Mixtures of Gases Daltons Law of Partial Pressures

1
Mixtures of GasesDaltons 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.

2
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.
3
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

4
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.
5
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?
6
Molecular SpeedsDerivation 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
  
  

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• 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

8

• 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) a nT we can write the equation as
• 2
• PV a ? ? nT
• 3
• Using a constant R the proportionality will give
  an equation
• PV nRT

9
Molecular Speeds

• Previously we derived the expression
• 1
• PV ? ? N mv2
• 3
• For one mole of gas, the number of molecules, N,
  is Avogadros 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 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).

10

• The root- mean- square speed of H2 (g) molecule
  at 0 C is
• v v 3RT/M
• v v 38.3143. 103 g.m2/(s2. K.mol).273 K/2.016
  g.mol-1 1.84 . 103 m.s-1
• and at 100 C
• v v 38.3143. 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?
• Answer 471 K

11
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

12
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 earths 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.

13
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 Avogadros
  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.

14
Molecular MotionDiffusion 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.

15
Grahams 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
• v
• vB mA
• The ratio of the molecular masses, mB / mA, is
  the same as the ratio of the molecular
• weights MB / MA, therfore

16

• vA MB
• v
• 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
• v
• 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
• v

17

• Hydrogen will effuse four times more rapidly than
  oxygen, because its lighter than oxygen
• rH2 MO2 32
• v v v
  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.

18

• Grahams 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
• 92235UF6 (A) and 92238UF6 (B). This mixture as a
  gas and at low temperature is allowed to
• 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
• v v
  1.004
• rB MA
  349

19
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.

20
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.

21

• 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.