物理化学英文3The Second Law
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DS for Phase Changes
• Melting
• Boiling
• T = Tm = constant • q = DHfus
• \ DS = DHfus / Tm
• T = Tb = constant • q = DHvap
• \ DS = DHvap / Tb
DS for a Temperature Change
Variation of G with Temperature
GT p S
G
Sgas Sliquid Ssolid
gas
liquid solid
T
Variation of G with Pressure
G p T
V
G
Vgas Vliquid Vsolid
gas
liquid solid
\ DA = wmax • dAT,V i.e. for constant T and V, A is a
minimum at equilibrium.
Gibbs Free Energy
• Definition: G U - TS + pV
• dGT,p = dU - TdS + pdV • dGT,p dU - dq - w • dGT,p dwe
y(x)
y2 y1
x2 - x1
Debye Extrapolation
For low temperatures, Cp T3
Therefore,
T0
0
Cp dT T
1 3
Cp
(T0 )
Helmholtz Free Energy
• Definition: A U - TS
• dAT = dU - TdS • dAT dU - dq • dAT dw
p
Gibbs-Helmholz Equation
G T
p
S
G
T
H
1 G T T
p
G T2
H T2
(G /
T
T )
p
H T2
Temperature dependence of DG
(DG /
T
T )
p
DH T2
T2
T2 DHdT
d (DG / T )
T1
T1
T2
DG(T2 ) T2
• Corollary: For an isolated system
• dS = 0 for a reversible change • dS > 0 for an irreversible change
p / atm
Carnot Cycle
3 2 1
isothermal: 100 K adiabatic isothermal: 200 K adiabatic
The Fundamental Equations
• dU = TdS - pdV • dH = TdS + Vdp • dA = -SdT - pdV • dG = -SdT + Vdp
Maxwell Relations - Derivation
2U SV
2U V S
dU
U S
Th qh
Engine w qc Tc
Efficiency of Heat Engines
• An engine operates in a cycle; \ DU = 0 and |w| = qh - |qc|
You can’t get something for nothing.
• efficiency is given by e = |w| / qh \ e = 1 - |qc| / qh
\ DG = wmax,e • If the only work is expansion/compression,
dGT,p i.e. for constant T and p, G is a minimum at equilibrium.
The Second Law
The Machinery
50 100 150 200 250 300
T/K
Numerical Integration
• Trapeziodal Rule:
• divide total area into trapezoids
• trapezoid area =
½ (y2 + y1) (x2 - x1)
• integral equals sum of all trapezoid areas
• |qc|>0 implies e < 1
You can’t even break even.
The Second Law (Kelvin)
• No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work.
Physical Chemistry
The Second Law
The Second Law
The Concepts
The Laws of Thermodynamics
• First Law You can’t get something for nothing.
• Second Law You can’t even break even.
• In an isolated system, any change for which dS>0 will occur spontaneously.
• In an isolated system at equilibrium, the entropy is at a maximum.
Spontaneous Heat Transfer
pure substance G
mixture G
n
ni
i is the slope of Gi vs. ni
• for a pure substance G = nGm and Gm=constant
• for a mixture, Gi vs. ni is not linear and is not constant
V
dS
U V
S
dV
TdS
pdV
\
T V
S
p S
V
Maxwell Relations - Summary
T V
S
p S
V
T p
S
V S
p
p T
V
S V
T
V T
p
S p
T
Thermodynamic Eq’n. of State
T
U V
T
T
T
p T
V
p
dU T dV CV dT
2. This ratio may be written: q1/q2 = T1/T2 3. The entropy, S, defined by dS=(dq/T)rev is
a function of state.
4. In an isolated system, DS is zero for a reversible process, and positive for a spontaneous one.
DG(T1 ) T1
DH
1 T2
1 T1
Pressure dependence of DG
DG
p2
p2
p
T
DV
dDG
p1
DVdp
p1
p2
DG( p2 ) DG( p1) DVdp
p1
for gases: DG( p2 ) DG( p1) nRT ln( p2 / p1)
Third Law of Thermodynamics
• Nernst Heat Theorem: DS 0 as T 0. • Arbitrary convention: The entropy of an
element in its pure crystalline form is zero at T=0. • Third Law: Therefore the entropy of any perfect crystalline substance is zero at T=0.
Assume a reversible path at constant pressure. Then,
DS
Tf
dq
/
T
Tf
Cp dT
Ti
T Ti
DS
Cp
ln
Tf Ti
DS for an Ideal Gas Isotherm
dq dw pdV
dS pdV nR dV
T
V
DS nR ln Vf Vi
Ideal Gas Mixture Model
for a pure substance : G( p) = G( p0 ) nRT ln( p / p0 ) Gm ( p) Gm ( p0 ) RT ln( p / p0 )
( p) 0 RT ln( p / p0 )
for a substance in an ideal gas mixture :
0
0
10
20
30
40
50
V/L
Carnot Efficiency
DS = DSengine + DSsurr 0 but DSengine = 0 \ DSsurr = - (qh / Th) + (qc / Tc) 0 \ qc / qh Tc / Th e = |w| / qh = 1 - |qc| / qh
\ e 1 - (Tc / Th)
Spontaneity and Equilibrium
• A spontaneous change is an irreversible one.
• Therefore any change for which dS>dq/T will occur spontaneously
Third Law Entropies
S 0 (T) S 0 (0) Tf CpdT DH fus
0T
Tf
Tb
Cp dT
DHvap
T
Cp dT
Tf T
Tb
Tb T
Third Law Entropy of Pb(s)
0.60 0.50 0.40 0.30 0.20 0.10 0.00
0
dS dSh dSc
dS dq dq
Th
Th
Tc
dq
1 1
dS dq Tc Th dS 0 Th Tc
Tc
Entropy at the Molecular Level
Winitial = 6
Wfinal = 28
DS>0 reflects an increase in W, the weight of the configuration
The Entropy Function
• The entropy S is defined by
dS = dqrev/T
• dS dqirrev/T
reversible: dS = dq/T
state B
state A
irreversible:
dS dq/T
The Second Law (Clausius)
Heat Engines
• A heat engine converts heat into work.
• Th = temperature of heat source
• Tc = temperature of heat sink
• qh = heat supplied • qc = heat released • w = work produced
• In other words, |w| < qh and the efficiency of a heat engine is less than 100%.
Propositions
1. For a reversible cycle, absorbing q1 at T1 and q2 at T2, the ratio q1:q2 depends only on T1 and T2.
for solids and liquids, DG( p2 ) DG( p1)
Composition Dependence
dG
Gp
T ,n's
dp
GT
p,nBaidu Nhomakorabeas
dT
G n1
p,T
dn1
dG Vdp SdT 1dn1 2dn2
dG Vdp SdT i dni
i
Chemical Potential
• The entropy S is a function of state.
• dS dq / T (Clausius inequality)
• dS = dq / T for a reversible change • dS > dq / T for an irreversible change