Transcript Gases, Liquids, and Intermolecular Forces

Gases and Liquids
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Topics
 Properties
of gases
 Gas
laws
 Gas Stoichiometry
 Diffusion / Effusion, defined
 Liquids - properties
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The Properties Of Gases
Gases exert pressure
Gases can be compressed
A gas takes up the volume of
the container it is in.
Gases have no free surfaces
Gases are fluid
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Free Surface...
A free surface is one that
does not have to be
contained by a wall of a
vessel. A gas must be
contained on all sides or it
will flow out of the container.
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Gases exert pressure
Pressure is defined as the force
per unit area.
Gas particles are always
colliding with each other and
with the walls of the container
in which they are held.
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The collisions with the walls of
the container produce pressure.
Gas also gets less dense when heated
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Why does a gas get less dense
when heated ?
When matter is heated, its
particles begin to vibrate
more. They have more
kinetic energy.
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The particles get farther apart when
they possess more kinetic energy.
The more kinetic energy they
possess, the more the particles
will overcome the intermolecular
forces that hold them close to
each other.
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Since density = mass per unit of
volume
The density decreases because
the particles move farther apart,
thus decreasing the amount of
mass in a given unit of volume.
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Gases can be compressed
Gas is compressible because the
particles are far apart.
Ex. A scuba tank may have 58 L of
O2 and He compressed into a 5 L
tank. The pressure in the tank
would then be 11.7 atm.
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Atmospheric pressure is measured
with a barometer.
Some of the units associated with the
measurement of pressure are:
 mm Hg (millimeters of mercury)
 torr
 atm. (atmospheres)
 bar (we will not use this one)
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Converting from one unit to another
1 atm. = 760 mm Hg = 760 torr
Ex.1 Convert 752.1 mm Hg to atm.
752.1 mmHg x 1 atm. =0.989= 0.99
760 mmHg
atm.
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Boyle’s Law This gas law states that as the
pressure on a gas is increased, its
volume decreases.
Pressure is inversely proportional
to volume.
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P1V1=P2V2
If the pressure of a gas is
changed while the temperature is
held constant, the new volume
can be calculated, using the
above equation, (Boyle’s law).
Or…
If the volume is changed the new
pressure can be calculated.
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Ex. 2
Boyle’s Law
A sample of Chlorine gas occupies
946 mL at a pressure of 726
mmHg. What is the pressure of the
gas if the volume is reduced to 154
mL at constant temperature ?
946 mL x 726 mmHg = P2 x 154 mL
P2 = 4.46 x 10 3 mm Hg
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Charles’s Law This gas law states that as the
temperature of a gas is increased,
the volume of the gas increases
when pressure is constant.
Volume is directly proportional
to temperature.
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V1/T1= V2/ T2
If the temperature of a certain
volume of gas is changed while the
pressure is held constant, the new
volume can be calculated using
the above equation, (Charles’s
law).
If the volume is changed the new
temperature can be calculated.
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Ex. 3
Charles’s Law
452 mL of F2 is heated from 22.0o C
to 187o C at constant pressure.
What is the final volume ?
PROBLEM ! ALL GAS
CALCULATIONS MUST BE DONE
USING ABSOLUTE
TEMPERATURE, KELVIN.
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Convert the
oC
to kelvin...
K = 22.0 oC + 273 = 295 K
K = 187
oC
+ 273 = 460 K
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V1/ T1= V2/ T2
452 mL / 295 k = V2 / 460 k
V2 = 705 mL
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The combined gas law This is a combination of the last
two laws and one other law.
P1 V1 / T1 = P 2 V 2 / T 2
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Ex. 4
Combined gas law -
A small bubble with volume of 2.1
mL at the bottom of a lake where
the pressure is 6.4 atm and the
o
temperature is 8.0 C rises to the
surface. The surface pressure is
1.0 atm. and the temperature is
25oC. Calculate the new volume.
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It is helpful to write out the
variables...
P1 = 6.4 atm.
P2 = 1.0 atm.
V1 = 2.1 mL
V2 = ?
T1 = 8.0 + 273
T2 = 25 + 273
T1 = 281 k
T2 = 298 k
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Rearrange to solve for V2
V 2 = V 1 x P 1 x T2
P2x T1
V2 = 2.1 atm. x 6.4 atm x 298 k
1.0 atm. x 281 k
V2 = 14 mL
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THE IDEAL GAS LAW This equation explains how a
gas would behave if it were an
ideal or perfect gas. Real
gases are not ideal but at high
temperatures and low pressure
they follow this equation fairly
closely.
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The Ideal Gas Law -
The ideal gas law accounts for
moles, pressure, volume,
temperature, and uses a
proportionality constant called
the gas constant, R.
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PV=nRT
The gas constant, R, has the
value, 0.0821 L atm / mole k
These units will cancel with the
undesired units in the calculation.
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Ex. 5
Ideal Gas Law -
Calculate the pressure (in atm)
exerted by 1.82 moles of SF6 gas
in a steel vessel with a volume of
o
5.43 L at 69.5 C.
Note: When using the ideal gas
law, you must use units of L, atm.,
and K to match the units in R.
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Ex. continued...
Write out the variables...
P=?
V = 5.43 L
n = moles = 1.82 moles
T = 69.5oC +273
=342.5 k
R = 0.0821 L atm / mole k
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Rearrange to solve for P.
P = nRT
V
P=(1.82 mol)(0.0821 L atm /mol k)(342 k)
5.43 L
P = 9.42 atm
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Dalton’s Law Of Partial Pressures
This law states that the sum of the
individual pressures of the gases in
a mixture of gases is equal to the
total pressure of the mixture.
PT = P1 + P2 +... Pn
and...
Pi = xi (PT)
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P i = xi (P T )
In this equation the ‘ i ’ stands for
the individual gas, the ‘ x ’ stands
for the mole fraction of the
individual gas in the mixture.
mole fraction is the desired gas
moles, divided by the total moles of
gas.
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Ex. 6 Dalton’s Law of Partial
Pressures
P i = x i (P T )
and
Xi = moles i
moles i + j + k +…
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Ex. 6 continued...
A mixture of gases contains
4.46 moles of Ne, 0.74 moles
Ar, and 2.15 moles Xe.
Calculate the partial pressures
of the gases if the total
pressure is 2.00 atm.
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First calculate the mole fraction
XNe = nNe
= 0.607
nNe +n Ar + n Xe
The same procedure is followed
to find the mole fractions of Ar and
Xe. They are 0.10 and 0.293
respectively.
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Finally, Pi = Xi (PT)
PNe = 0.607 x 2.00 atm = 1.21 atm
PAr = 0.10 x 2.00 atm = 0.20 atm
PXe = 0.293 x 2.00atm = 0.586 atm
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Ex. 7
PT = P1 + P2 + ...Pn
Calculate the partial pressure of
oxygen gas collected through
water if the total pressure is 762
mm Hg and the water vapor
pressure is 22.4 mm Hg.
POxygen = PT - P water = 740 mm Hg
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Gas Stoichiometry
An important piece of information;
At STP, 1 mole of gas
occupies 22.4 L
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STP
STP is defined as ...
STANDARD TEMPERATURE
AND PRESSURE
STANDARD PRESSURE IS
EQUAL TO 1 atm. STANDARD
TEMPERATURE IS EQUAL TO
0oC OR 273 k.
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22.4 L x grams
mole
liter
With this information, the molar
mass of a gas at STP can be
calculated, given the density.
or
The density can be calculated,
given the molar mass.
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Ex. 8
Molar Mass of a Gas
The density of a gas at STP is
equal to 0.761 g/L. Calculate its
molar mass.
0.761 g x 22.4 L = 17.04 g / mole
Liter
mole
This gas would be ammonia.
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Molar Mass and Density
If given the quantity, in grams, the
Ideal gas law can be used to
calculate the molar mass of a gas
when not at STP.
Rearrange... n = PV / RT
Solve for n, then divide grams /n
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n = PV / RT
Ex. 9 Calculate the molar mass of
a 0.100 g sample of a gas that fills
0.0470 L at 298 k and 0.993 atm.
n = 0.993 atm x 0.0470 L =0.00191
0.0821 L atm
298 k
mol
mole k
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Ex. 9 continued...
0.100 g
=
0.00191 mole
52.4 g/mol
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Gas Stoichiometry
At STP, the coefficients in a
balanced chemical equation
are related to the volumes of
gas that react.
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Ex. 10
Stoichiometry
Calculate the volume of O2 at STP,
required for the complete
combustion of 2.64 L of acetylene
(C2H2) at STP.
Remember the first step in
stoichiometry is to write the balanced
equation.
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2C2 H2 (g) + 5 O2 (g) ----->4CO2 (g) + 2H2O(l)
volume of O2 =
2.64 L C2 H2 x 5 L O2
2 L C 2 H2
= 6.60 L O2
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Ex. 11
Stoichiometry
Sodium Azide (NaN3) is used in
some automobile air bags. The
NaN3 is decomposed when the
collision impact triggers the
reaction. Calculate the volume of
o
N2 produced at 21 C and 1.08 atm
for 60.0 g of NaN3.
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2 NaN3 (s) ----> 2 Na (s) + 3 N2(g)
Since the reaction is NOT at
STP we must use the Ideal Gas
Law.
We were given T and P and
grams, but not moles, n.
We must first find the moles in
order to use the Ideal gas law to
find volume.
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moles N2 =
=60.0 g NaN3 x 1 mol NaN x
65.02g NaN3
3 molN2
2mol NaN3
= 1.38 mol N2
continued..
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continued...
V = nRT / P
= (1.38 mol)(0.0821Latm)(294 k)
1.08 atm mol k
= 30.8 L N2
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Diffusion and Effusion
Diffusion is the mixing of the
molecules of one gas with the
molecules of another gas.
Ex. An open bottle of ammonia in
a room will soon diffuse through
the air in the room. The vapors will
be noticeable in a short time.
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Rate of Diffusion
A lighter gas will diffuse more
quickly than a heavier gas. For
instance, NH3 gas will diffuse
about twice as fast as HCl gas.
mass of NH3 = 17.04 g / mol
mass of HCl = 36.46 g / mol
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Effusion
Effusion is defined as the
process in by which a gas
under pressure escapes out of
a container through a small
opening.
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Rates of both diffusion and effusion
can be determined by...
Graham’s Law of Diffusion -
The rates of diffusion for gases
are inversely proportional to the
square roots of their molar
masses.
r 1 / r 2= M 2 / M 1
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Ex. 12
Rate of Diffusion
Compare the effusion rates of
helium and oxygen gas at the
same temperature and pressure.
rHe / rO =
2
32.00 g / mol 4.003 g/mol
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Ex. 12
continued...
= 2.827:1 is the rate of He
compared to oxygen.
The Helium effuses 2.827 times
faster than the oxygen.
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Liquids
Some properties of liquids are;
 They are not very compressible
 liquids have one free surface
 liquids are fluid
 liquids have a definite volume
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Liquids are not compressible...
This is because though their
particles are loosely held
enough to be able to flow past
one another, they are still very
close together.
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Liquids have one free surface
A liquid must be surrounded
on all sides but the top or it
will flow out of its container.
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Surface Tension
Liquids have surface tension; some
more so than others. Water, for
instance, seems to almost have a
skin on the surface. This is due to
the very strong intermolecular
forces that hold like molecules
close together, (COHESION).
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Comparing surface tensions...
Water has a relatively high
surface tension. Methylene
chloride, (CH2Cl 2) a non polar
molecule has a relatively low
surface tension. Why ?
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It all has to do with intermolecular
forces.
Water has hydrogen bonding, a
very strong type of intermolecular
force of attraction holding its
molecules together. Methylene
chloride is non polar and thus has
only London Dispersion forces
which are very weak forces.
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Vapor Pressure
The pressure of a gas above a
liquid is the vapor pressure.
The weaker the intermolecular
forces of the liquid, the more
molecules will be able to escape as
vapor, thus, the higher the vapor
pressure.
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The Boiling point of a liquid
As temperature increases, the
kinetic energy of the liquid
increases. More and more
vapor molecules escape, thus
increasing the vapor pressure.
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Boiling Point
When the vapor pressure
above the liquid is equal
to the atmospheric
pressure, it is at the
boiling point.
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General Rules On Comparing
Boiling Points
Particles with the stronger
intermolecular forces have higher
boiling points:
Ionic > Hydrogen bonded > Polar
Covalent > Non polar covalent
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If the intermolecular forces are the
same type...
Then the heavier particle will
have the higher boiling point.
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If they also have the same
weight...
...as in two hydrocarbons in
which one of them is straight
and one is branched, the
straight chain will have the
higher boiling point.
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Why ???
Because the straight, less
bulky particles can get closer
together to have more surface
contact for the London forces
to take effect.
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Circle the one with the higher b.p.
and explain why ?
 HF
or HCl
 HCl or C 2 H 6
 C 2 H6 or C 3 H 8
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Answers...
 HF - Hydrogen bonding vs.
polar covalent (dipole - dipole)
 HCl - dipole dipole vs. non polar
covalent
 C 3 H 8 - Heavier molecule wins
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Remember, or else !
 Properties
of gases and liquids - explain
 Gas laws - define and put to use
 Gas stoichiometry
 Diffusion and Effusion - define and use
 Vapor pressure - define
 Boiling point - define, compare
substances for higher boiling point
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