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Physical Chemistry

Dae Yong JEONG

Inha University

1

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Energy?

What is energy? A feeling of possessing such strength and

vitality.(dictionary) Energy may be defined as the ability to do work.

It is a scalar physical quantity. Although energyis conserved, there are many different types ofenergy, such as kinetic energy, potential energy,light, sound, and nuclear energy.

Information ??

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2.1 Work and Heat

If heating the Balloon with gas (supplying the thermal

energy), what happens? , ? (

?

For individual gas?

, () ? ?

: ? : ?

Macroscopically, which change? Volume expansion

Temp. change vs Physical change

HEATvs WORS

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Work

Work done by an expanding gas

4

cosfLLfw

Force (f) : mass x acceleration : vector quantity

N (kgm/s-2)

Pressure (Force/area) = N/m2

amf

f

L f

Lcos( )

SI unit: Joule

A scalar quantity defined by

Fluid

Piston

Fext

+dz

-dz

(System)

(Surrounding)PdVPAdzdzFdw ext

As system does work on the system. () Negative work (Pext)

dVPdw ext

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5

syssurr dwdw

Works done by system = works given to surrounding

2807.9 smg

mghw

Works done against gravity force (lift the mass from ground)

Total work on a system when there is a finite change in volume.

2

1

dVPw ext Quasistatic

slow change keeping the P of gas uniform and equal to Pext

Maximum work

In case, Pext = P and P is function of T and V

Works depend on path.

2

1

),( dVVTPw

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6

V2 1V

P1

P2

V1P1( , )

2VP2( , )

V1P2( , )

2VP1( , )

Pressure

Volume

(P1,V1)(P2,V2)

Case1) (P1,V2)

,(P1)V1V2()(V

2)P

1P

2()

w1= -P1(V2-V1) = P1(V1-V2)

Case 2) (P2,V1)

,(V1)P1P2(= 0)

(P2)

V1V2

w2= (-P2(V2-V1)) = P2(V1-V2)

,

Adiabatic: no heat transfer bw system and surrounding

Isobaric: constant pressure

Isothermal: constant temp.

Constant volume

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Heat vs Work

James P. Joule(1818-1889) ''

.

Born in Salford, Lancashire

worked in a brewery Hero in Thermodynamics

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Joules Experiment

:

,

:

, ,,

1 cal=4.184 J)

Q: (), ? ?, ? ?

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Internal energy (U)

Heating the balloon with gases, Temp increases gas movement, collision, electron movement? , volume

change (work) dont know how much energy system has exactly!!! Letsconsider this energy in system as internal energy!!

Internal energy of a system can be changed by either heat or work!!

We dont know how much absolute energy does system has!!

We just know the how much energy changedoes system make!!

For gas system, we just consider two kinds of energy (internal energyand work)!! (Simple system)

In an adiabatic process: (for ex.) The work done on a closed system in an adiabatic process is equal to the

increase in internal energy of the system.

When no work is done for the system, The heat absorbed by a closed system in a process in which no work is done

is equal to the increase in internal energy of the system.

9

wU

qU

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Heat

calorie

1 cal = 4.184 J

1 Cal = 1 kcal

Erg 1 J = 107erg

BTU (British thermal unit) 1 Pound(F) 1

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2.2 1st law of thermodynamics and internal energy system ,

system(work). , .

Work and heat are path dependent.

Work (w) and heat (q) are not exact differentials.

The internal energy of an isolated system is constant.

Internal energy is a function of state of a system.

The cyclic integral equal to zero. (dq or dw may not be zero.)

, , system ()(strain)internal energy.

, system, .

(), ideal gas, ()internal energy.

PV=nRT

, U = U(T, P, n1, n2)

dwdqdU

wqU

0dU

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Internal Energy

().

, , ,

.: - .

.

, 0.

.

. .

: -0oC:

(Entropy)

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Internal Energy

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2.3 Exact and inexact differentials

TD ~ math description

State function ~ exact differential eq.

Note: we can just calculateU not U!!

Path function ~ inexact differential eq. Note: we can calculate w.

UUUdU

b

a

ab

)(ab

b

a

wwwdw

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Move from a to b

15

I

II

a

byb

ay

x

xa xb

y

)(xyy

)(xydxdyydxdz is an exact differential eq.

aabb

b

a

b

a

yxyxxydzdz )(

ydxdz is not an exact differential eq.

Iareaydxzdz

b

a

b

a

Area I is dependant on path from a to b.

Path independent

U, H, S and G

III areaareaydxxdyzdz

b

a

b

a

b

a

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For dzto be the Exact differential eq.

Eq. should satisfy the Eulers criterion forexactness. (Mathematically)

dyyzdx

xzdz

yxfz

xy

),(

For the system with two independent degree of freedom, dzmay be determined by the differentials dx and dz in two otherquantities x and y. (physical meaning)

dyyxNdxyxMdz ),(),( Math. description

xy y

M

x

N

yxxy y

z

xx

z

y

x

y

y

zyxN

x

zyxM

),(

),(

(:,

.,exact differential,

.

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Ex. 2.1

Meaning

If a property of system can be expressed in the form ofexact eq., then this property is independent of path.

Whenever a property is not state function, if you just

multiply the path function (a part of an inexact eq.),

then a property could become a state function.

17

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2.4 Work of compression of a gas at constant temperature

At constant T, Compressing the gas (V1 V2)

What happens? (Imagine) Volume shrinks

P increases

Description with Math. (TD deals with Eq. state!!)

Final pressure (at Eq.) work done on the gas (positive : as V2 < V1)

m

TP2,V2,

m

TP1,V1,h

(a) (b)

P2

P1

V2 V1(c)

Amg

P 2

Frictionless, weightless piston

Ideal gas inside

Cylinder at temp. T

Above the cylinder : vacuum () stopperP2m

Put m on top of cylinder

Removed the Stopper

Due to mass m cylinder goes down reach eq. state, compressed to V2.

)VV(PAhPmghw1222

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Infinitesimal step

2

1

V

V

PdVdww

: 2 m() (V1+V2)/2

mV2 .

V2 V1

P P P

P2

P1

P2

P1

P2

P1

V2 V1 V2 V1

Work in one step > Work in two steps

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2.4 Work of expansion of a gas at constant temperature

At constant T, expanding the gas (V1 V2)

What happens? (Imagine) Volume expands

P decreases

Description with Math.

Final pressure (at Eq.) work done on the gas (negativeas V2 > V1)

Amg

P 2

Frictionless, weightless piston

Ideal gas inside

Cylinder at temp. T

Put m on top of cylinder Removed the Stopper

cylinder goes up reach eq.state, expands to V2.

(compression) m, .

)( 122 VVPmghw

m

TP1,V1,

m

TP2,V2,h

P2

P1

V2V1(c)(a) (b)

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Infinitesimal step

2

1

V

V

PdVdww

expansion in two or more steps

Work in one step < Work in two steps

V1 V2

P P P

P1

P2

P1

P2

P1

P2

V1 V2 V1 V2

Cycle (expansion: V1 V2) + (compression: V2 V1)

02

1

2

1

1

2

2

1

V

V

V

V

V

V

V

V

cycle PdVPdVPdVPdVw

R ibl

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Reversible process(Infinitesimal change Integral calculation)

Reversible

A change that can be reversed By an infinitesimal change in a variable

Maximum work is possible

PV line. ()

1

2

1

2lnln 2211

2

1

2

1P

PnRT

V

VnRTdV

V

nRTPdVw

VPVP

V

V

V

V

rev

For ideal gas at isothermal process

212

1

2

2

211

ln

2

1VV

an

nbV

nbVnRTdV

V

an

nbV

nRTw

V

Vrev

Van der Waals gas at isothermal process

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23

Irreversible process

Real situation Irreversible process friction

Balancing of internal and external P ( fluctuation.)

),().

!!

VPVVPdVPdVPw extextV

V

ext

V

V

extirrev 12

2

1

2

1

For ideal gas at isothermal process, constant P and irreversible process

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2.5 Various kinds of work

PV workenergy.,()PV work

,internal energy.

Type of WorkIntensive

variable

Extensive

variableDifferential Work

Hydrostatic Pressure, P Volume, V -PdV

Surface Surface tension, Area, A dAs

Elongation Force, F Length, L fdL

ElectricPotential

difference, Electric charge, Q dQ

... dQfdLdAdVPdqdU sext

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Scenario

System

PV work.

PV work

Constant V PV work = 0 Constant T

Constant P

Adiabatic (dq = 0) PV work

() Constant T (dq = 0)

Constant P Cp Enthalpy

Constant V Cv

25

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2.6 Change in state at constant volume

At constant Volume (V = 0) no PWwork, w = 0

,?

()

? (

)

vs Temp

dT

dqC Heat capacity: C

1

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Mathematical description on system

27

]/[

]/[

0as

KmolJC

KJdT

dU

dT

dqC

dT

dT

dUdq

dV

dVdV

dUPdT

dT

dUdq

dVPdqdU

dVdV

dUdT

dT

dUdU

V

V

VV

V

V

T

ext

V

ext

TV

V

T

T

VV qdTCU 2

1

Over a small temp. range

TCTTCdTCUVV

T

T

VV )( 12

2

1

TV

U

Can you guess? At const T, how much energy change

is associated by the volume change.

J l i t f iU

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28

Vacuum

Water bath

Valves

ThermometerGas inlet Vacuum pump

Gas

Joules experimentfor measuringTV

U

Thermally isolated

,valve open gas

system

(dT = 0)

As Pext= 0 (vacuum)

w= 00 dwdqdU

Example: 2-5: you can know the difference on heat involved in reversible and irreversible process.

Question: For real gas? dT = 0 ??

Interaction (intermolecular force, collision) bw gas temp change.

Van der Waals gas

2

V

a

V

U

T

Analysis:Volume for 1 mole increase less collision (less force bw

gases) Volume effect on internal energy decreases.

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2.7 Enthalpyand change of state at constant pressure

Const P is more common condition. Ex: Reaction under 1 atm (in open vessel)

? ,()

dHdq

PVUH

HHPVUPVUq

VPqUUU

P

P

P

Enthalpy:

121122

12 For the const P condition, enthalpywas derived from internal energy. State function

Exact differential eq.

Mathematical description on system

]/[

]/[

KmolJC

KJdT

dH

dT

dqC

dTdT

dHdq

dPdP

dHdT

dT

dHdH

P

P

p

P

P

P

TP

2

1

T

T

PP dTCH

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2.8 Heat capacities / table

C is not a constant but a fxn of T. .

, .

Einstein and Debye considered thequantum effect to explain the C vsT curve.(-)

2

1

2

1

12

2

...T

T

P

H

H

P

dTCHdHH

TTC

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Relationship bw Cvand CP

Cv: . , CP: , (), . CV.

CP-CVin solid: can be expressed in terms of cubic expansion coefficient and theisothermal compressibility

31

0

Pconstfor

VconstAt

PTP

VP

P

PT

VP

T

VP

T

ext

V

dT

dV

dV

dU

dT

dVPCC

C

dT

dV

dV

dUPC

dT

dq

dVdV

dUPdTCdq

dVdV

dUPdT

dT

dUdq

0

0

TdV

dU

PT

nRTPV

P

dT

dV

dV

dU

nRdT

dVP

RCC

nRCC

VP

VP

,

Mi i ll (Ki ti th )

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Microscopically (Kinetic theory)

Translational energy of monoatomic ideal gas

x, y, z movement 3/2RT CV=3/2RT = 12.472 J/(K-mol)

CP = 3/2RT + RT=5/2RT RT (= PV) for one mole

CP=5/2RT = 20.786 J/(K-mol)

32

2 9 Joule Thomson Expansion; dT

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2.9 Joule-Thomson Expansion;

, ?

L L RR

Porous plug

Manometer

Thermometer

T P PT

Adiabaticinsulation

Adiabatic

insulation

HdP

dT

2

chamber, push,

pull

(PL& PR)

PLPR TLTR

Work done to left-side

1,1,1,2, )0()(

2,

1,

LLLLLLL

constP

V

V

LLL

VPVPVVP

dVPw LL

L

Work done to surr by Right-side

2,2,1,2, )0()(

2,

1,

RRRRRRR

constP

V

V

RRR

VPVPVVP

dVPw RR

R

chamber

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chamber

34

RRLLRL VPVPwww

Adiabatic condition: dq = 0

0)()( 111222

12

HVPUVPU

VPVPwqUUU RRLL

Joule-Thomson coefficient JT

PTP

T

PP

TT

JT

HP

JT

12

12

0

lim

JT

Ideal gas: JT = 0

Real gas at low T JT > 0

at high T JT< 0

heating effect above the inversion T

Cooling effect below the inversion T

N2

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2.10 Adiabatic processes (dq = 0) with gases

Work done on the surrounding = internal energy change

Pext = 0 PV work = 0 No change in internal energy

Work done on surroundingwith the expense of internal energy temp drop in system

wUUU

dVPdU

dVPdwdU

V

V

ext

U

U

ext

12

2

1

2

1

,(ideal gas)

)( 1212

2

1

2

1

TTCUUU

dTCdUdTCdU

V

T

T

V

U

U

V

For ideal gas, Cvis independent onT (3/2R).

wTTCV

)( 12

CvV, V, w.??

, V()dq=0() .

1+1 = 4-2 ()

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1

2

1

2lnln

2

1

2

1

V

VR

T

TC

VVdR

TdTC

V

VdR

T

dTC

VdV

RTVPddTC

V

V

V

T

T

V

V

V

2211

/1

1

2

1

2

1

2

1

1

2

VPVPP

P

T

T

V

V

T

T

V

PVP

C

CRCC ,

As >1,, (PV=const)

2 11 Th h i t

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2.11 Thermochemistry

In the point of system Lose heat (-q): exothermic

surrounding

Chemical rxn

Phase change

In the point of system gain heat (+q): endothermic

surrounding

Chemical rxn

Phase change

OH1O2

1H1 222 222 O

2

1HOH

Stoichiometric number (i)

reactant product productreactant

222 O21H1OH10 - OH1O21H10

222

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38

3H2, 2O2Initial state:

xH2, yO2,zH2O

0H2, 1/2O2,3H

2O

final state:

OH1O2

1H1

222

H2 O2 H2O total mole

3 2 0 5

3 -1 21/2 1 5 -1/2

: extent of rxn

ni+ i :

ni:

2H O H 1/2O

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39

2H2, O2Initial state:

2H2Ofinal state:

OH1O2

1H1

222

: extent of rxn

H2, 1/2O2

H2O

2 -1 11/2 1

Total mole: 3 -1/2

1-1 1/21/2 1

Total mole: 3/2 -1/2

H2O-2mole222 HOH2OH2 H2O-1mole222 HOHO21H

OHrH22OH-1mole

H

OH-1moleOH-2mole22

H2H

Rxn enthalpy for 1moleH2O (J/mol)

N

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Chemical RXN will happen under the various

conditions, Lets define the standard condition. With the standard data in table, we could calculate the

energy change for various rxn.

TD standard states Puregaseous substance (g) under 1 bar at a certain temp. Pure liquid (l), under 1 bar at a certain temp.

Pure x-talline (s), under 1 bar at a given temp.

Ideal solution (1mol/kg) under 1 bar at a given temp.

40

N

i

o

ii

o

r HH

1

H t d l l t th th l ?

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How to use and calculate the rxn enthalpy?

Enthalpy : state function Once initial and final state are same. Then energy change bw two

states (rxn enthalpy) is same.

Hess : Law of constant heat summation

enthalphy()

41

)()(2

1)( 2 gCOgOgraphiteC ?

o

rH

)()()( 22 gCOgOgraphiteC molkJHo

r /509.393

)()(2

1)( 22 gCOgOgCO

molkJHo

r /984.282

)()(2

1)( 2 gCOgOgraphiteC molkJH

o

r /525.110

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42

)()()( 22 gCOgOgraphiteC

molkJHo

r

/509.393

)()(2

1

)( 22 gCOgOgCO

molkJHor /984.282

2 12 Enthalpy of formation

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2.12 Enthalpy of formation

Since absolute enthalpy (energy level) are now known,enthalpies relative to a defined reference stateare used instead.

Can you know how much energy H2O has? (NO)

H2O can be made by reacting the H2(g) and 1/2O2(g). Can youmeasure the rxn enthalpy for the formationof H2O? (YES)

If the rxn enthalpy for the formation of substance from the elements isprovided as a table, the formation enthalpy will be used for the

reference data. rxn enthalpy is temp dependant (note: CPis temp dependent.)

At 25oC, there are one more forms. For example, H2(g), H(g).Lets consider the reference form as the most stable substanceat 25 oC and 1 bar.

H2(g) is most stable at 25o

C H2(g) is the reference state. For Carbon, graphite is the reference state.

Different reference state may be used in various temp. range.

The enthalpy of formation of an element in its standard state is

zero at every temperature. (PLS see Table C.2)

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condition)standardaatelementsfromsubstanceofenthalpy(rxn

contionstandardaatformationofEnthalpy:o

fH

N

i

oifi

or HH

1

Calculation of rxn enthalpy at T

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Calculation of rxn enthalpy at T (Enthalpy of formation at 298.15 K) : given in Table.

Remember, TD is dealing with equilibrium state. Enthalpy is a state function.

ProductsTK

H298

Reactants

ProductsReactants

THo

o

r

rT=298.15K

298

,

T

reaxtP dTC

T

pro dP dTC298

,

o

iPi

To

Pr

o

r

T

reaxtPprodP

o

r

T

prodP

o

r

T

reaxtP

o

Tr

C

dTCH

dTCCH

dTCHdTCH

,o

Pr

298

298

298

,,298

298

,298

298

,

Cwhere,

)(

Think about What is the enthalpy of formation for water at 5oC?

What is the enthalpy of formation of ice at -5oC?

2 13 Calorimeter

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2.13 Calorimeter

In real experiment, it is relatively easy tomeasure the heat transfer with temp

change. How about PV work?

Adiabatic bomb calorimeter {adiabatic andno volume change (ie. No PV work)}

Measure the temp change (internal energy)

Consider the PV work with the eq. for ideal

gas {You dont need to consider the liquid and

solid. Gas phase mainly contributed the PVwork. V(g) >> V(l), V(s)}

Solve Ex: 2.10

grr RTUH

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47

Born Haber cycle : 1 mol of NaCl

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Born-Haber cycle : 1 mol of NaCl

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1840?

.

.

.

, .

Friction.

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