWA
so that
2.30 x 10 4 mol / s
1.96 x 10
4
mol / s
=
MWB
= 1.173
32.0 g
MW B = 44.1 g



64. What is the molar mass of thionyl chloride if gaseous thionyl chloride at 100 oC effuses 1.72 times
faster than a sample of uranium hexafluoride?
65. A physicist used an effusion apparatus to separate a sample of methane, CH4 , made from a naturally
occurring mixture of carbon (98.90% C–12 and 1.10% C–13). 12CH4 has a molar mass of
16.03176 g/mol and 13CH4 has a molar mass of 17.03512 g/mol. Which molecule will effuse faster
and how many times faster than the other molecule does this molecule effuse?
I. THE KINETIC MOLECULAR THEORY OR “COLLISION THEORY”
The basic postulates of the Kinetic Molecular Theory are as follows.
1. A gas consists of a large number of tiny particles that are in constant, random motion.
2. The net volume of the gas particles is negligible with respect to the total volume of the gas; that is, the
gas particles are assumed to have ZERO volume.
3. The particles are hard spheres which undergo perfectly elastic collisions with themselves and the
walls of their container, and move in a straight line between collisions. (“ELASTIC” means no energy
is lost during a collision.)
The various gas laws can be interpreted in terms of the Kinetic Molecular Theory as follows.
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Hebden : Chemistry AP
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a) Kinetic Theory and Temperature
——————————————————————————————————————————————
It can be shown that the units of P•V are energy units as follows. (It helps if you know a little physics!)
and
1 Pa = 1 N/m2 and 1 L = 10–3 m3
1 kPa•L = 1 N•m = 1 J = energy
1 kPa•L = 1 J
Therefore
but
so that
P•V = f(KE)
P•V = f(T) ,
T = f(KE).
Remember this! It will be needed for Thermodynamics.
according to the Ideal Gas Law
This latter relationship is summarized by the behavior of the Boltzmann distribution of kinetic energies.
(Recall that the “peak” (average KE) moves to higher energies as temperature increases.)
# o f pa rticles
Kine tic ene rgy
b) The Relationship between Pressure and Temperature
——————————————————————————————————————————————
If T increases then KE increases. Therefore, particles transfer more energy when they strike the walls of
a container, increasing the pressure on the walls.
c) The Relationship between Temperature and Volume
——————————————————————————————————————————————
As temperature increases, gas particles exert more pressure on the walls of a container. If the wall of the
container is movable, the increased pressure pushes out the wall and increases the volume occupied by
the gas. (Charles' Law)
d) The Relationship between Pressure and Volume
——————————————————————————————————————————————
If a given amount of gas is compressed, all the particles are confined to a smaller volume and more
particles are close to the wall. This leads to more frequent collisions with the wall and therefore an
increased pressure. (Boyle's Law)
e) The Relationship between Moles and Pressure
——————————————————————————————————————————————
Since gas particles are assumed not to interact with each other (no attraction or repulsion between
particles) then they must act INDEPENDENTLY of each other. Therefore the more moles of gas present,
the more collisions with the container walls and the greater the pressure produced. (Dalton's Law of
Partial Pressures) Note that the fact that Dalton's Law holds very well is strong evidence for the non–
interaction between particles.
J. SYSTEMS OF REAL GASES: DEVIATIONS FROM IDEALITY
AP : GAS LAWS
17
————————————————————————————————————————————————–
Some problems exist with the "IDEAL" gas law.
• Gas particles have a definite volume.
• Gas particles attract each other to a certain extent (van der Waals forces).
J.D. van der Waals concluded that a better description of gas behaviour can be given by the van der Waals
Equation.
(P +
n2  a
)•(V – n•b) = n•R•T
V2
where “a” and “b” are called van der Waals constants, and reflect the extent to which a gas deviates from
ideality

NOTE: In order to help keep the following arguments clear, define
VMEAS = V = the measured volume of the container
PMEAS = P = the measured pressure in the container
n, R and T have their usual meanings.
If “a” and “b” are very small, the van der Waals equation reduces to the Ideal Gas Law, PV = nRT. The
manner in which “a” and “b” arise is shown below.
The origin of “b”
The Ideal Gas Law assumes molecules are mathematical points having ZERO VOLUME. An allowance must
be made for the fact that atoms and molecules do not have ZERO VOLUME. When two atoms touch during
a collision, the situation is shown below. Atoms A and B are impenetrable spheres with a diameter “d” and a
radius “r”. There is a certain volume, VX , called the “excluded volume”, which the centre of atom B can't get
into. Note that this volume is actually 8 times greater than the volume of atom A.
B
volu me o f B = VB = (4/ 3)r3 = (1/ 6)d3
X
exclude d volu me = VX = (1/ 6)(2d )3 = (8/ 6)d3
A
an d th erefo re th e exclude d volu me i s 8
ti mes the vo lume of B (and A)
Definition: b = the excluded volume per mole.
Since “b” is proportional to the size of the particles in the gas,
the larger the particle, the greater the value of “b”.
Since the Ideal Gas Law assumes zero particle volume, the “excluded volume” is subtracted from the
MEASURED VOLUME OF THE CONTAINER, VMEAS , in order to calculate the somewhat smaller VOLUME
IN WHICH A GAS CAN ACT AS IF IT WERE “IDEAL”, VIDEAL .
VIDEAL = VMEAS – n•b
You can think of the above expression in the following way. VMEAS is the volume of the container, n•b is the
small volume unavailable for molecules to move into, and VIDEAL is the volume available to the molecules and
in which they are free to act in an ideal manner.
To summarize:

Actual volume available  Actual volume available 

 

to a real gas

  to an IDEAL GAS 
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Hebden : Chemistry AP
————————————————————————————————————————————————–
When dealing with a real gas, the measured volume has the value “n•b” subtracted out to leave the volume
actually available to molecules that are effectively acting in an ideal fashion.
The ideal gas equation now can be modified as follows.
PIDEAL =
where:
n R  T
(VMEAS  n b)
PIDEAL is the pressure after correcting for the non–zero volume of the molecules, but before
correcting for the attractive forces between the molecules.

The origin of “a”
As can be seen at the right, molecules near the wall feel a net attractive pull
toward the main body of gas and AWAY FROM THE WALL. On the other
hand, molecules in the main body of gas feel NO net pull; that is, the
attractions are “balanced”. In addition, molecules moving through a real gas
“veer” around and towards neighbouring molecules being approached. This
deviation from a straight-line trajectory means the molecules travel a greater
distance than would a similar ideal gas before colliding with the container’s
walls. As a result, there are fewer collisions with the container walls in a
given time and the measured gas pressure decreases.
The attractions between the molecules make the measured pressure, PMEAS , smaller than it would otherwise
be if molecules didn't attract each other.
PMEAS = PIDEAL – (attractive force correction term)
and
PMEAS = PIDEAL –
n2  a
2
VMEAS
(don't worry about how this was derived)
Overall, the value of “a” is related to the strength of the ATTRACTIVE FORCES BETWEEN
PARTICLES.

It follows that
The larger the particle, the greater the van der Waals forces between particles and the greater
the value of “a”.
Also, if a molecule experiences intermolecular dipole–dipole or hydrogen bonding forces, which are stronger
than simple London forces, the value of “a” is additionally increased. (See the table below.)
Substituting the expression found for PIDEAL in The origin of “b” (above) gives
PMEAS =

n R  T
n2  a
– 2
(VMEAS  n b)
VMEAS

AP : GAS LAWS
19
————————————————————————————————————————————————–
and this is rearranged to give the van der Waals equation.
( PMEAS +
n2  a
) ( VMEAS
2
VMEAS
corrected pressure
–
n•b ) = nRT
corrected volume
The following
table gives the values of “a” and “b” for some common gases.

Gas
a
b
He
0.03412
0.02370
Ne
0.2107
0.01709
Ar
1.345
0.03219
Kr
2.318
0.03978
H2
0.2444
0.02661
O2
1.360
0.03183
N2
1.390
0.03913
NH3
4.170
0.03707
CH4
2.253
0.04278
CO2
3.592
0.04267
H2O
5.464
0.03049
CH3COOH
17.59
0.1068
In Summary
Real gases deviate most from ideal gases when the gases are close to condensing into a liquid. Under such
conditions the particles are closest together, interact with each other to the greatest degree, and have a small
enough volume that the excluded volume is a significant fraction of the total, measured volume. Getting the
gas particles as close as possible together requires
and
HIGH PRESSURES
LOW TEMPERATURES
(pack closely)
(close to condensing; energy of attractive forces are a significant fraction
of KE)
On the other hand, real gases approach ideal gas conditions at HIGH TEMPERATURES and LOW
PRESSURES. Under these conditions, molecules are moving sufficiently fast (“high temperatures”) that the
intermolecular forces become unimportant. Also, the molecules are far enough apart (“low pressure”) to make
the excluded volume negligible with respect to the measured volume.
EXERCISES:
66. Determine the units of “a” and “b” if P is measured in atmospheres and V is measured in litres.
67. The van der Waals equation makes corrections for the fact that two hypotheses in the Kinetic Theory
of Gases are not strictly correct. Which hypotheses are incorrect?
68. What does a small value of “b” indicate about a molecule?
69. What does a large value of “a” indicate about a molecule?
70. Is the ideal volume of a gas greater or smaller than the volume of the container holding the gas?
71. Is observed gas pressure greater or smaller than the pressure calculated from the Ideal Gas Law?
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Hebden : Chemistry AP
————————————————————————————————————————————————–
72
(a) Why does O2 have a greater “a” and “b” value than H2?
(b) Why does H2O have a greater “a” value and a lower “b” value than CH4?
(c) Given that MP(NH3) = –78oC and MP(C3H8) = –190oC, which substance should have the greater
value of “a”: NH3 or C3H8? Which substance should have the greater value of “b”?
73. Which of “a” and “b” is related to the boiling temperature of a substance? Why does this occur?
74. The kinetic energy of a gas is decreased. What physical properties are affected and in what way?
K. PHASE DIAGRAMS
Definition: A PHASE DIAGRAM is a graph of the phases existing at a given temperature and pressure.
EXAMPLE:
The phase diagram for CO2 is shown below. Note: scales are not linear.
B
72.9
C
L
S
P (atm)
5.2
T
G
1.0
A
Р78.5
Р57
31
T ( oC)
and
T = the triple point, where all 3 phases can co–exist in equilibrium.
G
S
L
The line C–T = the region of liquid–gas equilibrium: l
g
The line B–T = the region of solid–liquid equilibrium: s
The line A–T = the region of solid–gas equilibrium: s
EXERCISES:
(that is, the boiling point line)
l
g
(that is, the melting point line)
(that is, the sublimation point line)
The following questions are based on the phase diagram of CO 2 , above.
75. What phases exist along the line A–T? Along B–T? Along C–T?
76. What is the sublimation temperature of solid CO2 at 1 atm?
77. According to the diagram, what is the standard state of CO2?
78. Below what temperature can liquid CO2 NOT exist.
79. What phases are observed if a sample of CO2 at 10 atm pressure goes from –80oC to +100oC?
80. What phases are observed if a sample of CO2 at 2 atm pressure goes from –80oC to +100oC?
81. If a sample of CO2 at 0oC goes from 0.5 atm to 80 atm pressure, what phases are observed?
Special Note: Water exhibits very rare behaviour in that the melting point decreases as the pressure
increases. (NH3 also exhibits such behaviour.) THIS BEHAVIOUR OCCURS IN BOTH
CASES BECAUSE OF THE HYDROGEN–BONDING THEY EXPERIENCE.
AP : GAS LAWS
21
————————————————————————————————————————————————–
The phase diagram for water is as follows.
B
S
P (torr)
C
L
76 0
4.58
th is i s th e no rmal boil ing
po int of water
T
G
Note tha t th e scale s are
lo garithmic, whi ch is to sa y ,
ea ch power o f 10 invol ve s an
even scale incremen t.
A
0.01
10 0
T (oC)
EXERCISES:
All of these exercises are based on the phase diagram for water.
82. If the atmospheric pressure has the value "A" (shown below), what happens if the temperature is
raised to point B?
P
A
B
T
83. What phases exist at the triple point?
84. Starting at point A, what happens when the temperature remains constant and the pressure is steadily
increased?
P
A
T
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Hebden : Chemistry AP
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85. Substances such as H2O and NH3 are extensively hydrogen–bonded. In the liquid state, molecules
of H2O (say) are packed relatively close together. On the other hand, during the phase change from
liquid water to solid ice, the molecules must move farther apart to allow a network of low energy
hydrogen bonds to form, as shown below.
Note:
H — O = a covalent bond ; H … O = a hydrogen bond
O
H
H
O
H
H
H
O
H
H
H
H
H
O
O
H
H
free ze
O
O
O
H
H
O
H
H
H
H
O
H
H
When water freezes in a crack in a rock, the water increases in volume.
(a) What happens to the pressure exerted by the water as it freezes?
(b) Why does an increase in pressure during the freezing process require the solid– liquid phase
boundary to have a “backwards slope”?
THE CRITICAL POINT
The concept of “critical point” behaviour is illustrated with the phase diagram for CO 2 (below) although most
substances exhibit similar behaviour.
thi s li ne continue s
A
B (se e te xt)
L
PC
C
S
ТCRITICAL POINTУ
P
G
T
G
T
Definitions:
thi s li ne e nds at ТCУ
TC
Tc = the CRITICAL TEMPERATURE = the highest temperature at which a gas may be
liquified by the application of pressure alone. Above this temperature the substance can
only exist as a gas.
Pc = the CRITICAL PRESSURE = the minimum pressure required to liquify a substance at its
critical temperature.
CRITICAL POINT = the point at which a substance is at its critical temperature and critical
pressure (see point C on the diagram).
When increasing the temperature of a liquid in a sealed container, two things happen simultaneously:
i)
ii)
the liquid phase expands, causing a decrease in density, and
more vapour enters the gas phase, causing the density of the gas phase to increase.
The gas phase exists above the liquid phase due to the greater density of the liquid phase. As temperature
increases, at some point the increasing density of the gas phase becomes equal to the decreasing density of
AP : GAS LAWS
23
————————————————————————————————————————————————–
the liquid phase, so that the phases instantly mix and "blur". Suddenly there is only one phase present.
G
in crea se the tempe rature
L
Graphically, the density–versus–temperature is shown below.
liq uid
den sity
gas
cri tical tempe rature
tem perature
The density difference between liquid and gas ends at the critical point, and it becomes impossible to
distinguish liquid from gas. Examine the phase diagram introducing the concept of critical point. If a
substance starts at point A and the temperature is increased to point B, at some point the solid melts and
forms a dense fluid. There is no point trying to characterize this fluid as a liquid or a gas; it is simply a fluid.
EXERCISES:
86. What are the two ways in which a gas can be liquified?
87. Why might the critical temperature of one substance be greater than that of another?
88. A block of ice is suspended between two benches. A wire is placed on the top of the ice, extending
down each side. Each end of the wire has a weight attached. Gradually, the wire melts its way down
into the ice. The water above the wire continually refreezes as the wire moves downward.
Eventually, the wire and weights have cut completely through the ice and fall to the floor but the ice
remains in one piece. Explain why this series of events happen.
89. Xenon’s normal melting and boiling temperatures are –111.76oC and –108.10oC, respectively. The
triple point exists at –111.78oC and 81.6 kPa and the critical point exists at 16.6oC and 5840 kPa.
Use this information to sketch the phase diagram of xenon.
90. The normal melting and boiling temperatures of ammonia are –77.75oC and –33.35oC, respectively.
The triple point exists at –77.69oC and 0.0600 atm and the critical point exists at 132.4oC and
111.3 atm. Use this information to sketch the phase diagram of ammonia.
91. Why does a CO2 fire extinguisher “slosh” when it is left outside in the winter and shaken, but no
sound is heard when it is left outside in the summer and shaken?
92. Examine the phase diagram below.
D
P
CХ
BХ
A
B
DХ
AХ
T
C
Describe the phases observed when the following
changes are made.
(a) Temperature is increased from point A to point A’.
(b) Temperature is decreased from point B to point B’.
(c) Pressure is increased from point C to point C’.
(d) Pressure is decreased from point D to point D’.