# Mixtures of Gases Dalton's Law o

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```					                         Mixtures of Gases
Dalton’s Law of Partial Pressures
All gases respond in the same way to changes in pressure, volume and temperature.
Therefore for our calculations it is unimportant whether all the molecules in the
sample are the same.
A mixture of gases that do not react with one another behaves like a single pure gas.
John Dalton showed how to calculate the pressure of a mixture of gases (1801). His
reasoning was something like this:

The total pressure of the mixture of gases was the sum of the pressures of the
gases A and B if they were alone in the container.
Dalton summarised his observations in his
Law of Partial Pressures:
The total pressure of a mixture of gases is the sum of the partial pressures of
its components:
P = PA + PB +     …

He called the partial pressure of each gas:
The pressure it would exert if it occupied the container alone.
According to kinetic theory, molecules of gas A have the same average kinetic energy the
molecules of gas B, since they have the same temperature. Furthermore, they have no
physical or chemical interaction on each other. Consequently, the act of mixing two or
more gases does not change the average kinetic energy of any of the gases, so each gas
will exert the same pressure that it would exert if it were the only gas in the container.
but
The amount of each gas in the mixture (the number of its moles) will affect its
participation in the total pressure.
If nA mol of gas A and nB mole of gas B are mixed, the total number of moles in
the mixture is (nA+ nB). The ratio of the number of moles of A to the total number
of moles present is called the mole fraction of A, XA:

nA               nA
XA=                     =
nA + nB             ntotal

The fraction of the total pressure that is due to gas A is given by the mole fraction
of A. The partial pressure of A, therefore is

nA
PA= ﴾                   ﴿ Ptotal = XA Ptotal
nA + nB

The partial pressure for gas B

nB
PB= ﴾                   ﴿ Ptotal = XB Ptotal
nA + nB
Notice that the sum of the mole fraction is:
XA + XB = 1

nA                 nB              nA + nB
+                 =                   = 1
nA + nB            nA + nB            nA + nB

Example: A mixture of 40.0 g of oxygen and 40.0 g of helium has a
total pressure of 0.900 atm. What is the partial pressure of oxygen?
Mw of O2 = 32.0 g.mol-1 Mw of He = 4.00 g.mol-1
If it is not very soluble in water, a gas evolved in in the course of a
reaction is frequently collected over water. It is conducted into an
inverted bottle that has been filled with water. The gas displaces the
water, and the collected gas is mixed with water vapour. The total
pressure of the mixture is the sum of the partial pressure of the gas
and the partial pressure of the water vapour.
Example: A 370.0 mL sample of oxygen is collected over water at 23º and a
barometric pressure of 0.992 atm. What volume would this sample occupy dry
and at STP?
Molecular Speeds
Derivation of the Ideal Gas Law from the Kinetic Theory

Consider a gas sample that contains N molecules, each having a mass of m. If this
sample is enclosed in a cube a cm on a side, the total volume of the gas is: V = a3 cm3
The pressure of the gas on any wall is due to the impacts of the molecules on that wall.
The force of each impact can be calculated from the change in momentum per unit time.

a cm

Momentum before the collision : mv
Momentum after the collision : - mv (negative because the direction is changed)
Change in the momentum in each collision: 2mv
Consider the shaded wall and take into account only those molecules moving in the
direction of x axis. A molecule moving in this direction will strike this wall every 2a cm
of its path . If its velocity is v cm/s, the number of collisions with the wall in every
second: v/2a
The total change in momentum per molecule in one second:
the number of collisions . the change in momentum
mv2
(v/2a﴿ 2mv =
a
The total changes in momentum (force) for all the molecules striking the wall in
one of the three possible directions, x in one second:
N              mv2              Nmv2
﴾          ﴿.﴾             ﴿=
3              a                 3a
Pressure is force per unit area. The area of the shaded wall is a2.
The pressure on the wall:
force             Nmv2                  Nmv2
pressure =               =                 / a2 =
area                3a                    3a3

Nmv2                   1
Since V = a3                 P=                or PV =          Nmv2
3V                   3
It can be written:
2         1
PV = ﴾       N﴿﴾       mv2﴿
3       2
Since KE = 1/2mv2 therefore: PV= 2/3N(KE)
The average molecular kinetic energy, KE, is directly proportional to the absolute
temperature, T, and the number of the molecules is directly proportional to the
number of moles, n.
Since N(KE) α nT we can write the equation as:
2
PV α ﴾ ﴿ nT
3
Using a constant R the proportionality will give an equation:

PV = nRT
… Molecular Speeds
Previously we derived the expression
1
PV = ﴾      ﴿ N mv2
3
For one mole of gas, the number of molecules, N, is Avogadro’s number; and N times the mass
of a single molecule, m, is the molecular weight, M:
1
PV = ﴾       ﴿ Mv2
3
For one mole, PV = RT; thus,
1
RT = ﴾       ﴿ Mv2
3
Rearranging and solving for the molecular speed, we obtain
v = √ 3RT/M
The speed, v, in this equation, is the root-mean-square speed. It is the speed of a molecule that
has the average kinetic energy of a collection of molecules at the temperature under
consideration.
To obtain v in m/s, R must be expressed in appropriate units. If M is expressed in g/mol, the
appropriate value of R is 8.3143. 103 g.m2/(s2. K.mol).
The root- mean- square speed of H2 (g) molecule at 0 °C is:
v = √ 3RT/M
v = √ 3[8.3143. 103 g.m2/(s2. K.mol)].273 K/2.016 g.mol-1 = 1.84 . 103 m.s-1
and at 100 °C :
v = √ 3[8.3143. 103 g.m2/(s2. K.mol)].373 K/2.016 g.mol-1 = 2.15 . 103 m.s-1

This speeds are high 4.12 . 103 mile.hr-1 and 4.81 . 103 miles.hr-1. Although a given
molecule travels at a high speed, its direction is continually being changed through
collisions with other molecules.
At STP conditions a H2 molecule, on average, undergoes about 1.4 . 1010 collisions in
one second.
The average distance travelled between the collisions is 1.3 . 10-5 cm; this value is
called mean free path of hydrogen.

Example: At what temperature the root- mean- square speed of N2O (g) equals to the
root- mean- square speed of N2 (g) at 300 K?
The Maxwell Distribution of Speeds

Not all of the molecules of a gas have the same kinetic energy and speed. The speed and
the direction of molecules in a sample of gas change continually. So the molecular speeds
and kinetic energy are distributed over a range.
Distribution of molecular speeds of different gases in different temperatures are
investigated by Scottish scientist James Maxwell. His conclusions are summarised in
some graphs:
Heavy molecules like CO2 travel with speeds close to their average values.
Light molecules such as H2 not only have a higher average speed in the same
temperature, but the speed of many of them are very different from their average speed.

Why hydrogen and helium are very rare in the earth’s atmosphere, but are abundant on
massive planets like Jupiter?

The spread of speeds widens as the temperature increases. At low temperatures, most
molecules have speeds close to their average speed. At high temperatures, a high
proportion have speeds widely different from their average speed.

The molecules of all gases have a wide
range of speeds. As the
temperature increases, the average
speed and the range of speeds
increase.
The average Molecular Kinetic Energy
We previously obtained this equation:
2
PV =     N(KE)
3
So we have:

3 PV
KE=
2N
For one mole of gas, PV = RT and N is Avogadro’s number:
3 RT
KE=
2N
To obtain KE in Joules R = 8.314 J.K-1.mol-1
Example: What is the average Kinetic Energy of a H2 molecule at 0 °C?
Distribution of molecular kinetic energies over a range gives typical curve like that for molecular
speeds.
Molecular Motion
Diffusion and Effusion
The gradual dispersal of one substance
through another is called diffusion like
the spread of pheromones and perfume
through air (fig 05, a). It helps to keep the
the composition of the atmosphere
approximately constant, because
abnormally high concentrations of one gas
diffuse away and disperse.
The escape of one substance (particularly a
gas) through a small hole into vacuum is
called effusion (fig 05, b) . Effusion occurs
whenever a gas is separated from a
vacuum by a porous barrier or a single pin
hole. A gas escapes through the hole from the
high concentration to the low concentration
region.
Graham’s Law of Effusion
Suppose that samples of two gases, A and B, are confined separately in identical
containers under the same conditions of temperature and pressure.
According the kinetic theory they have the same average kinetic energy:

KEA = KEB
1/2 mAvA2 = 1/2 mBvB2
mAvA2 = mBvB2

vA2        mB
=
vB2         mA

vA          mB
=√
vB          mA
The ratio of the molecular masses, mB / mA, is the same as the ratio of the molecular
weights MB / MA, therfore:
vA        MB
=√
vB        MA

Imagine that each container has an identical, extremely small opening ( an orifice) in it.
Gas molecules will escape through these orifices; molecular effusion.
The rate of effusion, r, is equal to the rate at which molecules strike the orifice, which in
turn is proportional to molecular speed, v. Molecules that move rapidly will effuse at a
faster rate than slower moving molecules. So the ratio vA / vB is the same as the ratio of
effusion rates, rA / rB :
rA       MB
=√
rB        MA

It can be expressed in terms of gas densities. Since the desity of a gas, d, is proportional
to the molecular weight of the gas, M:

rA        dB
=√
rB        dA
Hydrogen will effuse four times more rapidly than oxygen, because its lighter than oxygen:

rH2        MO2        32
=√         =√        = √ 16 = 4
rO2        MH2        2

In the figure below, the plug on the left is soaked in HCl and that on the right in aqueous
ammonia. Formation of ammounium chloride occurs where the two gases meet. The
reaction occurs closer to the HCl plug because this gas has the greater molar mass and thus
diffuses more slowly.
Graham’s law of effusion has been used for separation of isotopes. Uranium occurs in
nature as 0.72% 92235U and 99.28% 92238U. Of the two isotopes, only 92235U will undergo
nuclear fission. It is necessary to separate 92235U from 92238U. By converting uranium to
uranium hexafloride, which boils at 56 °C. The uranium hexafloride actually is a mixtur of
235UF (A) and 238UF (B). This mixture as a gas and at low temperature is allowed to
92      6          92      6
effuse through a porous barrier. The lighter 92235UF6 effuses 1.004 times faster than the
heavier 92238UF6 . Hence the emerging gas has a higher 92235UF6 content than the original
mixture. By repeating this effusion procedure for thousands of times a significant
separation will accomplish.

rA          MB         352
=√         =√           = 1.004
rB          MA          349
Liquefaction of Gases

Liquefaction of a real gas is the consequence of intermolecular attractions. If the pressure
is high, the molecules are close together, and intermolecular forces are appreciable.
This forces are opposed by the motion of the molecules; thus; liquefaction is favoured by
low temperatures, where the average kinetic energy of the molecules is low.
The behaviour of a gas deviates more and more from ideality as the temperature is
lowered and the pressure is raised. At extremes of these conditions, gases liquefy.
The higher the temperature of the gas, the more difficult it is to liquefy and the higher
pressure that must be employed.
For each gas there is a temperature above which it is
impossible to liquefy the gas no matter how high the applied pressure. This temperature is
called the critical temperature.
The critical temperature of a gas gives an indication of the strength of the intermolecular
forces. The weaker the intermolecular forces, the lower the critical temperature. As can be
seen in Table 10.6, helium with the weakest intermolecular forces, liquefies below 5.3 K.
Water with strong forces is liquid.
The minimum pressure needed to liquefy a gas at its critical temperature is called the
critical pressure.
The Joule-Thomson Effect
It is necessary to cool many gases below room temperature (295 K) before
these substances can be liquefied.
When most compressed gases are allowed to expand to a lower pressure,
they cool. In the expansion work is done against the intermolecular
attractive forces. The energy used in performing this work must be taken
from the kinetic energy of the gas molecules themselves; hence; the
temperature of the gas decreases.
This effect was studied by James Joule and William Thompson (Lord Kelvin)
during the years 1852-1862.
A Linde refrigerator for liquefying gases. The compressed gas gives up heat to the
surroundings in the heat exchanger and is cooled further as it passes through
the coil past which cooled gas circulates. The gas is cooled still further by the
Joule-Thompson effect as it emerges through the throttle. This gas cools the
incoming gas and is recirculated through the system. Eventually, the
temperature of the incoming gas is so low that it condenses to a liquid.

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