Properties of gases
Chapter 6
Gases expand to fill and take the shape
of the container they are in.
Gases can diffuse into each other and
mix in all proportions.
We can fully describe the physical state
of a gas by knowing any three of four
variables related to each other by a gas
equation of state.
Gases
Dr. Peter Warburton
[email protected]
http://www.chem.mun.ca/zcourses/1050.php
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State variables of a gas
State variables of a gas
Amount - we express this in moles (mol)
which is related to the number of gas
molecules through Avogadro’s number
(NA = 6.022142 x 1023 mol-1)
Volume - this is the size
of the container holding
the gas. While we often
express volume in liters
(L), this unit for volume is
derived from the base SI
unit for distance, the
meter (m) where
1 L = 1 dm3.
Recall that the molar mass (M) relates the
mass of the gas to the number of moles.
Equal masses of different gases
represent a different number of moles!
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1
State variables of a gas
State variables of a gas
Pressure – the molecules of the gas are in
constant motion and often collide with the
walls of the container and each other.
The collisions exert a net force on the wall
of the container.
The SI unit of force, the newton (N) is
another derived unit based on
force = mass x acceleration
1 N = 1 kg m s-2
Temperature - this is a
reflection of the “average”
energy available to a
molecule in the system.
Recall that in chemistry that
we ALWAYS express
temperature in kelvin (K).
T(K) = t(ºC) + 273.15
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State variables of a gas
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Liquid pressure
Pressure – is then the net force the gas
molecules exert on a given area of the
container wall. The SI derived unit for
pressure is the pascal (Pa)
Pressure = force / area
1 Pa = 1 N m-2 = 1 (kg m s-2) m-2
1 Pa = 1 kg m-1 s-2
Most of the time we would talk about
pressure in kilopascal (kPa)
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We can compare gas
pressure to the pressure
exerted by a liquid,
since assessing the force
exerted by a column of
liquid is easy.
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2
Liquid pressure
Liquid pressure
Imagine a cylinder of liquid, like a
drinking straw.
The liquid in the cylinder will have
a certain total mass (m)
determined by the density (d) of
the liquid and the total volume (V)
of the liquid.
m=dxV
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Now, the force (F) exerted at the
bottom of the cylinder by the liquid
is the mass multiplied by the
acceleration due to gravity (g).
F=mxg
so F = d x V x g
This force is what we think of as the
“weight” of the liquid.
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Liquid pressure
10
Liquid pressure
:and since P = F / A
then P = (d x h x A x g) / A
so P = d x h x g
The pressure depends on
the density of the liquid and
the height it rises up the
cylinder, but NOT on the
area (shape) of the cylinder.
However, the volume of the liquid
is related to the cross-sectional
area (A) of the cylinder and the
height (h) it rises up from the
bottom of the cylinder
V=hxA
so F = d x h x A x g
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3
Measuring gas pressure by liquid pressure
Measuring gas pressure by liquid pressure
A barometer can be used to measure the
barometric pressure (Pbar) of the atmosphere
when it forces a liquid up a cylinder with a
closed end.
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Most barometers use mercury as
the liquid because it is quite
dense. Even so, the atmosphere
of the Earth can push about 760
millimeters of mercury (mmHg)
up the tube.
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Measuring gas pressure by liquid pressure
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Drawback of using liquid pressure
The density of a liquid changes with
temperature and the acceleration of
gravity can change with your
specific location on Earth. We
define 1 mmHg of pressure as the
liquid pressure of
exactly 1 mm of Hg at 273.15 K
and g = 9.80655 m s-2
The atmosphere
pushes mercury
over two and a half
feet up the tube. If
we used water as
the liquid, it would
rise about 34 feet
up the tube!
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4
Problem
Problem answer
Pbar = 760.7 mmHg
The answer should have 3 significant figures
(s.f.) since the liquid height had 3 s.f.
(multiplication/division rule - text page 20).
The subscript 7 is called a “guard digit”. It IS
NOT significant, but would be used to help
avoid rounding errors in later calculations
involving this calculated value of Pbar.
We would report the answer to the correct
number of s.f. as 761 mmHg.
A barometer is filled with diethylene glycol
(d = 1.118 g cm-3) instead of mercury (d =
13.5951 g cm-3). The liquid height is found
to be 9.25 m. What is the barometric
pressure in mmHg?
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Units of pressure
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Units of pressure
Some textbooks treats 1 atm as the commonly
used “standard” unit of pressure.
IT ISN’T!
The standard unit of pressure in chemistry is
1 bar = 105 Pa (exactly) ≈ 0.98692 atm
Pressure can be measured in many
different units. You should practice unit
conversion calculations.
One atmosphere of pressure is 1
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atm
1 atm = 760 Torr ≅ 760 mmHg (to 4 d.p.)
1 atm = 101,325 Pa = 101.325 kPa
(these equalities are “exact” by def’n)
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5
Measuring gas pressure by liquid pressure
Measuring gas pressure by liquid pressure
A manometer can be used to measure a gas
pressure (Pgas) compared to the atmospheric
pressure (Pbar) when it forces a liquid up a
cylinder with an end open to the atmosphere.
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Pgas = Pbar + ∆P
∆P = g x d x ∆h
∆h = hopen - hclosed
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Problem
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Problem answer
Suppose we have a mercury manometer
(d = 13.6 g cm-3) like in Figure 6.5(b) of the
previous slide where the height of mercury
in the open arm is 7.83 mm higher than in
the closed arm. If Pbar is 748.2 mmHg,
then what is Pgas in the manometer?
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Pgas = 756.03 mmHg
The answer should have 4 significant figures
(s.f. to the 1st decimal place) since Pbar had
significant figures to the least decimal places
(addition/subtraction rule - text page 20).
The subscript 3 is a “guard digit”. It IS NOT
significant. It’s good for avoiding rounding
errors in later calculations using Pgas.
We would report the answer to the correct
number of s.f. as 756.0 mmHg.
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6
Boyle’s Law
Boyle’s Law
Therefore for constant n and
constant T
P1V1 = P2V2
For a fixed amount of gas at a constant
temperature, the gas volume is inversely
proportional to the gas pressure.
P Ñ 1/V or PV = a
a is a constant
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Charles’s Law
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Charles’s Law
Therefore for constant n and
constant P
V1/T1 = V2/T2
For a fixed amount of
gas at a constant gas
pressure, the gas
volume is directly
proportional to the
absolute (Kelvin)
temperature.
V Ñ T or V = bT
b is a constant
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7
Standard conditions
Equal volumes – equal numbers hypothesis
To be sure people are talking about the
same conditions when discussing gases,
it’s often useful to define a standard.
When discussing gases, the set of
standard conditions often used are
standard temperature and pressure (STP)
where T = 0 ºC = 273.15 K and P = 1 bar
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Equal numbers
of molecules of
different gases at
the same T and
P occupy equal
volumes.
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Avogadro’s Law
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Molar volume at STP
At a fixed temperature and pressure, the
volume of a gas is directly proportional to
the amount of gas.
If we choose our constant temperature and
pressure to be STP (273.15 K and 1 bar)
then Avogadro’s Law tells us the volume of
some given amount of gas will be a
constant. If we choose 1 mole of gas as
this amount, then the molar volume (Vm)
of a gas at STP is
V Ñ n or V = c n
c is a constant
Therefore for constant T and constant P
V1/n1 = V2/n2
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Vm = 22.711 L mol-1 at STP
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8
Gas law assumptions
Properties of an ideal gas
Charles’s Law ignores the fact that all
gases will eventually condense to a liquid if
the T is low enough, and then eventually
form a solid if the T is lowered further.
Therefore, the gas law behaviors are
actually reflecting an idealized picture of
a gas – the ideal gas!
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33
Combining the gas laws
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34
Value of the gas constant R
If V Ñ 1/P (since P Ñ 1/V)
and V Ñ T and V Ñ n
then
V Ñ nT/P
or V = RnT/P where R is gas constant
Ideal gas law!
PV = nRT
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1) The molecules of the gas occupy no
volume. (They can occupy the same space!)
2) There are no intermolecular forces
between gas molecules. (No attractive or
repulsive interactions!)
3) The gas molecules are in constant random
motion and collide elastically. (No net change in
total energy of the gas!)
4) The gas has achieved a state of equilibrium
(n, P, T and V are all constant!).
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We always measure T in kelvin
and n in moles, but P and V can
often be expressed in many
different units, therefore the gas
constant R can APPEAR to have
many different values depending
on the units of P and V.
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Value of the gas constant R
Value of the gas constant R
Consider the two statements:
I have 1 dozen eggs
AND
I have 12 individual eggs
Both statements say the SAME THING,
but have different numbers appearing in
them due to different units.
UNITS MATTER!
ALWAYS SHOW THEM!
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Therefore, R
always
represents the
SAME constant,
even if the
numbers don’t
always look the
same.
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R in J K-1 mol-1
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R in J K-1 mol-1
We can convert any value of R to any other by
unit conversion:
101325
1
0.0820574
1
1000
= 8.31447 The first four values
of R look straight
forward. We have
just measured P
and V in various
units.
P in atm or bar or
kPa or Pa
V in L or m3
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R in J K-1 mol-1
R in J K-1 mol-1
What about R in J K-1 mol-1?
How do we get from P and V to energy
units?
Recall that
1 Pa = 1 kg m-1 s-2
and
volume can be expressed as a cubic
distance like m3
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Therefore PV can have units of
kg m-1 s-2 m3 = kg m2 s-2
The formula for kinetic energy (Ek) is
Ek = ½ m v2
If we measure
mass in kg and velocity in m s-1
then Ek has units of
kg (m s-1)2 = kg m2 s-2 = J (joule)
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R in J K-1 mol-1
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General gas equation
Since PV = nRT then
R = PV/nT = constant
For any sample of gas undergoing any
sort of change in any of these variables
A pressure multiplied by a volume has
units equivalent to energy units!
1 Pa m3 = 1 J
We’ll see more in the next chapter!
For now consider blowing up a balloon. To
increase the volume of the balloon against
the external atmospheric pressure requires
us to blow air into the balloon.
This takes energy to accomplish.
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= f is “after” the change – final state
i is “before” the change - initial state
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Problem
Problem answer
M = 17.02917 g mol-1 n = 1.186 mol
P = 0.9895 atm T = 248.15 K
R = 0.082057 L atm K-1 mol-1
V = 24.41 L = 24.4 L
In this problem we see the importance of
keeping track of units so we can convert
when necessary, using guard digits to avoid
rounding errors and choosing an appropriate
value of R to use based on the units of our
variables.
What is the volume occupied by 20.2 g of
ammonia [NH3 (g)] at -25 ºC and 752
mmHg?
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Problem
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Problem answer
How many molecules of N2 (g) remain in
an ultrahigh vacuum chamber of 3.45 m3 in
volume when the pressure is reduced to
6.67 x 10-7 Pa at 25 ºC?
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NA = 6.022 x 1023 (N2 molecules) mol-1
T = 298.15 K
R = 8.3145 Pa m3 K-1 mol-1
n = 9.283 x 10-10 mol
= 5.590 x 1014 N2 molecules
= 5.59 x 1014 N2 molecules
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12
Problem
Problem answer
Ti = 309.35 K and Tf = 310.95 K
Vf = 2.109 mL = 2.11 mL
A 1.00 mL sample of N2 (g) at 36.2 ºC and
2.14 atm is heated to 37.8 ºC and the
pressure is changed to 1.02 atm. What
volume does the gas occupy at this final
temperature and pressure?
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Using the ideal gas equation
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Using the ideal gas equation
Finding molar mass (M) of a gas –
since PV = nRT
and M = m/n (or n = m/M)
then PV = mRT/M
leading to
Finding gas density (d) –
since M = mRT/(PV) = (RT/P)(m/V)
but since m/V = d
then M = (RT/P)(m/V) = dRT/P
leading to
d = PM/RT
M = mRT/(PV)
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Problem
Problem answer
MNO = 30.0061 g mol-1
MN2O = 44.0128 g mol-1
Mcalculated = 29.94 g mol-1 = 29.9 g mol-1
The oxide of nitrogen must be NO.
A 1.27 g sample of an oxide of nitrogen,
believed to be either NO or N2O occupies
a volume of 1.07 L at 25 ºC and 737
mmHg. Which oxide is it?
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Problem
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Problem answer
M = 31.98 g mol-1 = 32.0 g mol-1
L-1
The density of a gas sample is 1.00 g
at 109 ºC and 745 mmHg. What is the
molar mass of the gas?
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Gases in chemical reactions
Problem
If we treat any gases in a chemical
reaction of interest as ideal gases, then we
can use the ideal gas equation and the
balanced equation of the reaction to solve
reaction stoichiometry problems, if we
know something about the state variables
of the gas.
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Problem answer
∆
2 →2 + 3" ($)
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Law of combining volumes
n = 0.0403 mol of N2 (g)
in 1 L at the given T and P
m = 0.618 g of Na (l)
per liter of N2 (g) produced
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How many grams of Na (l) are produced
per liter of N2 (g) at 25 ºC and 1.0 bar.
In reactions involving ALL gases (treated
ideally) at a given T and P,
since V Ñ n
then
the ratios of volumes of the gases
match the stoichiometric ratios for
the same gases
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Law of combining volumes
Law of combining volumes
If 2 NO (g) + O2 (g) 2 NO2 (g)
then 2 moles of NO react with 1 mole of O2
to form 2 moles of NO2
(2:1:2 stoichiometry)
but the law of combining volumes tells us
that 2 L of NO will react with 1 L of O2 to
give 2 L of NO2
(still 2:1:2, but in terms of volume)
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Problem
More generally,
if 2 NO (g) + O2 (g) 2 NO2 (g)
then
2Vm NO (g) + 1Vm O2 (g) 2Vm NO2 (g)
where Vm is the molar volume of an ideal
gas at the given T and P of the reaction
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Problem answer
Since balancing the equation shows us that 3
moles of H2 are required to produce 2 moles
of NH3, then the law of combining volumes
tells us that 3 L of H2 will produce 2 L of NH3.
Since we used 225 L of H2 we produced
If all gases are measured at the same
temperature and pressure, what volume of
NH3 (g) is produced when 225 L of H2 (g)
are consumed in the reaction
2/3 (225 L)
= 150 L NH3
∆
" $ + &" ($)→& ($)
not a balanced equation!
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16
Mixtures of gases
Law of partial pressures
Dalton’s law of partial pressures The total pressure of a mixture of (ideal)
gases Ptot is the sum of the pressures of
each gas treated individually (partial
pressure) at the same temperature and
volume as the mixture.
In a mixture of gases A, B, C and D
Ptot = PA + PB + PC + PD
here PA is the partial pressure of A
If we treat all gases in a mixture as ideal,
then by the properties of an ideal gas,
none of the gases behave any differently
from any of the others, and each gas can
be treated separately and the properties of
mixture can be treated as a sum of the
same properties of the individual gases.
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Law of partial pressures
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Law of partial pressures
In a mixture of gases A, B, C and D
Ptot = PA + PB + PC + PD
here PA is the partial pressure of A
Be careful here! Since the mixture
occupies a fixed volume, then
Vtot = VA = VB = VC = VD
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Mixtures of gases
Mixtures of gases
In a mixture of gases A, B, C and D
Vtot = VA + VB + VC + VD
so percent volume of A is
VA/Vtot x 100%
Be careful here! The mixture is at a
fixed total pressure so
Ptot = PA = PB = PC = PD
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In a mixture of gases A, B, C and D
ntot = nA + nB + nC + nD
leads to
mole fraction of A (xA)
nA/ntot = xA
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Mole fraction in ideal gas mixture
70
Problem
)(
,(
(
'( =
=
=
)*+* ,*+* *+*
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A gas mixture is made when 2.0 L of O2
and 8.0 L of N2, each at 0.00 ºC and 1.00
atm, are mixed together. This non-reactive
mixture is then compressed to occupy a
total volume of 2.0 L at 298 K. What is the
pressure of the mixture at these new
conditions?
71
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18
Problem answer
Collecting a gas over water
We can collect a gas
over water (or another
liquid) via a setup like
we see here.
As more gas is
collected, the liquid is
displaced downwards
as the pressure of the
gases in the container
must match Pbar.
nO2 = 0.0892 mol
nN2 = 0.357 mol
ntot = 0.446 mol
Ptot = 5.45 atm = 5.5 atm
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73
Collecting a gas over water
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Collecting a gas over water
We must be careful
though! The gas we’re
interested in IS NOT
THE ONLY GAS IN
THE CONTAINER.
The liquid water can
evaporate to give
gaseous water in the
container as well!
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Therefore
Ptot = Pgas + Pwater
= Pbar
Therefore, for the gas
we’re collecting
Pgas = Pbar - Pwater
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So what is Pwater?
Problem
The partial pressure of
the water that must be
subtracted from Pbar to
get Pgas is called the
vapour pressure of
water.
Vapour pressure
changes as a function
of temperature and
identity of the liquid.
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The reaction of aluminum with hydrochloric
acid produces hydrogen gas:
2 Al (s) + 6 HCl (aq) 2 AlCl3 (aq) + 3 H2 (g)
If 35.5 mL of H2 is collected over water at
26 ºC and a barometric pressure of 755
mmHg, how many moles of HCl must have
been consumed?
Pwater = 25.2 mmHg at 26 ºC
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Problem answer
78
Kinetic molecular theory of gases
nHCl = 2.777 x 10-3 mol = 0.00278 mol
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This is a theory we
use to explain the
behaviour of gases,
based on a model with
certain properties
(compare these to the
description of an
ideal gas in slide 34).
79
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20
Properties of the model
Properties of the model
i. A gas is made up of an extremely large
number of very small particles
(molecules or atoms) moving in
constant, random, straight-line motion.
ii. The distance between any two
molecules is vast in comparison to the
size of the molecules that most of the
space is empty and the molecules can
be treated as point masses (no volume)
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81
Deriving Boyle’s Law
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82
Deriving Boyle’s Law
If a given molecule of mass m is
travelling with a speed ux in the x
direction (speed is just the size
[magnitude] of the vector quantity
velocity) towards a wall perpendicular
to it’s path (that is, the yz plane), then
the force this molecule exerts on the
wall depends on 2 factors:
We’ve seen PV = a or P = a / V
Here the constant a will depend on the
number of molecules N = n x NA
(extremely large number of molecules! –
see note i. on slide 81)
and
the temperature
(see note v. on slide 82)
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iii. Molecules collide fleetingly with each
other and the container walls.
iv. No intermolecular forces EXCEPT
during collisions (no “action at a
distance” of one molecule on another).
v. A given molecule may gain or lose
energy during collisions, but if the
temperature is constant then the
TOTAL energy is ALSO constant.
83
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21
Deriving Boyle’s Law
Deriving Boyle’s Law
Frequency of collisions - measures the
number of collisions of molecules with the
walls of the container in a given time
period. More collisions per unit time
implies a greater net force from collisions:
collision frequency Ñ molecular speed x
Momentum transfer (impulse) – in a
collision the molecule transfers momentum
impulse Ñ molecular speed x mass
impulse Ñ ux x m
Now P Ñ collision frequency x impulse
molecules per unit volume
collision frequency Ñ ux x (N/V)
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85
Deriving Boyle’s Law
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86
Deriving Boyle’s Law
Remember, we have a very large number
of molecules, each with its own ux2. So it
makes sense to talk about the
average mean-square speed in the xdirection is -. " and
Molecules travel in 3 dimensions! So
average mean-square speed is
= 01 2 + 03 2 + 04 2
5
and if 6 02 = 01 2 since01 2 = 03 2 = 04 2 1
, = -"
3
∝ -. "
,
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P Ñ ux x (N/V) x ux x m
Ñ (N/V) m ux2
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22
Distribution of molecular speeds
Distribution of molecular speeds
Here we have a plot of
the percentage
(fraction) of molecules
that have a given
speed.
Notice that no
molecules have zero
speed.
We’ll talk about other
features soon.
We can statistically predict what fraction
F(u) of our N molecules have a given
speed u
9
7 - = 48
28:;
<
"
>?@<
" "A
- =
89
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Distribution of molecular speeds
7 -
90
Distribution of molecular speeds
This figure shows three
different
characteristic speeds
of the distribution.
The most probable
speed um is the speed
at the peak of the
distribution, and
more molecules have
this speed compared
to any other speed.
<"
@
9
>? <"A
= 48
-" =
28:;
As temperature increases
the average speed of any
given molecule increases.
(Constant M)
As the molar mass
increases, the average
speed of any given molecule
decreases. (Constant T)
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Distribution of molecular speeds
Distribution of molecular speeds
The average speed
uav is literally the sum
of the speeds of all
molecules divided by
the total number of
molecules.
um ≠ uav
The root-meansquare speed urms is
found by taking the
square root of the
average mean-square
speed.
because the
distribution curve is
not symmetric
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-BCD =
93
Deriving Boyle’s Law
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94
Deriving Boyle’s Law
Consider
But
1
, = -"
3
, = :; for 1 mol
So
1
:; = ( - "
3
But NA m is the mass of one mole of
molecules – the molar mass M!
3:; = 9- "
For N = NA molecules (so we have 1 mol)
1
, = ( -"
3
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-"
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96
24
Deriving Boyle’s Law
Deriving Boyle’s Law
3:;
= -"
9
Taking the square root of both sides
3:;
= - " = -BCD
9
Heavier molecules (larger M) have a
smaller urms than lighter molecules at the
same T.
The urms of a sample of molecules
increases with the square root of
temperature.
3:;
= -" = -BCD
9
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Problem
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Problem answer
Which has a greater urms at 25 ºC, NH3 or
HCl? Calculate urms for the gas with the
higher urms.
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NH3 has the smaller
molar mass, so it will
have the larger urms.
At 25 ºC
urms (NH3) = 660.8 m s-1
= 661 m s-1
99
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100
25
The meaning of temperature
The meaning of temperature
The kinetic energy of a moving object is
Ek = ½ mv2
Here v is the velocity of the object, but
remember, the speed (u) is just the
magnitude (size) of the velocity, so for a
given molecule, the kinetic energy is
But if we look at the mean-square speed
u2 of a group of molecules then the
average kinetic energy of a molecule is
5
EF = G 02
2
But
1
2
1
2
:; = ( - " = (
- " = ( EF
3
3
2
3
Ek = ½ mu2
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101
The meaning of temperature
102
Diffusion and effusion
3 :;
= EF
2 (
This is why on slide 5 I said the
temperature was a reflection of
the average (translational
kinetic) energy available to a
molecule in the system!
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Because molecules are constantly colliding, it
takes time for them to travel “large” distances,
even if their average speed is “pretty fast”.
Diffusion is the name we give to this
molecular migration due to the random
motion of the colliding molecules.
Effusion is the escape of the randomly
moving gas molecules through a small hole in
the container.
103
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104
26
Diffusion and effusion
Rate of effusion
Since the rate of effusion is directly
proportional to molecular speed, which is
inversely proportional to the square root of
molar mass, lighter molecules diffuse
faster than heavier molecules
rateeffusionA -BCD,(
=
=
rateeffusionB -BCD,U
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105
In general
3:;<
9(
=
3:;<
9U
9U
9(
106
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Time of effusion
Since many gas properties are directly
proportional to molecular speed, which is
inversely proportional to the square root of molar
mass, the ratio of those properties for two gases
is
propertyofA
=
propertyofB
9U
9(
The properties include molecular speed, effusion
rate, distance travelled and amount of gas
effused.
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107
Since the time it takes for a specific amount of
gas to effuse is inversely proportional to the rate
of effusion AND rate is inversely proportional to
the molar mass, then the time of effusion must
be directly proportional to the molar mass. That
is
timeofeffusionA
=
timeofeffusionB
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9(
9U
108
27
Problem
Problem answer
If 2.2 x 10-4 mol of N2 effuse through a hole
in 105 s, then how long would it take for
the same amount of H2 to effuse through
the same hole?
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109
Problem
time = 28.2 s
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110
Problem answer
A sample of Kr(g) effuses through a hole in
87.3 s. The same amount of an unknown
gas escapes in 131.3 s through the same
hole. What is the molar mass of the
unknown gas?
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111
M = 189.5 g mol-1 = 1.90 x 102 g mol-1
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112
28
Nonideal (real) gases
Compressibility factor
The properties of an ideal
First consider the
gas on slide 34 include
molar volume at STP
some properties that
cannot be really true for
gases in reality – namely
that real molecules must
have size and must have
intermolecular interactions
(attractions and
repulsions) with each
other.
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Recall the ideal gas law is
PV = nRT
So for an ideal gas
(PV)/(nRT) = 1 = z
for ALL conditions
We call z the compressibility factor.
However, for real gases at almost any set
of conditions z ≠ 1 due to real behaviour!
113
Compressibility factor
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114
Compressibility factor
Real gases show very close to ideal gas
behaviour at high T and low P (or large V)
where the chance for intermolecular
interactions (attractions and repulsions) is
fleeting due to “fast” moving molecules that
are very far apart on average.
When T is low and/or P is high (or V is low),
the chance for interactions is much greater,
leading to nonideal behaviour.
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115
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116
29
Molecular volume
Attractive intermolecular forces
The volume of the
Here b represents the
container available to a
excluded volume
occupied by
given molecule is
1 mol of the real gas
actually the volume of
molecules
the container MINUS
the total volume
occupied by all the
other molecules.
Vavailable = Vcontainer - nb
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117
The van der Waals equation
)"
+ "
,
B]`?aD[\]
, − )_ = ):;
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Here a represents the
attractive forces There
is a dependence on n2
and V2 because both the
rate and force of
collisions are affected
Plowering = an2/V2
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118
The van der Waals equation
If the ideal gas law is an equation of state for
an ideal gas, then real gases cannot be
described adequately by this equation of
state.
One equation of state used to describe real gas
behaviour is the van der Waals (vdW) equation
Y**BYZ*[\]
Real molecules can
attract each other over
reasonable distances.
These attractions serve
to decrease the rate
AND force of collisions
with the walls, lowering
pressure
119
Rearranging the vdW equation makes it a bit
easier to see the result of intermolecular forces
):;
)"
=
−
, − )_ , "
The attractive forces reduce the pressure, while
the excluded volume (repulsive forces) increase
the pressure compared to an ideal gas at the
same temperature and volume.
On average, ONLY attractive OR repulsive
forces dominate at a given set of conditions.
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120
30
vdW factors a and b
Problem
We see that the values
of a and b increase
with molar mass, for
the most part.
For molecules with
similar molar masses,
like N2 and CO or N2O
and CO2, the less
symmetric molecule
usually has a larger a.
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Given the vdW factors for CO2 from the
table on the previous slide, what is the
pressure and compressibility factor of 1.00
mol of CO2 in a 2.00 L container at 273 K?
121
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122
Problem answer
P = 10.68 bar
z = 0.941
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123
31