The Gaseous Phase
The three phases of matter, solids, liquids and gases, have
different characteristics.
A gas expands to fill any container it occupies
A liquid has a fixed volume but takes the shape of the volume
of the container it occupies
A solid has both fixed volume and shape.
These characteristics originate from the nature of the
interactions between the atoms or molecules
On a macroscopic scale, gases are distinguished from solids
and liquids by their much smaller values of density.
On the microscopic scale, the smaller values of density arise
due to the much lower NUMBER DENSITY (number of
molecules per cm3 of the sample) compared with liquids and
solids.
Understanding the behavior of gases and how reactions
occur in the gas phase is of practical importance
CH4(g) + O2(g) --> CO2(g) + H2O(g) - combustion of fuels
N2(g) + H2(g) --> NH3(g) - production of ammonia for fertilizers
2NO(g) + O2 (g) -> 2NO2 (g) - responsible for acid rain
Properties of Gases
- A gas will fill the volume of the container which contains it.
- The volume of the gas equals the volume of its container
- Gases are highly compressible; when pressure is applied
to a gas, its volume readily decreases
- Gases form homogenous mixtures with each other
regardless of their identity or relative proportions
These properties arise because the individual atoms/
molecules are relatively far apart
Three properties of gases that are used to describe gases are
pressure (P), volume (V) and temperature (T).
The volume of a gas is defined by the volume of the
container.
Typical units for volume of gases is the liter, L.
PRESSURE
The force exerted by a gas on a unit area of the walls of its
container is the pressure exerted by the gas.
Pressure = Force
Area
SI Units for pressure
Force is newton, N (=kg m/s2)
Area - m2
Pressure - N/m2 or pascal (Pa)
Atmospheric Pressure
Because of gravity, the earth’s atmosphere exerts a
downward force and consequently a pressure on the earth’s
surface.
Atmospheric pressure: pressure exerted by the atmosphere
around us
A column of air 1m2 in cross
section extending through the
atmosphere has a mass or
roughly 10,000 kg.
The acceleration produced by earth’s gravity is 9.8 m/s2
force = mass x acceleration
Force exerted by this air column is ~ 1 x 105 N
The pressure exerted by this air column ~ 1 x 105 N
1 m2
~ 1 x 105 Pa
More precisely, 1.01325 x 105 Pa = 1 atmosphere (atm)
A barometer operates on the principle that the height of a
liquid in a closed tube depends on the atmospheric pressure
Pressure = g h d
g ~ 9.8 m/s
h is the height of the liquid in the sealed tube
d is the density of the liquid
a) What is the height of a mercury column in a barometer at
atmospheric pressure?
b) What is the height of a water column in a barometer at
atmospheric pressure?
Explains why mercury is used in barometers and not water
Units of pressure
1 atm = 760 mm Hg = 760 torr = 1.01325 x 105 Pa
There are other units of pressure (lbs/in2, bar) but we will
typically deal with atm, mm of Hg or torr and Pa.
The Gas Laws
Through experimental observations, relationships have been
established between the pressure (P), temperature (T) and
volume (V) and number of moles (n) of gases.
These relationships are called the GAS LAWS.
Having defined P, V, T, and n for a gas, this information
defines the physical condition or state of a gas.
The relationships between P, V, T and n that will be discussed
hold for IDEAL gases
(or for low pressures; “ideal” conditions)
Relationship between Pressure and Volume: Boyle’s Law
Boyle noted from the experiments he performed that at a fixed
temperature and for a fixed amount of gas, as pressure on a
gas increases, the volume occupied by a gas decreases.
Pa 1
V
P = constant
Volume
Boyle’s Law
The product of pressure and volume of a sample of gas is a
constant, at constant temperature and for a fixed amount of
gas.
P V = constant
At a fixed temperature and
for a fixed amount of gas
Plot illustrating P-V relationship
The conditions of 1.00 atm pressure and 0oC are called
standard temperature and pressure (STP).
Under STP conditions, the volume occupied by the gas in the
J-tube is 22.4 L.
Since, PV = constant
P1V1 = P2V2
Temperature-Volume Relation - Charles Law
The volume of a fixed quantity of gas at constant pressure
increases linearly with temperature.
V = V0 + a V0 t
V0 is the volume of the gas at 0oC
t is the temperature in oC
a is the coefficient of thermal expansion
Volume
V = V0 + a V0 t
y = mx + b
From a plot of V vs t we can determine V0 from the yintercept.
From the slope = a V0, a can be determined
Volume
Since gases expand by the same relative amount when
heated between the same two temperatures (at low
pressure) implies that a is the same for all ideal gases.
For gases, at low pressure
a=
1
(oC-1)
273.15
For liquids and solids a varies from substance to substance
Re-writing the expression connecting V and t:
t=
V
V0
-1
a
t = 273.15 oC [
V
V0
-1 ]
Gas thermometer: By measuring the volume of a gas at 0oC
and measuring the volume change as temperature
changes, the temperature can be calculated
Absolute temperature - Kelvin Scale
V = V0 [ 1 +
t
]
o
273.15 C
At t = -273.15 oC => volume of gas is zero
temperatures < -273.15oC => negative volume which is
physically impossible.
Hence 273.15oC is the lowest temperature that can be
physically attained and is the fundamental limit on
temperature.
This temperature is called ABSOLUTE ZERO and is
defined to be the zero point on the Kelvin scale (K)
T (Kelvin) = 273.15 + t (Celsius)
If we substitute the above expression in
V = V0 [ 1 +
t
]
o
273.15 C
and solve for V
V0
T
V=
273.15
V0
is a constant
273.15
Hence, V a T
Charles’ Law
In other words, on an absolute temperature scale, at a
constant pressure and for a fixed amount if gas, the volume
of the gas is proportional to the temperature
Hence,
V1
T1
=
V2
T2
at a fixed pressure and for a
fixed amount of gas
Note: T is temperature in Kelvin
Quantity-Volume relation - Avogadro’s Law
Volume is affected not just by pressure and temperature, but
also by the amount of gas.
Avogadro’s hypothesis - Equal volumes of gases at the same
temperature and pressure contain the same number of
molecules.
Avogadro’s law: the volume of a gas maintained at constant
pressure and temperature is directly proportional to the
number of moles of gas.
V = constant x n
Hence, doubling the moles of gas will cause the volume to
double (as long as T and P remain constant)
The Ideal-Gas Equation
Boyle’s law: V a P-1
(constant n, T)
Charles’ law: V a T
(constant n, P)
Avogadro’s law: V a n
(constant P, T)
Putting the three laws together:
Va
nT
P
V=R nT
P
P V = n R T IDEAL GAS EQUATION
An ideal gas is a gas whose pressure, volume and
temperature behavior is completely described by this
equation.
R is called the universal gas constant since it is the same for
all gases.
Note: The ideal gas equation is just that - ideal. The equation
is valid for the most gases at low pressures. Deviations from
“ideal” behavior are observed as pressure increases.
The value and units of R depends on the units of P, V, n and T
Temperature, T, MUST ALWAYS BE IN KELVIN
n is expressed in MOLES
P is often in atm and V in liters, but other units can be used.
Values of R
Units
L-atm/(mol-K)
cal/(mol-K)
J/(mol-K)
m3-Pa/(mol-K)
L-torr/(mol-K)
Numerical value
0.08206
1.987
8.314
8.314
62.36
Example:
Calcium carbonate, CaCO3(s), decomposes upon heating to
give CO2(g) and CaO(s). A sample of CaCO3 is decomposed,
and the CO2 collected in a 250. mL flask. After the
decomposition is complete, the gas has a pressure of 1.3
atm at a temperature of 31oC. How many moles of CO2 were
generated?
Given:
Volume of CO2 = 250 mL = 0.250 L
Pressure of CO2 = 1.3 atm
temperature of CO2 = 31oC
First convert temperature to K
T = 31 + 273 = 304 K
To calculate n:
n= PV
RT
Based on the units given for P and V, use appropriate value
for R
R = 0.08206 L-atm/(mol K)
n=
(1.3 atm)(0.250 L)
(0.08206 L-atm/(mol-K)) (304K)
n = 0.013 mol CO2
Problem
The gas pressure in a closed aerosol can is 1.5 atm at 25oC.
Assuming that the gas inside obeys the ideal-gas equation,
what would the pressure be if the can was heated to 450oC?
Since, the can is sealed, both V and n stay fixed.
P (atm)
t (oC)
T(K)
Initial
1.5
25
298
Final
?
450
723
P = n R = constant
T
V
Pi =
Ti
Pf
Tf
Pi Tf = Pf
Ti
Pf = 3.6 atm
Molar Mass and Gas Density
The ideal gas law, P V = n R T can be used to determine the
molar mass of gaseous compounds.
The number of moles of a compound =
mass of gas sample (m)
Molar Mass (M)
n= m
M
Substituting this in the ideal gas equation
PV= m RT
M
Solving for M, the molar mass
M=
mRT
PV
The ideal gas equation can be also be used to determine
the density of the gas
PV= m RT
M
d=
PM
RT
Gas Stoichiometry
If the conditions of pressure and temperature are known, then
the ideal gas law can be used to convert between chemical
amounts i.e. moles, and gas volume.
Hence in dealing with chemical reactions involving gases, we
can deal with volumes of gases instead of moles of gases,
being that volume is usually an easier quantity to measure.
Problem
Dinitrogen monoxide, N2O, better known as nitrous oxide or
laughing gas, is shipped in steel cylinders as a liquid at
pressures of 10 MPa. It is produced as a gas in aluminium
trays by the decomposition of ammonium nitrate at 200oC.
NH4NO3(s) --> N2O(g) + 2H2O(g)
What volume of N2O(g) at 1.00 atm would be produced from
100.0g of NH4NO3(s) after separating out the H2O and cooling
the N2O gas to 273K. Assume a 100% yield in the production
of N2O(g) .
Assuming a 100% yield => all the NH4NO3(s) is converted to
N2O(g)
Moles of NH4NO3(s) decomposed = 100.0 g/80.04g/mol
= 1.249 mol
Hence, 1.249 mol of N2O(g) formed.
To calculate the volume of N2O(g) produced, use the ideal gas
equation.
V = n R T/ P
V =[ (1.249 mol) (0.08206 L-atm/(mol-K)) (273)]/(1.00 atm)
= 28.0 L N2O(g)
Mixtures of Gases
Can we use the ideal gas equation to determine the
properties of gases in a mixture?
Dalton observed that the total pressure of a mixture of gases
equals the sum of the pressures that each would exert if each
were present alone.
The partial pressure of a gas in a mixture of gases is
defined as the pressure it exerts if it were present alone in
the container.
Dalton’s law states that the total pressure is the sum of the
partial pressures of each gas in the mixture.
For example, consider a mixture of two gases A and B in a
closed container
Assuming that the pressure is low enough, A and B obey the
ideal gas equation.
The fact that A and B behave as ideal gases implies that A
and B do not interact with each other.
The pressure exerted by A, PA is then
PA = nA RT
V
and that exerted by B is:
PB = nB RT
V
From Dalton’s law:
Ptotal = PA + PB
nA RT
nB RT
=
+
V
V
= (nA + nB) RT
V
= ntotal
RT
V
Where ntotal = nA + nB is the total number of moles
Mole Fraction
What is the fraction of the number of moles of A in the
mixture?
To find this out, we need to divide the number of moles of A
by the total number of moles of gases in the mixture
The quantity
nA
is called the MOLE FRACTION of A, XA
ntotal
Note: mole fraction is unitless since it is a fraction of two
quantities with the same unit.
Also, sum of mole fractions of the components in a mixture
=1
For the component A in the mixture, we can write its pressure
as
PA = nA RT
V
The total pressure is Ptotal = ntotal
Dividing PA/Ptotal
PA
nA
= n
= XA
total
Ptotal
Hence, PA = XA Ptotal
RT
V
A study of the effect of certain gases on plant growth
requires a synthetic atmosphere composed of 1.5 mol
percent of CO2, 18.0 mol percent of O2 and 80.5 mol percent
Ar.
a) Calculate the partial pressure of O2 in the mixture if the
total pressure of the atmosphere is to be 745 torr?
b) If this atmosphere is to be held in a 120-L space at 295 K,
how many moles of O2 are needed?
a) PO2 = XO2 Ptotal = (0.180) (745 torr) = 134 torr
b) PV = n R T
(134 torr) (1 atm) (120 L) = n (0.08206 L-atm/molK)(295 K)
(760 torr)
n = 0.872 mol
Kinetic Theory of Gases
The ideal gas equation describes how gases behave.
In the 19th century, scientists applied Newton’s laws of
motion to develop a model to explain the behavior of gases.
This model, called the kinetic theory of gases, assumes that
the atoms or molecules in a gas behave like billiard balls.
In the gas phase, atoms and molecules behave like hard
spheres and do not interact with each other.
Assumptions of Kinetic Theory of Gases
1) A gas consists of a large number of particles that are so
small compared to the average distance separating them, that
their own size can be considered negligible.
2) The particles of an ideal gas behave totally independent,
neither attracting nor repelling each other.
3) Gas particles are in constant, rapid, straight-line motion,
incessantly colliding with each other and with the walls of the
container. All collisions between particles are elastic.
4) A collection of gas particles can be characterized by its
average kinetic energy, which is proportional to the
temperature on the absolute scale.
Gas particles are constantly colliding with each other and
the walls of the container.
It is the collisions between the gas particles and the walls of
the container that define the pressure of the gas.
Every time a gas particle collides with the wall of the
container, the gas particle imparts its momentum to the wall
momentum = mass x velocity
The pressure exerted by the gas is proportional to the
momentum of the particle and the number of collisions per
unit time, the collision frequency.
Pressure a (momentum of particle) x (rate of collisions with
the wall)
The rate of collision is proportional to the number of particles
per unit volume (N/V) and the speed of the particle (u).
P a (m x u) x [ N x u]
V
P a N m u2
V
P V a N m u2
The speed, u, is the average speed of the particles, since not
all the particles move with the same speed.
Replace u2 with the mean-square speed, u2
P V a N m u2
Particles are moving in a 3-D space:
P V = 1 N m u2
3
Comparing this equation with the ideal gas equation
PV=nRT
n R T = 1 N m u2
3
This equation relates the speed of the gas particles with the
temperature of the gas
where No is Avogadro’s number
n No = N
n R T = 1 n No m u2
3
RT =
1 N m u2
o
3
(m No) is the molar mass of the gas, M
RT =
u2
1 M u2
3
3RT
=
M
The mean square speed depends on
T and M
Temperature and Kinetic Energy
The kinetic energy of a particle is = 1 m u2
2
The average kinetic energy of a mole of particles is =
1 No m u2
2
From the equation: R T =
1 N m u2
o
3
3 R T = No m u2
2
1
N
m
u
o
Average kinetic energy per mole of particles =
2
Average kinetic energy per mole of particles = 3 RT
2
Hence the average kinetic energy of the molecules in a gas
depends only on the temperature of the gas and is
independent of the mass or density of the gas.
The kinetic theory of gases explains the observed behavior of
ideal gases; i.e. it explains, the three gas laws
Boyle’s Law:
P V = 1 N m u2
3
The pressure exerted by a the gas is due to the collisions
between the particles and the walls of the container.
If the temperature stays the same, then the average speed of
the particles is the same
The pressure will depend on the number of collisions per unit
area of the wall per unit time.
Reducing the volume of the container, will result in more
frequent collisions and hence a higher pressure.
Charles’ Law:
P V = 1 N m u2
3
u2 depends on T. As T increases, u2 increases.
With increasing temperature, the number of collisions with
the walls of the container must increase as u2 increases.
If the P and n are kept constant, the volume of the container
must increase with increasing T.
Avogadro’s Law:
P V = 1 N m u2
3
At fixed pressure and temperature, the volume of the gas is
proportional to the number of particles.
This is explicit in the equation above, for fixed P and T.
Distribution of Molecular Speeds
Atoms or molecules in a gas do not all travel at the same
speed.
There is a distribution of speeds, with some travelling slow,
others fast, and the majority peaked about a value called the
most probable speed.
A plot of the number of molecules travelling at a given speed
versus speed, at a fixed temperature is called the MAXWELLBOLTZMANN distribution.
m 3/2 2
mu2
f(u) = 4(
) u exp(
)
2kT
2kT
ump u
urms
ump =
u=
2RT
M
8RT
M
urms =
3RT
M
The root mean square speed, is the square root of the
square of the average speed
Root mean square speed : sum the squares of the speeds,
divide by the number of particles and then square root the
resulting number.
Average speed: divide the sum of the speed of all particles
by the number of particles.
1, 2, 3, 4, 5, 6
Average = (1+2+3+4+5+6)/6 = 3.5
RMS value = ((1+4+9+16+25+36)/6)1/2 = 3.89
The rms value is always slightly higher than the average
value.
For the same molecule, as temperature increases the most
probable speed shifts to higher values; the distribution also
broadens.
For two gases at the same temperature, but different masses:
the rms speed for the lighter gas is higher than that for the
heavier gas particles.
2
urms = u =
3RT
m
The average kinetic energy of all gases at the same
temperature is the same, regardless of mass.
Temperature describes a system of gaseous molecules only
when their speed distribution is represented by the MaxwellBoltzmann distribution.
A collection of gas molecules whose speed distribution can
be represented by a Maxwell-Boltzmann distribution is said
to be at thermal equilibrium.
Motion of Gas Molecules- Diffusion & Effusion
Gas molecules do not travel in a straight line, but undergo a
more random type motion.
Each time a gas molecule collides with another its direction
changes.
The average distance covered by a gas molecule between
two collisions is the mean free path.
Lower the gas pressure, longer is the mean free path.
If the pathway of a gas molecule from point A to B is tracked,
its path would look like this:
This type of irregular motion is called DIFFUSION and is
responsible for gases mixing; like an the odor filling up a
room,
The rate of diffusion depends inversely on the mass of the
molecule; heavier molecules diffuse more slowly than
lighter molecules.
rate of diffusion of A = √MB
rate of diffusion of B √M
A
Effusion
Effusion is the motion of gas molecules through a small hole.
Within the container, each gas molecule undergoes the
random motion, colliding with other gas molecules.
During this process, if the gas molecule encounters the hole
in the container it will emerge out of the container.
Although each particle traces its own unique path to the hole,
the faster the molecules move, the more quickly will they
emerge from the hole.
The rate of effusion is proportional to urms.
For a mixture of two different gases, A and B, in the same
container, and hence at the same temperature and pressure,
the rate of effusion for each depends on urms of each.
(3 R T/MA)1/2
rate of effusion of A = urms(A) =
rate of effusion of B
urms(B)
(3 R T/MB)1/2
rate of effusion of A = √MB
rate of effusion of B √M
A
Isotope separation by Diffusion
Rate of diffusion of a gas is inversely proportional to the
square root of the mass
Light atoms diffuse through a porous barrier faster than
heavier atoms.
235U
is fissile, not 238U
Natural U is about 99.28% 238U and 0.72% 235U
For a nuclear reactor ~ 10% 235U
For a weapons-grade ~ 90% 235U
To enrich U in 235U one of the ways is to take advantage of
the different diffusion rates of 235U vs 238U
Need a gas; use gaseous UF6
The enrichment factor, is theoretically 0.43%, but in practice
only about 0.14%
To produce 99% uranium-235 from natural uranium ~ 4000
stages are required. The process requires the use of
thousands of miles of pipe, thousands of pumps and motors,
and intricate control mechanisms.
The biggest obstacle was finding a suitable material for the
"porous barrier" that was able to withstand the corrosive
properties of the uranium gas - one of the contributions of
the Manhattan Project at Columbia
Note: other methods of enriching U with 235U were also used
K-25, Oak Ridge National Lab
4 stories high and almost a half mile long; enclosed ~2
million square feet of space, making it the largest building
in the world at the time. The eventual cost of the K-25
complex > $500 million.
Real Gases: Deviations from Ideal Behavior
The fact that gases can liquefy at low temperatures or high
pressures, indicate that gases do not behave “ideally” over
all ranges of temperature and pressure.
Gases liquefy because of interactions between molecules
become important as the molecules come closer together.
For Boyle’s law to hold a gas must never liquefy; it must
remain a gas at all pressures.
This means that there must be no interactions between gas
molecule.
Temperature at which
He(g) condenses to He(l) ~ 4 K
Ar(g) condenses to Ar(l) ~ 87 K
When a gas is compressed by an increase in pressure and
corresponding decrease in volume, gas molecules are forced
closer together.
As the pressure increases, the amount by which the gas can
be compressed decreases because of the finite volume
occupied by each gas molecule.
For Boyle’s law to be valid over all ranges of pressure means
that gas molecules must have zero volume
The ideal gas equation is important in determining limiting
values of pressure, volume, and is a useful description of the
behavior of gases at low pressures and high temperatures.
Deviations from ideal behavior can be quantified by a
compressibility factor, Z:
PV
=Z
nRT
If Z = 1, the gas behaves as an ideal gas
The further the compressibility factor is from 1, the greater
the deviation of the gas from an ideal gas
Nitrogen
Equation of State for Real Gases: van der Waals Equation
Real gases:
1) Particles of a real gas occupy space
2) Attractive and repulsive forces do exist between gas
molecules.
The van der Waals equation of state accounts for the real
behavior of gases.
The ideal gas equation: P V = n R T
must be modified to account for the non-zero volume of each
gas molecule, and the interactions between gas molecules.
Accounting for Volume
Because of the non-zero volume of each gas molecule, the
volume available to a gas molecules is less than the volume
of the container by V- nb
b is the volume occupied by 1 mole of gas molecules (L/mol)
n is the number of moles of gas (mol)
V is the volume of the container.
Accounting for Pressure
Since real gas molecules interact with each other, the
observed pressure is lower than the ideal gas pressure.
n2a
The term P +
accounts for the effect of interactions
V2
between molecules on the pressure of the gas
Accounting for real gas behavior results in van der Waal’s
equation of state
n2a
(P +
) (V - n b) = n R T
2
V
The van der Waal’s constants, a and b, are different for
different gases.
The constant b is related to the size of the gas particle.
Larger the value of b, larger is the particle.
Atom/Molecule
b (L/mol)
Ar
0.03219
Cl2
0.05622
He
0.02370
H2
0.02661
The magnitude of the constant a is a measure of the
attractive forces between molecules.
Gases with larger a values liquefy or solidify more easily
than gases with smaller a values since the attractive forces
between molecules are strong.
Atom
a (L2atm/(mol2))
boiling pt (K)
Ar
1.345
87.3
He
0.03412
4.2
In general atoms or molecules like He and H2 which have
small a and b values exhibit behavior fairly close to ideal.