engineering chemistry 1
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Gases
Matter exists in three distinct physical phase: gas, liquidand solid.
Relatively few substances exist in the gaseous state,gases are very important.
Gas is easily compressed, mixes completely with any
other gases and uniformly fills any container.
One of the gas properties that it exerts pressure on its
surroundings. 2
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Pressure is defined as force per unit area
Units of pressure
The most commonly used unit is based on the heightof mercury column that the gas pressure can support.Thus the unit is mm Hg. Anther unit for pressure is
the standard atmosphere which is atm.
1 atm = 760 mm Hg (torr) = 101325 Pa = 29.92 in Hg= 14.7 lb/in2.
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The atmospheric pressure is the result from the massof the air being pulled by earth gravity .... In otherwords its the result from the weight of air.
The height of the column of Hg supported byatmosphere at sea level is 760 mm. But this would
vary according to the weather.
In a manometer, which is a device for measuring thepressure of gas in a container, the pressure is given bythe difference in mercury level in unites of torr (mmHg).
Pgas = Patm h
Pgas = Patm + h 4
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Manometer
5
Pb
Pa
750 mm HgPa =
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Gas pressure (Pgas)
Pgas = Patm h
If Patm = 750 mmHg and h = 150 mm then
Pgas= 750 150 = 600 mmHg
6
lower
pressure
Pa
height
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Gas pressure (Pgas)
Pgas = Patm + h
If Patm = 750 mmHg and h = 150 mm then
Pgas= 750 + 150 = 900 mmHg
7
higher
pressure
Pa
height
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Boyles law
The relation between the pressure of the tapped gas and its
volume was studied.
The product of the pressure of trapped gas and its volume
sample is constant within the accuracies of Boyles
measurements. This behaviour is given by Boyles lawPV = k
where kis a constant for a given sample of air at a specific
temperature.
Data can be presented by plottingPversus V. Anther type of
plotting is by rearranging Boyles law as follows
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where the equation is for a straight line of the type
y = mx + b
The Boyles law hold accurately at very low pressures,
according to the results of highly accurate measurements.
At high pressures, the measurements show thatPVis not
constant but varies as the pressure varies.
The changes inPVat small pressure values is small, while
the changes inPVvalues at higher pressures is more
significant where the nature of dependence ofPVbecomes
more obvious. 9
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To calculate the ideal value ofk, a plot forPV
versusPis plotted then extrapolate the line back to
zero pressure where the gas behaves most ideally.
Extrapolate: extends the line beyond the
experimental point.
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Example
Consider a 1.53 L sample of gaseous SO2 at a pressure
of 5.6103 Pa. If the pressure is changed to 1.5104 ata constant temperature, what will be the new volume
of the gas?
SolutionP1 = 5.610
3 Pa P2 = 1.5104 Pa
V1 = 1.53 L V2 = ??
From Boyles lawPV = k, then it can be written asP1 V1 = P2 V2
V2 = 0.57 L
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Example
If a balloon is designed to be inflated to a volume
not more than 2.5 L. If the balloon is filled with2.0 L at sea level then released where it rises toaltitude at which the atmospheric pressure is only500 mmHg, will the balloon burst or not?
Solution
From Boyles lawPV = k, then it can be written as
P1 V1 = P2 V2
760 mm Hg 2.0 L = 500 mm Hg V2
V2 = 3.04 L
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Example
For gaseous ammonia several volume measurements were
done at different pressures to see how closely does theammonia obeys Boyles law. For 1.o mol NH3 at 0
calculate Boyles constant at various pressure values?
What is the ideal value fork?
P (atm) Volume (L)
0.13 172.1
0.25 89.28
0.3 74.35
0.5 44.49
0.75 29.55
1.0 22.08 13
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Solution
By applying Boyles law PV = k
P(atm) V(L) k0.13 172.1 22.373
0.25 89.28 22.32
0.3 74.35 22.3050.5 44.49 22.245
0.75 29.55 22.1625
1.0 22.08 22.08
The deviation from Boyles law is small, and kchanges
regularly. In order to calculate the ideal value fork, plot
PVversus Vand extrapolate the line back to zero
pressure (intercept with y-axis). 14
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Charles law
Charles found that the volume of a gas at constant pressure increases
linearly with the temperature of gas. The plot of the volume of a gas
at constant pressure versus its temperature ( ) give a straight line.
The slopes of the line are different cause there different number of
moles of gases.
If the volume lines for all gases were extrapolated to a zero volume
at the same temperature, the temperature would be -273 (0 K).
At Kelvin scale this point is defined as 0 K, this leads to the relation
between the Kelvin and Celsius scales
K = + 273
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5
10
15
20
25
30
Volum
e(mL)
Temperature (C)
0 100 273
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If the volumes of gases in the previous figure are
plotted versus temperature on the Kelvin scale, then
the volume of each gas is directly proportional to
temperature and extrapolate to zero when the
temperature is 0 K.
Charles law V T (temperature is expressed in K) of
constant pressure.
This behaviour is known as Charles law
V = bT
where Tis in Kelvin and b is a proportionality constant.
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Charles law also can be written as
0 K is known as the absolute temperature. The
extrapolated volume of gases below 0 K would benegative which can not happen.
ExampleA sample of gas at 15 and 1 atm has a volume of
2.58 L. What volume will this gas occupy at 38
and 1 atm? 19
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Solution
T1 = 15 + 273 = 288 K.
T2 = 38 + 273 = 311 K.
V1 = 2.58 L.
V2 = ??
Since the pressure is constant, then from Charles law
Then
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Example
A balloon is filled to a volume of 7.00 102 mL at a
temperature of 20.0 . The balloon is then cooled atconstant pressure to a temperature of 1.00 102 K.
What is the final volume of the balloon?
Solution
Since the pressure is constant, then from Charles law
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Avogadros law
Avogadro proposed that equal volumes of gases at the
same temperature and pressure contain the same numberof particles, this is known as Avogadros law.
Avogadros law is given as follows
V = a n
where Vis the volume of gas, n is the number of moles of
gas particles and a is a proportionality constant.
Avogadros equation states that for a gas at constant
temperature and pressure, the volume is directly
proportional to the number of moles of gas. 22
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Example
If 12.2 L sample containing 0.5 mol oxygen gas (O2)
at a pressure of 1 atm and temperature of 25 . Ifall O2 is converted to ozone (O3) at the same
temperature and pressure what would be the
volume of the ozone?
Solution
n1 = 0.5 mol O2
n2 = ?? mol O3
V2 = 12.2 L O2
=23
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Find how many moles of O3 by 0.5 mol O2
3O2 (g) 2O3 (g)
From the mole ratio, the number of moles of O3
formed are
Avogadros law can be re-written as because a is
constant
Thus
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The ideal gas law
Three laws described behaviour of gases based on the
experimental observation
Boles law V = k/p (at constant T and n)
Charless law V = bT (at constant P and n)
Avogadros law V = an (at constant T and P)
From these laws it was noticed that the volume of a gas
depends on pressure, temperature and number of moles of
gas present, by combing these relationships as follows
where R is the universal gas constant. 25
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By rearranging the previous equation gives the ideal gas law
PV = nRT
where R is the universal gas constant (0.08206 L.atm/K.mol).
The ideal gas law is an equation of state for a gas. The state ofa gas is its condition at a given time.
The state of gas is described by its pressure, volume,temperature and number of moles. Thus knowing any three ofthese properties is enough to define the state of the gas.
The ideal gas law expresses behaviour that the real gasesapproach at low pressures and high temperatures.
The ideal gas law applies best at pressure smaller than 1 atm.26
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Example
A sample of H22
has a volume of 8.56 L at a temperature of 0
and a pressure of 1.5 atm. Calculate moles of H 22
molecules present in this gas sample?
Solution
PV = nRT
(1.5 atm) (8.56 L) = n (0.08206 L.atm/K.mol) (0+273 K)
n = 0.57 mol
ExampleThe steel reaction vessel of a bomb calorimeter, which has a
volume of 75.0 mL, is charged with oxygen gas to apressure of 145 atm at 22 . Calculate the moles of oxygen
in the reaction vessel? 27
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Solution
PV = nRT
(145 atm) (0.075 L) = n (0.08206 L.atm/K.mol) (295 K)
n = 0.449 mol
Example
A sample of ammonia gas with a volume of 7.0 mL at apressure of 1.68 atm. The gas is compressed to a volume of2.7 mL at a constant temperature. Use the ideal gas law tocalculate the final pressure?
Solution
The variables are P and V, while n, R and T are constant. Thuswrite the ideal gas law by collecting the variables on one
side and the constants on the other side. 28
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P11V
11= nRT = P
22V
22
P11V
11= P
22V
22
(1.68 atm) (7.0 mL) = P(1.68 atm) (7.0 mL) = P2 (2.7 mL)(2.7 mL)P
22= 4.4 atm
As the volume decrease at a constant temperature, the pressure
would increase .
Example
A gas sample containing 1.50 mol at 25 exerts pressure of400 torr. Some gas is added to the same container and the
temperature is increased to 50 . If the pressure increases to
800 torr, how many moles of gas were added to the container?
Assume a constant-volume container. 29
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Solution
Thus number of moles added = 2.77 - 1.50 = 1.27 mol
Example
A sample of methane gas that has a volume of 3.8 L at 5
is heated to 86 at constant pressure. Calculate it new
volume?
Solution 30
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T11 = 5 + 273 = 278 K.= 5 + 273 = 278 K.
TT22 = 86 + 273 = 359 K.= 86 + 273 = 359 K.
VV11 = 3.8 L.= 3.8 L.VV22 = ??
Substituting the variables into the ideal gas law
Thus V22
= 4.9 L.
The above problem can be described by Charless law, and
the previous one might be called Boyles law problem. In
both cases, however the solution started with ideal gas law,
were the advantage of the ideal gas law it can be applied to
any problem dealing with gases. 31
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Try to solve examples 5.9 and 5.10 in chapter 5 and exercises
5.59 to 5.61.
Gas stoichiometry
The molar volume of an ideal gas is 22.42 L at 0 and 1
atm.
The conditions 0 and 1 atm are called standard
temperature and pressure (STP), which are common
reference conditions for the properties of gases.
Example
A sample of N22has a volume of 1.75 L at STP. How many
moles of N22are present?? 32
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Solution
This can be solved using the ideal gas law and can be
solved using the molar volume of an ideal gas at STP.Since 1 mole of an ideal gas at STP has a volume of
22.42 L, thus the number of moles can be obtained
using the ratio of 1.75 L to 22.42 L as follows
Try to solve examples 5.11 to 5.13 in chapter 5 and
exercises 5.63 to 5.72.
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Molar mass of a gas
The ideal gas law is used in the calculation of the molarmass of a gas from the gas measured density.
Number of moles can be expressed as
Substituting the above equation into the ideal gas law gives
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Since the gas density (d) is m/V, then the previous equation
can be written as follows
or
Example
The density of a gas was measured at 1.50 atm and 27 and
found to be 1.95 g/L. Calculate the molar mass of the gas?
Solution
P = 1.50 atm, T = 27 + 273 = 300K and d = 1.95 g/L
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Y ld i h i i l i d i d
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You could memorise the equation involving gas density andmolar mass, but its better to remember the ideal gas equation, thedefinition of density and the relationship between number ofmoles and molar mass.
Daltons law of partial pressures
For a mixture of gases in a container, the total pressure exerted is
the sum of the pressures that each gas would exert if it werealone in the container.
Thus the Daltons law of partial pressures is
PTotalTotal = P= P11 + P+ P22 + P33 + ....
where subscripts (1, 2, 3 ....) refers to the individual gases (gas 1,where subscripts (1, 2, 3 ....) refers to the individual gases (gas 1,gas 2, gas 3 and so on) and Pgas 2, gas 3 and so on) and P11, P, P22, P33and so on represents the
partial pressure for each gas. 36
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It each gas was assumed to behave ideally , then the partial pressure of each
gas can be calculated from the ideal gas law as follows
The total mixture pressure (PTotalTotal
) can be represented as follows
where nTotalTotalis the sum of the numbers of moles of particles.
The pressure exerted by an ideal gas is not affected by the composition of the
gas particles which reveals that
1. The volume of the individual gas particles should not be important.
2. The force among particles should not be important. 37
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Example
Mixture of O22and He can be used in scuba diving tanks to help
prevent the bends. For a particular dive 46 L H22
at 25 and
1.0 atm and 12 L O22 at 25at 25 and 1 atm were pumped into a
tank with a volume of 5.0L. Calculate the partial pressure of
each gas and the total pressure in each tank at 25 ?
Solution
From the data we can calculate the number of moles for each gas
from the ideal gas law as follows
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From the ideal gas equation, the number of moles is directly
proportional to the pressure of the gas, thus the mole fraction () can
be presented as follows
Example
Partial pressure of O22was observed to be 156 torr in air with a total
atmospheric pressure of 743 torr calculate the mole fraction of O22
fraction ofOO22present?
Solution
The mole fraction of O22
can be calculated as follows
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The partial pressure can be calculated using the ideal gas law as
follows
The total pressure is the sum of the partial pressures
PTotalTotal = P= P11 + P+ P22 = 9.3 atm + 2.4 atm = 11.7 atmOr it can be calculated by finding the total number of moles and
substituting it in the ideal gas law.
The mole fraction () is the ratio of the number of moles of agiven component in a mixture to the total number of mole in the
mixture. Thus mole fraction () is
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Example
The mole fraction of N22
in the air is 0.7808. Calculate the
partial pressure of N22
in air when the atmospheric pressure is
760 torr?
Solution
Since
Rearranging the above equation gives the following
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Th ki i l l h f
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The kinetic molecular theory of gases
At pressure less than 1 atm most gases approaches the behaviour described
by the ideal gas law.
Kinetic molecular theory (KMT) is a simple model that attempts to explain
the properties of an ideal gas.
KMT is based on speculations about the behaviour of the individual gas
particles (atoms or molecules).
Postulates (assumptions) of the kinetic molecular theory
1. The particles are so small compared with the distance between them that
the volume of the individual particles can be assumed to be negligible.
2. The particles are in constant motion. The collusions of the particles with the
walls of the container are the cause of the pressure exerted gas.
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3 Th i l d f h h h
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3. The particles are assumed to exert no forces on each other; they are
assumed neither to attract not to repel each other.
4. The average kinetic energy of a collection of gas particles isassumed to be directly proportional to the Kelvin temperature of
the gas.
For the real gases case, the molecules have finite volumes and doexert forces on each other. Thus real gases do not confirm to these
assumptions.
The assumptions of kinetic molecular model picture
1. An ideal gas consisting of particles no volume and no attractions
for each other.
2. The gas produces pressure on its container by collisions with the
walls. 43
KMT t f th ti f di t th id l l
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KMT accounts for the properties of gases according to the ideal gas law as
follows
a. Pressure and volume (Boyles law)
For a gas sample at a given temperature (n and T are constant) thus if the volume of
a gas decrease then the pressure would increase
According to KMT this makes sense because the decrease in volume means that the
gas particles will hit the surrounding walls more often resulting in increasing thepressure
b. Pressure and temperature
From the ideal gas of law, it can be predicted that the pressure is directly
proportional to temperature if the volume is constant
This behaviour is counted for in the KMT because when the temperature of a gas
increases then the speed of its particles increases. Where the particles hit the wall
of the container with greater force and frequency, since the volume is constant it
would result in increasing the pressure, the volume is constant. 44
c Volume and temperature (Charless law)
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c. Volume and temperature (Charles s law)
According to the ideal gas law for a gas sample at a constant pressure, the volume of thegas is directly proportional to the temperature (where T is in K)
According to the KMT, the gas pressure would increase as a result of heating the gaswhere its temperature would increase. Thus to maintain the pressure at a constantvalue then the volume of the container would be increased.
d. Volume and number of moles (Avogadro's law)
According to the ideal gas law, the volume of a gas at a constant temperature and
pressure depends directly on the number of particles present
In terms of KMT this makes sense because an increase in number of gas particles at thesame temperature would increase the pressure if the volume was held constant.Thus to maintain the pressure at a constant value, the volume would be increased.
At a constant P and T, the gas volume depends only on the number of gas particles.
For a gas behaving ideally, the individual volume of particles are not a factorbecause the particles volume is small when compared with distance between theparticles.
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e Mixture of gases (Dalton's law)
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e. Mixture of gases (Dalton s law)
Since KMT assumes that all gas particles are independent of each other and thevolumes of the individual particles are unimportant, then the observation of thetotal pressure exerted by a mixture of gases is the sum of the individual gases isexpected.
Effusion and diffusion
Diffusion is the mixing of two or more gases. The rate of diffusion is the rate of
the mixing of gases.
Effusion is the process in which a gas passes through a small hole into anempty chamber. The rate of effusion measure the speed at which the gas istransferred into the chamber.
The rate of effusion of a gas is inversely proportional to the square root of the massof the gas molecules.
The relative rates of effusion of two gases at the same temperature and pressureis given by Grahams law of effusion as follows
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where M11
and M22
are the molar masses of the gases (g/mol or kg/mol).
Example
Calculate the ratio of the rates of hydrogen gas (H22) and uranium hexafluoride
(UF66)?
Solution
First the molar masses is needed. The molar mass for H22
= 2.016 g/mol, and the
molar mass of UF66
= 352.02 g/mol. Using Grahams law
The effusion rate of H22 molecules is about 13 times that of UF66.
Real gases
The ideal gas can best be thought of as the behaviour approached by real gases
under certain conditions. 47
Real gas typically exhibits behaviour that is closest to ideal behaviour at low
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Real gas typically exhibits behaviour that is closest to ideal behaviour at low
pressures and high temperature.
In the ideal gas law it was assumed that the gas consists of volumeless entities
that do not interact with each other. Thus the volume of the real gas is less thanthe volume of the container because the particles themselves takes place.
The modification of the ideal gas equation was done by van der Waals, where
he represented the actual volume as the volume of the container minus a
correction volume of the molecules (nb), where n is the number of moles of gas
and b is an empirical constant. By taking into account the volume of the gas
particles, then the ideal gas equation becomes as follows
The next step is considering the attraction that occurs between particles in a real
gas. This is done by making the observed pressure (Pobsobs
) smaller than it would be
if the gas particles did not interact
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The size of the correction factor depends on the concentration of gas molecules
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The size of the correction factor depends on the concentration of gas molecules
defined in terms of moles of gas particles per litter (n/V).
The higher the concentration means that a pair of particles will be close enough
to attract each other.
For large number of particles, the number of interacting pairs of particles
depends on the square of the number of particles and thus on the square of the
concentration ((n/V)22). Thus the pressure is corrected as follows
where a is the proportionality constant, where can be determined for a real gas from
observing the actual behaviour of that gas.
Inserting the corrections for volume of the particles and the attraction of the
particles gives
where Pobsobs
is the observed pressure, V is the volume of the container, nb is the