engineering chemistry 1

8/2/2019 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

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