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Chapter 3 The second lawChapter 3 The second law A spontaneous direction of change: the direction of change that does not require work to be done to bring it about.
Clausius statement: No cyclic process is possible in which the sole result is the transfer energy from a cooler to a hotter body.
All real process is irreversible process.
3.1 The second law
Second kind of perpetual motion machine
A machine that converts heat into with 100 percent efficiency.
It is impossible to built a second kind of perpetual motion machine.
Kelven statement : No cyclic process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work.
3.1.1 Carnot principle
1. Efficiency of heat engine
def work performed
heat absorbed h
w
Q
1 2
1 1
def W Q Q
Q Q
2. Carnon cycleProcess AB : (i)
A
B
V
VnRTQ ln11
CD : (ii)2 2lnD
C
VQ nRT
V
BC : (iii)1 1
1 2B CTV T V p
V
C
C
DA : (iv)
1 11 2A DTV T V
(iii),( iv)
CB
A D
VV
V V
therefore (v)1 2 1 2( - )ln B
A
VQ Q nR T T
V
1 2
1 1
def W Q Q
Q Q
1
21
1
21
1 T
TT
Q
Q
W
0
2
2
1
1 T
Q
T
Q
3. Carnot principle
No heat engine can be more efficient than a reversible heat engine when both engines work between the same pair of temperature T1 and T2.
1 2r
1
T T
T
Reversible engine (=)
Irreversible engine(<)
Reversible engine (=)
Irreversible engine(<)
1
21
T
TT
02
2
1
1 T
Q
T
Q
3.1.2 Entropy
1. Definition of entropy
0≤+2
2TQ
1
1TQ
Reversible engine (=)
Irreversible engine(<)
≤0δ
exT
Q
Reversible engine (=)
Irreversible engine(<)
≤0δ
exT
Q
Reversible engine (=)
Irreversible engine(<)
Clausius inequality
Reversible process
Irreversible process
0ex
ir T
Q
0r T
Q
Entropy S
(1) S is a state function
(2) S is an extensive property
(3) unit is J·K - 1
T
QS rδ def
d * Reversible process
2
1
2r
2 1 1
δdS
S
QS S S S
T * Reversible process
Reversible process0r T
Q
2. Clausius inequality
Bir ir
Aex ex
δ δor d
Q QS S
T T
B Air r
A Bex
δ δ0
Q Q
T T (Irreversible
cycle)B B
ir r
A Aex
δ δQ QS
T T therefore
B
Aex
δ
T
QS Irreversible
Reversible
ex
δd
T
QS Irreversible
Reversible
3 . The principle of the increase of entropy
For an adiabatic process , δQ =0,
B
Aex ex
irreversible irreversibleδ δor d
reversible reversible
Q QS S
T T
The entropy of a closed system must increase in an irreversible adiabatic process.
0ad S
0d ad S IrreversibleReversible
IrreversibleReversible
4 . The entropy criterion of equilibrium
For an isolated system , δQ =0,
B
Aex ex
irreversible irreversibleδ δor d
reversible reversible
Q QS S
T T
reversible
leirreversib0dor
reversible
leirreversib0 isoiso SS
Since all real process is irreversible, when processes are occurring in an isolated system, its entropy is increasing.
Thermodynamic equilibrium in an isolated system is reached when the system’s entropy is maximized.
5. Calculation of entropy change of surrounding
Note: δQsu = - δQsy
sysu
ex
δQS
T
sysu
ex
(-δ )d
QS
T
Tex=constantsy
suex
QS
T
3.1.3 Calculation of entropy change of system
T
Q
T
QS rrδ
S=0
Reversible adiabatic process.
2
1
r12
δd2
1 T
QSSSS S
S Reversible process
Reversible isothermal process.
T
VpU
T
QS
ddδd r
V
VnR
T
TnCS V dd
d m, (perfect gas)
)lnln(
)lnln(
)lnln(
1
2m,
1
2m,
2
1
1
2
m,
1
2
1
2
m,
V
VC
p
pCn
p
pR
T
TCn
V
VR
T
TCnS
pV
p
V
IsochoricIsobaricIsothermalAdiabatic reversible irreversibleliquid and solid
1. p, V, T change
(1). p, V, T change
(2) mixing of different perfect gas at constant temperature and constant pressure
mixing at constant T, p
2
212
1
211mix lnln
V
VVRn
V
VVRnS
at constant T, p 221
21
21
1 , yVV
Vy
VV
V
mixS = - (n1 Rlny1 + n2Rlny2)
y1 < 1 , y2 < 1 , so mixS >0
n1
T, p, V1
n2
T, p,V2
n1 +n2
T, p, V1+V2
2. The phase transition
(1) The phase transition at transition temperature
At constant T, p , the two phase in the system are in equilibrium,
the process is reversible, and W′ =0, so Qp = H ,
T
HnS
)transitionphase ofenthalpy(m
fusHm >0, vapHm >0,
Sm(s) < Sm(l) < Sm(g)
(2) irreversible phase transition
Irreversible B(,T1,p1) B(,T2,p2)
S=?
B(,Teq,peq) B(, Teq,peq) reversible
S2
S1 S3
S = S 1+ S 2+ S 3
T/TnCST
T p d)l,OH( 2
m,1
eq
1
T/TnCST
HnS
T
T p d)g,OH( , 2
m,3mvap
2
2
eq
S = S 1+ S 2+ S
3
For example
Irreversible
S =?H2O ( l, 90 ,100kPa)℃ H2O ( g, 90 , ℃
100kPa)S1 S3
Reversible
S2H2O ( l, 100 , 100kPa)℃ H2O ( g, 100 , ℃
100kPa)
3.2 The third law
3.2.1 The third law
Nernst heat theorem: The entropy change accompanying any physical or chemical transformation approaches zero as the temperature approaches zero: S0, T0.
The entropy of all perfect crystalline substance is zero at 0 K.
-1
0KJ0)(lim
TS
T
S *(perfect crystalline , 0 K) =0J · K -
1
3.2.2 Conventional molar entropy and standard molar entropy
Second law T
T
QSTS
0K
rδ)K0(*)(*
Third law S *( 0K )= 0
T
T
QTS
0K
mr,*m
δ)B,(
Sm* ( B , T)— conventional molar entropy of substanc
e at temperature T 。
S m(B,,T) —standard molar entropy
(p = 100kPa)
3.2.3 The entropy change of chemical reaction
For reaction aA + b B →yY + zZ
rS m(T) = Σν B S m ( B, ,T )
rS m (298.15K) = Σν B S m ( B, ,298.15K )
rS m (T) = yS m (Y, ,T ) + z S m (Z, ,T )
- a S m (A, ,T ) - b S m (B, ,T )
1S
2S 4
S3
S
)( 1mr TS
aA + bB yY + zZ
)( 2mr TSaA + bB yY + zZ
r S m (T1) = S 1 + S 2 + r S m (T2) + S 3 + S 4
T p
T
TCSTS
298.15K
m,Bmrmr
d)B().15K298()(
3.3 Helmholtz and Gibbs energies
3.3.1 Helmholtz energy A
reversible
leirreversibδd
exT
QS
reversible
leirreversib
exT
QS
Tex ( S2 - S1 )≥ Q
at constant T , T2S2 - T1S1 = Δ(TS)
Q = ΔU - W
Δ(TS )≥ ΔU - W
- Δ(U - TS )≥-W
The maximum work done by a closed system in an isothermal
process is obtained when the process is carried out reversible.
TSUA def
Helmholtz energy A is an extensive state function
reversible
irreversible A T≥ W
reversible
irreversible A T ≤ W
reversible
irreversible dA T ≤ W
In a closed system doing volume work only at constant T and
V, the Helmholtz energy A decrease in a spontaneous change.
At constant T and V W = 0
δW′ = 0
reversible
irreversible dA T, V ≤ 0 reversible
irreversible A T, V ≤ 0
In a closed system capable doing only volume work, the constant temperature and volume equilibrium condition is the minimization of the Helmholtz energy A. dA T, V = 0
Helmholtz energy criterion
3.3.2 Gibbs energy G
at constant T Δ(TS)≥ΔU - W
Δ(H - TS )≤ W′
at constant p p1 = p2 = pex
W =- pex ( V2 - V1 )+ W ′
=- p2V2 + p1V1 + W′
=- Δ(pV) + W′
Δ(TS)≥ΔU + Δ(pV) - W′
-[ ΔU + Δ(pV)-Δ(TS) ]≥- W′
- Δ(U+pV-TS)≥ - W′
G H - TS = U + pV - TS = A + pV def
Gibbs energy G is an extensive state function
reversible
irreversible G T ,p≤ W
reversible
irreversible dG T, p ≤ W
The maximum possible non-volume work done by a closed
system in a constant temperature and pressure process is
equal to the G
reversible
irreversible G T ,p≤ W
reversible
irreversible dG T, p ≤ W
W′ = 0, δW′ = 0
reversible
irreversible G T ,p≤ 0
reversible
irreversible dG T, p ≤ 0
In a closed system doing volume work only at constant T
and p, the Gibbs energy G decrease in a spontaneous change.
In a closed system capable doing only volume work, the constant temperature and pressure equilibrium condition is the minimization of the Gibbs energy G . dG T, p = 0
Gibbs energy criterion
3.3.3 Calculation of A and G
(1).change p ,V at constant temperature
at constant T, reversible, dAT = δWr =- pdV + δW r '
If δWr ' = 0 , then dAT =- pdV
VpAV
VT d2
1 closed system, change p ,V at constant T,
W′ = 0 reversible process
reversible
irreversible dA T≤ W
1
2
1
2 lnlnp
pnRT
V
VnRTAT For perfect gas
ΔA= ΔU- TΔS ΔG= ΔH- TΔS
pVGp
pT d2
1
G = A + pV, dG = dA + pdV + Vdp
at constant T, reversible, δW′r = 0, dA =- pdV
1
2
1
2 lnlnV
VnRT
p
pnRTAG TT
dGT = Vdp
closed system, change p ,V at constant T, W′ = 0 reversible process
1
2
1
2 lnlnV
VnRT
p
pnRTGT For an perfect gas
(2) phase transition
(i) reversible phase transition
(ii) irreversible phase transition
reversible phase transition at constant T and p
ΔU = ΔH-Δ(pV) = ΔH- pΔV = ΔH- nRT
ΔA = ΔU -Δ(TS) = ΔH - nRT - TΔS = - nRT
G =H - (TS)
A = U - (TS) ΔG =ΔH-Δ(TS) = ΔH- TΔS
ΔA = ΔU -Δ(TS) = ΔU - TΔS
ΔH =TΔS
reversible vaporization or sublimation at constant T and p, and vapor is an perfect gas
ΔG= ΔH- TΔS = 0
3.3.4 Funtdantal relations of thermodynamic functions
U, H, S, A, G, p, V and T
H = U + pV
H
U pV
pVATS
TS G
A = U - TS
G = H - TS = U + pV -TS = A +pV
1. Master equation of thermodynamic
If δWr ′ = 0 , then δWr =- pdV ,
dU=δQr+δWr ,For a reversible change
δQr = TdS
dU = TdS pdV
dH = TdS + Vdp
dA = SdT pdV
dG = SdT + Vdp
H =U+pV
A =U-TS
G =H-TS
closed system of constant composition , reversible process , volume work only.
Master equation of thermodynamic
dU = TdS - pdV
dH = TdS + Vdp
dA = - SdT - pdV
dG = - SdT + Vdp
TS
U
V
pV
U
S
TS
H
p
V
p
H
S
ST
A
V
p
V
A
T
ST
G
p
Vp
G
T
dyy
Zdx
x
ZdZ
xy
dZ = M dx + N dy
xy y
ZN
x
ZM
,
dZ = M dx + N dy
yxx
N
y
M
dU = TdS - pdVSV V
T
S
p
dH = TdS + VdppS
S
V
p
T
VT T
p
V
S
pTT
V
p
S
dA = - SdT - pdV
dG = - SdT + Vdp
2. Maxwell’s relation
3. Gibbs - helmholtz equation
2T
TT
GTG
Tpp
2
)/(
T
H
T
TG
p
2
)/(
T
U
T
TA
V
2
T
GTS
2T
H
pT
TG
)/(
3.4 Chemical potential 3.4.1 Chemical potential
Total differential
B
)BC,C(,,B
A
)AC,C(,,A)B(,)B(,
d
dddd
nn
G
nn
Gp
p
GT
T
GG
npT
npTnTnp
G = f (T , p , nA , nB……)
Consider a system that consists of a single homogeneous phase whose composition can be varied
ST
G
np
)B(,
Vp
G
nT
)B(,
B
B
)BC,C(,,B
d d d- d nn
GpVTSG
npT
)BCC,(,,BB
def
nPTn
G
B
BBdddd npVTSG
B
BBdddd nVpTSA
B
BBdd-dd nVpSTU
B
BBdddd npVSTH
)BCC,(,,BB
npS
n
H
)BCC,(,,BB
nVT
n
A)BCC,(,,B
B
nVS
n
U
)BCC,(,,BB
nPT
n
G
Chemical potential is an intensive state function.
B of pure substance
G* (T , p , nB) = nB G *m,B (T , p , )
),(*B,mB
,B
pTGn
G
pT
α
αB
B
αB
α
αα
α
αα dddd nVpSTU
Heterogeneous system
α B
αB
αBdnpdVTdSdU
α B
αB
αBdnVdpTdSdH
α B
αB
αBdnpdVSdTdA
α B
αB
αBdnVdpSdTdG
(1) Condition of phase equilibrium
mequilibriu
sspontaneou0β
BαB
0ddd BαB
βB nnn
T , pB() B( )
dnB
0d αB
α B
αB n
β
B
α
B Condition of phase equilibrium
β
B
α
B spontaneous
3.4.2 Equilibrium criterion of substance
(2) Condition of chemical reaction equilibrium
Chemical reaction B0 B
B
BBB
BB mequilibriusspontaneou 0dd n
ΣBμB < 0 , dξ > 0 spontaneous
Homogeneous system dξ
reversible
leireverisibn 0d α
Bα B
αB
For example aA + bB = yY + zZ
aμA+bμB = yμY+zμZ
ΣBμB = 0 , equilibrium
BB
defA
A—potential of chemical reaction
A < 0 left spontaneous 。
A = 0 at equilibrium ;
A > 0 right spontaneous ;
A= (aμA+bμB) -( yμY+zμZ )
aA + bB = yY + zZ
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