Chapter 10 Spontaneity, Entropy, and Free Energy
Chapter 10
Spontaneity, Entropy, and Free Energy
Concept for second law of thermodynamic
熱機原理壓縮機原理
亂度
Isothermal expansion device
One-Step Expansion (No Work)
Mass M1 is removed from the pan, the gas will expand, moving the piston to the right end of the cylinder.
P1→1/4P1, V1→4V1,
No work is done. W0=0Free expansion
One-Step Expansion
M1 is replaced by M1/4.
11111
1
1
1
4
3)4)(
4(
4
VPVVP
W
VPW
PP
ex
ex
Two-Step Expansion
P1→1/2P1, V1→2V1
112
1111
1
1111
1
224
4
22
22
1
VPW
VP)VV(
PW
tep Work for s
VP)VV(
PW
tep Work for s
''
'
1/2P1→1/4P1, 2V1→4V1
PV diagram two-step expansion
The PV diagram six-step expansion
Infinite-Step Expansion
)V
VnRTln(Wq
V1.4P1.4nRTnRTln4W
)V
4VnRTln()
V
VnRTln()lnVnRT(lnVW
V
dVnRTW
dVPW
.increments smallmally infinitesiby changed is P
ΔVPW
1
2revrev
11rev
1
1
1
212rev
V
Vrev
V
V exrev
ex
i
n
1iin
2
1
2
1
(dV: V→0 )△
)(2
1
2
1
W
W
V
VPdVdWPdVdW
)(V
nRTP
1
2ln12
1 x
xdx
x
x
x
Reversible expansion
Reversible Process
Reversible process: the system is always infinitesimally close to equilibrium, and an infinitesimal change in conditions can reverse the process to restore both system and surroundings to their initial states.
Heat Engines
A heat engine converts some of the random molecular energy of heat flow into macroscopic mechanical energy.
qH: the working substance from a hot body
-w: the performance of work by the working substance on the surroundings
-qC: the emission of heat by the working substance to a cold body
The Second Law of Thermodynamics Kelvin-Planck statement for heat engine
It is impossible to extract an amount of heat qH from a hot reservoir and use it all to do work W. Some amount of heat qC must be exhausted to a cold reservoir.
This is sometimes called the "first form" of the second law, and is referred to as the Kelvin-Planck statement of the second law.
Heat Efficiency
100
1
0
0ΔU
0q and 0 wso
0q- and 0w- ,0q engineheat a
c
cw
e , qq
q
q
q
qqe
qq-wwqqwq
tion, e of operaFor a cycl
q
w
q
-w
utenergy inp
workiency eheat effic
For
Hc
cc
cc
HH
H
HH
HH
The Second Law of ThermodynamicsClausius statement for refrigerator
It is not possible for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. Energy will not flow spontaneously from a low temperature object to a higher temperature object.
The statements about refrigerators apply to air conditioners and heat pumps which embody the same principles.
Carnot’s Principle
No heat engine can be more efficient than a reversible heat engine when both engines work between the same pair of temperature H and C.
Isothermal Process: the temperature of the system and the surroundings remain constant at all times. (q=-w)Adiabatic: a process in which no energy as heat flows into or out of the system. (∆U=w)
Carnot cycle
four stage reversible sequence consisting of
1. isothermal expansion at high temperature T2
2. adiabatic expansion
3. isothermal compression at low temperature T1
4. adiabatic compression
V
dVnR
T
dq
T
dT(T)C
V
dVnR
T
dq
T
dT(T)C
dVV
nRTdq(T)dTC
PdVdqdwdqdU
(T)dTC and dUV
nRTP
V
V
V
V
1
4
3
2
4
3
2
1
)()()()(T
T V
T
T
T
T VV
T
T VV T
dTTC
T
dTTC
T
dTTC
T
dTTC
T
dT(T)C
0)(
)()(
)()()(
1
4
3
2
T
dTTC
T
dTTC
T
dTTC
T
dTTC
T
dTTC
T
dTTC
V
T
T V
T
T V
T
T V
T
T
T
T VV
H
C
H
C
C
H
0 0
H
CH
H
CHrev
C
C
H
HT
T
T
T
T
T
T
T
T
T
T
T
T
TT
q
qqe
T
q
T
q
T
dq
T
dq
T
dq
T
dq
T
dq
T
dq
T
dq
T
dq
0
0
4
3
2
1
1
4
3
2
4
3
2
1
0 0
Adiabatic Process
)( )()()(
R)C-C( )()()()(
)ln()ln()ln(11
0
1
2
1)1(
2
1
1
2
vp)(
2
1
1
2
2
1
1
2
2
1
1
2
1
22
1
2
1
v
pC
C
CCCRC
v
V
V
T
Tv
vvv
C
C
V
V
V
V
T
T
V
V
T
T
V
V
T
T
V
VR
V
VR
T
TC
VRdT
TC
dVV
RdT
T
CdV
V
nRTdTnCPdVdTnC
)qw (ΔU
v
p
vpvv
12 1
11 2
11 1 2 2 2 2 2 1
1 1 2 211 2 1 1 1 2
1 1 2 2
1 1 2 2
T V
T V
for ideal gas
PV PV T PV VPV PV
T T T PV V
For isothermal process PV PV
For adiabatic process PV PV
Isothermal process
1
2
1
2
ln
ln12
1
2
1
V
VnRTq
V
VnRTdV
VnRTdVPW
V
V
V
V
∆U=0, q=-w
Adiabatic process
1
)( 11
12
2
1
2
1
VVKW
dVV
KdVPW
KPVV
V
V
V
No heat transfer (q=0) , ∆U=w
Adiabatic Process
Process in which no heat transfer takes place
WTTnRU )(2
312
Application of Carnot Cycle
P (atm) V (L)
3 10
1.5 20
1 25.5
2 12.75
Calculate Q, U, W First law:
△U = QH – QL + W
W = QL - QH
Spontaneous Process and Entropy
Spontaneous Process: A process occurs without outside intervention.
Entropy: In qualitative terms, entropy can be viewed as a measure of randomness or disorder of the atoms or molecules in a substance.
Definition of Entropy
S=kBlnΩ
kB: Boltzmann’s constant
Ω : the number of microstates
corresponding to a given state
For one particle
S1=kBlnΩ1
S2=kBlnΩ2
∆S=S2-S1= kBlnΩ2-kBlnΩ1=kBln(Ω2/Ω1)
∆S= kBln(2Ω1/Ω1)=kBln2
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巨觀來看 微觀來看
Ω
4顆粒子都有 2種選擇,微觀態數: 24=16
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1 2
1 2
2 2
1 1
2
1
2
1
1 ln( )
ln( )
V V
V V
V
V
VFor mole of gas S R
V
VFor n mole of gas S nR
V
Definition of entropy in term of probability
Entropy for Isothermal Process
1
2
1
2
ln
ln12
1
2
1
V
VnRTq
V
VnRTdV
VnRTdVPW
V
V
V
V
T
qΔS
)V
V(nRT) and q
V
V(nRΔS
wqΔUwq
rev
rev
1
2
1
2 lnln
0
Entropy and Physical ChangesTemperature Dependence
2
1
2
121
2
1
2
121
2
1
2
121
1
2
1
2
ln
constant VFor
ln
constant PFor
T
T vv
T
T
p
TTvv
p
T
Tp
T
T
p
TTpp
T
T
revT
TTTrev
rev
T
TnC
T
dTnC
T
dqSdTnCdq
T
TnC
T
dTnC
T
dqSdTnCdq
T
dqdSS
T
dqdS
T
qS
Entropy and Physical ChangesChange of State
Change of state from solid to liquid
qrev=ΔHfusion
T=melting point in K
Change of state from liquid to gas
qrev=ΔHvaporization
T=boiling point in K
T
qS rev
The Second Law of thermodynamicsThe Third Statement
In any spontaneous process, there is always an increase in the entropy of the universe.
dq/T is the differential of a state function S that has the property ∆Suniv ≥ 0 for any process
Entropy and Second Law of Thermodynamics
ΔSuniv= ΔSsys+ΔSsurr
T
HSsurr
Gibbs Free EnergyAt constant T and P
0 spontaneous
0 equilibrium
0 non-spontaneous
surr surr sys univ
G HG H T S S
T TH G
S S S ST T
G
G
G
△Suniv>0, so G<0△
Free Energy and Chemical Reactions
0 0 0
0 0 0
reaction products reactantsS S S
G H T S
Third Law of Thermodynamics The entropy of a perfect crystal at 0 K is zero.
It is impossible to reach a temperature of absolute zero
It is impossible to have a (Carnot) efficiency equal to 100% (this would imply Tc = 0).
0lim 0T
S
(a) T=0 K, S=0
(b) T>0 K, S>0
The Dependence of Free Energy on Pressure
3 3 2 2 2 2
3 2 2 3 2 2
3 2 2
0
2 2 3
0 0 0
0 0 0
0 2 3
0
ln( )
3 2
2[ ln( )] [ ln( )] 3[ ln( )]
(2 3 ) [2 ln( ) ln( ) 3ln( )]
[ln( ) ln( ) ln( )]
NH NH N N H H
NH N H NH N H
reaction NH N H
reaction
G G RT P
N H NH
G G RT P G RT P G RT P
G G G RT P RT P P
G G RT P P P
G G
3
2 2
3
2 2
2
3
20
3
ln[ ]( )( )
ln( ) (Q= )( )( )
NH
N H
NHreaction
N H
PRT
P P
PG G RT Q
P P
Free Energy and Equilibrium
0
0
0 ln( )
ln( )
products reactants reaction
reaction
At equilibrium
G G G G RT Q
G RT K
The Temperature Dependence of K
0 0 0
0 0 0 0
0 0
22
0 0
11
02
1 2 1
ln( )
1ln( ) ( )
ln( )
ln( )
1 1ln( ) [ ]
G RT K H T S
H S H SK
RT R R T R
H SK
RT R
H SK
RT R
K H
K R T T
Free Energy and Work
max
P and Tconstant at process aFor
wG
dwdGPdVdwPdVdG
dwPdVdwdwdw
PdVdwVdPPdVdwdG
PVwTSPVwTSGTSPVUH-TSG
wTSUdwTdSdUTdSdwdUdq
TdSdqT
dq
T
dqdS
otherother
otherotherVP
irrev
revirrevirrevrev
rev