2-0-1st law of td
TRANSCRIPT
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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.
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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
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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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