Physical Chemistry (Part-2)

7/29/2019 Physical Chemistry (Part-2)

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Thermodynamic Systems

A thermodynamic system is acollection of matter which has

distinct boundaries. OR

A real or imaginary portion ofuniverse which has distinct

boundaries is called system. OR

A thermodynamic system isthat part of universe which isunder thermodynamic study.

Lagrange (Closed Systems)Vs.

Eulero (Open Systems)

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Definitions A system is the object or material(s) under study,

separated from its surroundings by some specifiedboundary in essence, some collection of atoms.

If neither matter nor energy crosses the boundary,

.

If only energy crosses the boundary, it is a closedsystem.

If both energy and material cross the boundary, it isan open system.

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Thermodynamic process

* The transformation of a system between two states describes

a path, which is called a thermodynamic process.

* There are infinite paths to connect two states.

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Type of processes

1- Isothermal process T = 0

2- Isobaric process P = 0

3- Isochoric process V = 0

4- Adiabatic process q = 05- Cyclic process = 0

Reversible and Irreversible Processes:

Reversible process:

Is a process which maybe reversed at any moment by changing an

independent variable by an infinitesimal amount.

Irreversible Process:

Is the process that will not reverse the system if any change has

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Extensive Properties( variables):

The property which depends on the amount of the substance, such as : ( S, m, U, V,

G .)

Intensive properties( Variables) :

The properties which are independent on the amount of substance, such as (

Temp, press, density, viscosity, chemical potential )

State of a system at equilibrium:

Is defined by the collection of all macroscopic properties that are described by

s a e var a e , n, , an n epen en on e s ory o e sys em.

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Work force applied over a distance against a load (e.g. foot-pounds, or

Newton-meters = joules)

W = F . dh dh : distance F: force

P =

F = P. A dh

pex

W = P . A. L V = A. L

Then W = P V

W = Pext. ( V2 V1)

The ( +ve ) sign for work means the system achieved work on the

surrounding, such as gas expansion.

The (-ve) sign for work means the surroundings achieved work on the

system, such as gas compression in a cylinder.

gas

A

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1- Expansion work

W = - Pex dv ( work done by the gas)

Work against a constant external pressure

Pex = constant P1 = P2

PW = - Pex .dV

= -Pex (V2 V1) W

= - Pex VV1 V2

2- Free Expansion ( Expansion against vacuum )

Pex = 0

W = - Pex .dV

W= 0

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3- Isothermal Reversible expansion work

Pgas = Pex

W = - Pex .dV

W = - Pgas .dV

for ideal gas Pgas = nRT/ V

PPff

P=nRT/VP=nRT/V

VolumeVolume

PPii

ww

If T = constant ( Isothermal) ( Isothermal reversible work )

ff

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Example:For evaporation of one mole of water at constant pressure at one atm and 100oC :

1- what is the work done 2- what is the change in internal energy?

1- V1 = Mwt / d = 18 ml = 0.018 lit V2 calculation according to STP

W = -Pex (V2 V1 )

W = -1 ( 30.6 0.018)

W = - 30.6 lit.atm = 30.6 *24.2 cal = 741.4cal

2- U = q + W

q = 540 * 18 g

U = 9720 741.4= 8979 cal

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Example :

what is the work of reversible expansion of a mole of ideal gas at (0oC) from ( 2.24 to 22.4)

??

Reversible and isothermal

V1 = 2.24

V2 = 22.4

T= 0oC = 273oK

W= ?

W= -1 x 1.987 x 273 ln 22.4/2.24

W = -1250 cal . Mol-1

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Ex: calculate the maximum work for a reversible expansion at constant temperature from an

initial volume of V1 to a final volume ( 10 V1) that will do 10000 cal of work, if the initial

pressure is 100 atm .

1- find V12- if 2 mole of gas their, what would be the temperature??

Wrev = 2.303 nRT log 10V1/V1Wrev = 2.303 nRT

P1 V1 = nRT = Wrev / 2.303

100 x V1 = 10000 / 2.303V1 = 1.78 lit

2- W = nRT ln V2/V1 = 2.303 nRT log 10 V1 /V1 = 2.303 nRT

RT = 10000/ 2x2.303T = 1100 oK

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Heat ( q) the energy that flows from a body at higher temperature to one at lowertemperature by virtue of the temperature difference.

Experiments have shown that every material has a characteristic heat capacityper gram, or specific heat capacity

heat capacity ( C ) amount of heat that is required to raise the temperature of thesystem 1oC .

If ( C ) is known , then we can determine the amount of heat required to raise thetemperature of a constant mass of a system from ( T1) to ( T2) by:

q = C ( T1 T2) = CT

The unit of heat capacity is ( joul / mole. Kelvin)

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1st Law of ThermodynamicsLaw of conservation of energy

(Energy can not be destroyed nor created)

statement of ener conservation for a thermod namic s stem

systemdonework:positive

systemaddedheat:positive

by

to

W

q

WqdE =

internal energy Eis a state variable

W, q process dependent

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Thus the first law may also be state as: the net energy change of a closed system is

equal to heat transferred to the system minus the work done by the system.

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Exact differentials:If a system is taken along path ( for example by heating it ), U changes from Ui to Uf ,

and the overall change is the sum (integral) of all the infinitesimal changes along the

path:

U = du the value of U depends on the initial and final states of the system

but independent on the path between them. (Exact differential )

and U is said to be ( state function) .

Units of energy are ( J, kj, cal, kcal)

1 j = 1kg.m2

.S-2

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Heat Content ( Enthalpy) ( H ):Is the thermal change at constant pressure. ( the amount of heat absorbed or emitted by

a reaction).

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H = E + P V + V P

at constant ( P ) P = 0

H = E + P V

E = q - w

From first law of thermodynamics

= q -

So... q = E + P V

H = qp

Substitute in the equation above :

At constant prerssure

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dE = Cv dT

dE = Cv dT if Cv = constant , then

E2 E1 = E = Cv dT

E = Cv ( T2 T1 )

E = Cv T ( for n= 1 )

E = n Cv T ( for n moles)

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dH = Cp dT

In the same way as for Cv :

when Cp is constant , then :

H = n Cp T

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The variation of heat capacity with temperature can sometimes be

ignored if the temperature range is small; this approximation is highly

accurate for a monatomic perfect gas (for instance, one of the noblegases at low pressure). However, when it is necessary to take the

variation into account, a convenient approximate empirical expression

is:

Cp,m = a + bT + cT2

The empirical parameters a, b, and c are independent of temperature

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Example :

Cp = a + bT + c T2 + ..

dH = n Cp dT n=1

H2 - H1 = n ( a + bT + cT2 ) dT

H = a .dT + bT .dT + c T2

. dT

= a ( T2 - T1) + ( T2 - T1 ) + ( T2 - T1 )b2

c3

2 2 3 3

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Example / What is the change in molar enthalpy of N2 when it is heated from 25C to

100C? Use the heat capacity information in Table 2.2-

Cp,m = a + bT + cT2

Method The heat capacity ofN2 changes with temperature, so we cannot use eqn 2.23b (which

assumes that the heat capacity of the substance is constant). Therefore, for the temperature

dependence of the heat capacity, and integrate the resulting expression from 25C to 100C.

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Example : -

The equation for the molar heat capacity for n-butane is ( Cp 4.64 + 0.0558 T ).Calculate the heat necessary to rise the temperature of 1 mole of the Butane from 25

to 300oC at constant pressure.

H = n Cp T

H2 - H1 = ( 4.64 + 0.0558 T ) dT

H2 - H1 = 4.64 ( T2 T1 ) + 0.0558/ 2 ( T22 - T1

2 )

H = (4.64 (300 -25) + 0.0558/ 2 ( 5732 - 2982 ) = 7959 Cal mol-1

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Th M l l I t t ti f th I t l E


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The Molecular Interpretation of the Internal Energy1- For ideal gas, the only energy is the translational kinetic energy of the molecules,

and it will be found that the energy of the model gas ofN molecules is

2- for Real gas , total Energy = Transitional energy + Internal Energy

Real atoms and molecules are not structureless particles. Real atoms and moleculeshave translational energy and electronic energy, and molecules also have rotational

and vibrational energy.

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3- Internal Energy = Erot + EvibSo Etotal = Etrans + Erot + Evib

Etot = 3/2 RT + 1/2 RT + 1/2RT

4- Evib consist of two type, each contribute by 1/2RT , so 2 x 1/2RT = RT

5- For linear molecules, there are two axis of rotation, each contribute by 1/2RT

For non-linear molecules, there are three axis of rotation, each contribute by 1/2RT

6 I t l 3N 3 N f t i th l l


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6- Internal energy = 3N 3 N : no of atoms in the molecule

7- Evib

= 3N 5 -------------- ( for linear molecules )

Evib = 3N 6 -------------- ( for non-linear molecules)

Example : calculate Cv for a linear diatomic molecule .

Etot = Etrans + Einternal

Etrans = 3 x 1/2 R T = 3/2 RT 1/2 R T for each axis

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Erot = 2 x 1/2 RT = 1 R T 1/2 RT each axis( linear molecule have 2 axis of rotation)

Evib = (3 N 5 ) x R T there are 2 types of vibrations ( 2 x 1/2 RT = R T)

Evib = (3 x 2 - 5 ) x RT = 1 R T

Etot = 3/2 RT + RT + RT

Etot = 7/2 RT Cv = dE / dT

Cv = 7/2 R

Adi b ti h


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Adiabatic changes

We are now equipped to deal with the changes that

occur when a perfect gas expands adiabatically. A

decrease in temperature should be expected: because

work is done but no heat enters the system, the

internal energy falls, and therefore the temperature

of the working gas also falls. In molecular terms, the

kinetic energy of the molecules falls as work is done,so their average speed decreases, and hence the

temperature falls. The change in internal energy of a

perfect gas when the temperature is changed from T

to Tf and the volume is changed from Vi to Vf can beexpressed as the sum of two steps (Fig. 2). In the first

step, only the volume changes and the temperature is

held constant at its initial value. However, because the

internal energy of a perfect gas is independent of the

volume the molecules occupy, the overall change ininternal energy arises solely from the second step, the

change in temperature at constant volume. Provided

the heat capacity is independent of temperature, this

change is E = Cv T 31


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Because the expansion is adiabatic, we know that q = 0; because E = q- w , it then

follows that E = - w . Therefore, by equating the two values we have obtained for E,

we see thatw = - Cv T

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Example :- A system consisting of 2.000 mol of argon expands adiabatically and


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reversibly from a volume of 5.000 L and a temperature of 373.15K to a volume of

20.00 L. Find the final temperature.

Assume argon to be ideal with CV equal to 3nR/2.

Example :- Find the volume to which the system of the previousexample must be adiabatically and reversibly expanded in order to

reach a final temperature of 273.15 K.

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Example: a sample of Ar at one atm press and 25oC expands reversibly and adiabatically from

( ) l l h f l d h k d d h


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(0.5 to 1 Lit). Calculate the final temperature and the work done during this expansion. Given

that the molar heat capacity of Ar= 12.48 J/K mol-1.

P1 =1atm T1 =25oC

Cv =12.48 T2 = ?

V1 =0.5 L V2 =1L

1) T2

=? 2) w = ? 3) E

V1T1Cv/R = V2 T2

Cv/R

= R/Cv

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.

T2 = ( 0.5/1)1/1.5 x 298 = 188oK

For adiabatic process w = CvdT = Cv T

P V = n R T

n = ( 1x 0.5) / (0.0821x298) = 0.02 mol

2) W = nCv T = 0.02 x 12.48 x (188 298)

W = -27 j

3) E = q w

q = 0E = W = -27 J


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Thermochemistry


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Thermochemistry

The study of the energy transferred as heat during the course of

chemical reactions is called thermochemistry. Thermochemistry is a

branch of thermodynamics because a reaction vessel and its contents

form a system, and chemical reactions result in the exchange of energybetween the system and the surroundings. Thus we can use calorimetry

to measure the energy supplied or discarded as heat by a reaction, and

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can identify q with a change in internal energy (if the reaction occurs at

constant volume) or a change in enthalpy (if the reaction occurs at

constant pressure). Conversely, if we know U or H for a reaction, we

can predict the energy (transferred as heat) the reaction can produce.


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The Truth about Enthalpy

1. Enthalpy is an extensive property.2. H for a reaction in the forward

direction is e ual in size, but o osite

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Thermochemistry

in sign, toH

for the reverse reaction.3. H for a reaction depends on the state

of the products and the state of thereactants.


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Standard enthalpy changes


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Changes in enthalpy are normally reported for processes taking place

under a set of standard conditions. In most of our discussions we shallconsider the standard enthalpy change, Ho, the change in enthalpy for a

process in which the initial and final substances are in their standard

states: The standard state of a substance at a specified temperature is its

pure form at 1 bar.

For exam le the standard state of li uid ethanol at 298 K is ure li uid

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ethanol at 298 K and 1 bar; the standard state of solid iron at 500 K ispure iron at 500 K and 1 bar. The standard enthalpy change for a reaction

or a physical process is the difference between the products in their

standard states and the reactants in their standard states, all at the same

specified temperature.

Different types of Enthalpies of physical change


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yp a p p y a a g

The standard enthalpy change that accompanies a change of physicalstate is called the standard enthalpy of transition and is denoted trsHo.

The standard enthalpy of vaporization, vapHo, is one example. Another

is the standard enthalpy of fusion, fusHo, the standard enthalpy change

accompanying the conversion of a solid to a liquid, as in

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Because enthalpy is a state function, a change in enthalpy is independentof the path between the two states. This feature is of great importance in

thermochemistry, for it implies that the same value of I'1H-e- will be

obtained however the change is brought about between the same initial

and final states. For example, we can picture the conversion of a solidto a vapour either as occurring by sublimation (the direct conversion

from solid to vapour),


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or as occurring in two steps, first fusion (melting) andthen vaporization of the resulting liquid:

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Because the overall result of the indirect path is the same as

that of the direct path, the overall enthalpy change is the

same in each case, and we can conclude that (for processes

occurring at the same temperature)


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Hess's law


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The standard enthalpy of an overall reaction is the sum of the standardenthalpies of the individual reactions into which a reaction may be

divided.

The individual steps need not be realizable in practice: they may be

hypothetical reactions, the only requirement being that their chemical

equations should balance.

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Standard enthalpies of formation


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Standard enthalpies of formation

The standard enthalpy of formation, fHo

, of a substance is the standardreaction enthalpy for the formation of the compound from its elements in

their reference states. The reference state of an element is its most stable

state at the specified temperature and 1 bar. For example, at 298 K the

reference state of nitrogen is a gas of N2 molecules, that of mercury isliquid mercury, that of carbon is graphite, and that of tin is the white

(metallic) form.

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Standard enthalpies of formation are expressed as enthalpies per mole ofmolecules or (for ionic substances) formula units of the compound. The

standard enthalpy of formation of liquid benzene at 298 K, for example,

refers to the reaction

6C(s, graphite) + 3H2(g) -- 7C6H6(l)and is +49.0 kJ mol-I. The standard enthalpies of formation of elements

in their reference states are zero at all temperatures.


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Physical Chemistry (Part-2)

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