Chapter 5 Simple Applications of Macroscopic Thermodynamics.
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Transcript of Chapter 5 Simple Applications of Macroscopic Thermodynamics.
![Page 1: Chapter 5 Simple Applications of Macroscopic Thermodynamics.](https://reader033.fdocuments.us/reader033/viewer/2022061510/56649d365503460f94a0edc2/html5/thumbnails/1.jpg)
Chapter 5Simple Applications of Macroscopic
Thermodynamics
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Preliminary DiscussionClassical, Macroscopic,
Thermodynamics• Now, we drop the statistical mechanics
notation for average quantities. So that now,
All Variables are Averages Only! • We’ll discuss relationships between
macroscopic variables using
The Laws of Thermodynamics
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• Some Thermodynamic Variables of Interest:Internal Energy = E, Entropy = S
Temperature = T• Mostly for Gases:
(but also true for any substance):
External Parameter = VGeneralized Force = p(V = volume, p = pressure)
• For a General System:External Parameter = xGeneralized Force = X
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• Assume that the External Parameter = Volume V in order to have a specific case to discuss. For systems with another external parameter x, the infinitesimal work done đW = Xdx. In this case, in what follows, replace p by X & dV by dx.
• For infinitesimal, quasi-static processes:
1st & 2nd Laws of Thermodynamics1st Law: đQ = dE + pdV
2nd Law: đQ = TdS
Combined 1st & 2nd LawsTdS = dE + pdV
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Combined 1st & 2nd LawsTdS = dE + pdV
• Note that, in this relation, there are
5 Variables: T, S, E, p, V• It can be shown that:Any 3 of these can always be expressed
as functions of any 2 others.• That is, there are always 2 independent
variables & 3 dependent variables. Which 2 are chosen as independent is arbitrary.
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Brief, Pure Math Discussion• Consider 3 variables: x, y, z. Suppose we
know that x & y are Independent Variables. Then, It Must Be Possible to express z as a function of x & y. That is,
There Must be a Function z = z(x,y).• From calculus, the total differential of z(x,y)
has the form:
dz (∂z/∂x)ydx + (∂z/∂y)xdy (a)
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• Suppose that, in this example of 3 variables: x, y, z, we want to take y & z as independent variables instead of x & y. Then,
There Must be a Function x = x(y,z).• From calculus, the total differential of x(y,z) is:
dx (∂x/∂y)zdy + (∂x/∂z)ydz (b)• Using (a) from the previous slide
[dz (∂z/∂x)ydx + (∂z/∂y)xdy (a)]& (b) together, the partial derivatives in (a) & those in (b) can be related to each other.
• We always assume that all functions are analytic.
So, the 2nd cross derivatives are equalSuch as: (∂2z/∂x∂y) (∂2z/∂y∂x), etc.
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Mathematics Summary• Consider a function of 2 independent variables:
f = f(x1,x2).• It’s exact differential is
df y1dx1 + y2dx2 & by definition:
• Because f(x1,x2) is an analytic function, it is always true that:
2 1
2 1
1 2x x
y y
x x
• Most Ch. 5 applications use this with theCombined 1st & 2nd Laws of Thermodynamics
TdS = dE + pdV
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Some MethodsMethods & Useful Math ToolsUseful Math Tools for Transforming DerivativesTransforming Derivatives
Derivative Inversion
Triple Product (xyz–1 rule)
Chain Rule Expansion to Add Another Variable
Maxwell Reciprocity Relationship
xx FyyF
1
TT SPPS
1
1
xFy Fy
yx
xF
1
THP HP
PT
TH
xxx yF
yF
TCTC
HT
TS
HS
P
P
PPP
11
y
x
x
y
x
yF
y
xF
yxxy FF
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Pure Math: Jacobian Transformations•A Jacobian Transformation is often used totransform from one set of independentvariables to another.•For functions of 2 variables f(x,y) & g(x,y) it is:
yxxy
xy
xy
xg
yf
yg
xf
yg
xg
yf
xf
yxgf
,,
Determinant!
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Transposition
Inversion
Chain Rule Expansion
yx
fgyxgf
,,
,,
gfyxyx
gf
,,
1,,
yx
wzwzgf
yxgf
,,
,,
,,
Jacobian TransformationsJacobian TransformationsHave Several Useful PropertiesUseful Properties
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• Suppose that we are only interested in the first partial derivative of a function f(z,g) with respect to z at constant g:
gz
gfzf
g ,,
yx
gzyxgf
zf
g
,,,,
• This expression can be simplified using the chain rule expansion & the inversion property
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dE = TdS – pdV (1)
First, choose S & V as independent variables:E E(S,V)
Properties of the Internal Energy E
dVV
UdS
S
UdU
SV
TS
U
V
p
V
U
S
Comparison of (1) & (2) clearly shows that
dE∂E
(2)
Applying the general result with 2nd cross derivatives gives:
VS S
p
V
T
Maxwell RelationMaxwell Relation I! I!
and∂E
∂E
∂E
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If S & p are chosen as independent variables, it is convenient to define the following energy:
H H(S,p) E + pV EnthalpyEnthalpyUse the combined 1st & 2nd Laws. Rewrite them in terms of dH: dE
= TdS – pdV = TdS – [d(pV) – Vdp] ordH = TdS + Vdp
Comparison of (1) & (2) clearly shows that
(1)
(2)
Applying the general result for the 2nd cross derivatives gives:
pSS
V
p
T
But, also:
and
Maxwell RelationMaxwell Relation II! II!
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If T & V are chosen as independent variables, it is convenient to define the following energy:
F F(T,V) E - TS Helmholtz Free Helmholtz Free EnergyEnergy• Use the combined 1st & 2nd Laws. Rewrite them in terms of dF:
dE = TdS – pdV = [d(TS) – SdT] – pdV or
dF = -SdT – pdV (1)
• But, also: dF ≡ (F/T)VdT + (F/V)TdV (2)
• Comparison of (1) & (2) clearly shows that (F/T)V ≡ -S and (F/V)T ≡ -p
• Applying the general result for the 2nd cross derivatives gives:
Maxwell RelationMaxwell Relation III! III!
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If T & p are chosen as independent variables, it is convenient to define the following energy:
G G(T,p) E –TS + pV Gibbs Free Gibbs Free EnergyEnergy• Use the combined 1st & 2nd Laws. Rewrite them in terms of dH:
dE = TdS – pdV = d(TS) - SdT – [d(pV) – Vdp] or
dG = -SdT + Vdp (1)
• But, also: dG ≡ (G/T)pdT + (G/p)Tdp (2)
• Comparison of (1) & (2) clearly shows that
(G/T)p ≡ -S and (G/p)T ≡ V• Applying the general result for the 2nd cross derivatives gives: Maxwell RelationMaxwell Relation IV! IV!
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1. Internal Energy: E E(S,V)
2. Enthalpy: H = H(S,p) E + pV
3. Helmholtz Free Energy: F = F (T,V) E – TS
4. Gibbs Free Energy: G = G(T,p) E – TS + pV
Summary: Energy FunctionsEnergy Functions
Combined 1Combined 1stst
& & 22ndnd Laws Laws
1. dE = TdS – pdV
2. dH = TdS + Vdp
3. dF = - SdT – pdV
4. dG = - SdT + Vdp
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dyy
zdx
x
zdzNdyMdx
xy
yxx
N
y
M
pSS
V
p
T
VS S
p
V
T
VT T
p
V
S
pTT
V
p
S
1. 2.
3. 4.
Another Summary: Maxwell’s Relations
(a) ΔE = Q + W
(b) ΔS = (Qres/T)
(c) H = E + pV
(d) F = E – TS
(e) G = H - TS
1. dE = TdS – pdV2. dH = TdS + Vdp3. dF = -SdT - pdV4. dG = -SdT + Vdp
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Maxwell Relations: “The Magic Square”?
V F T
G
PHS
E
Each side is labeled with an
Energy (E, H, F, G). The corners are labeled with
Thermodynamic Variables
(p, V, T, S). Get the
Maxwell Relations by “walking” around the square. Partial derivatives are obtained from the sides.
The Maxwell Relations are obtained from the corners.
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SummaryThe 4 Most CommonMost CommonMaxwell Relations:Maxwell Relations:
PTPS
VTVS
T
V
P
S
S
V
P
T
T
P
V
S
S
P
V
T
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Maxwell Relations: Table (E → U)
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InternalEnergy
HelmholtzFree Energy
Enthalpy
Gibbs FreeEnergy
Maxwell RelationsMaxwell Relations from dE, dF, dH, & dG
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Some Common Measureable PropertiesHeat Capacity at Constant Volume:
Heat Capacity at Constant Pressure:
∂E
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More Common Measureable PropertiesVolume Expansion Coefficient:
Isothermal Compressibility:
Note!! Reif’snotation forthis is α
The Bulk Modulus is theinverse of the IsothermalCompressibility!
B (κ)-1
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Some Sometimes Useful RelationshipsSummary of Results
Derivations are in the text and/or are left to the student!
Entropy:
dTRT
HdP
RT
V
RT
Gd
2
Enthalpy:
Gibbs FreeEnergy:
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Typical Example• Given the entropy S as a function of temperature
T & volume V, S = S(T,V), find a convenient expression for (S/T)P, in terms of some measureable properties.
• Start with the exact differential:
• Use the triple product rule & definitions:
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• Use a Maxwell Relation:
• Combining these expressions gives:
• Converting this result to a partial derivative gives:
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• This can be rewritten as:
• The triple product rule is:
• Substituting gives:
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Note again the definitions:• Volume Expansion Coefficient
β V-1(V/T)p
• Isothermal Compressibilityκ -V-1(V/p)T
• Note again!! Reif’s notation for theVolume Expansion Coefficient is α
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• Using these in the previous expression finally gives the desired result:
• Using this result as a starting point,
A GENERAL RELATIONSHIP between the
Heat Capacity at Constant Volume CV
& the Heat Capacity at Constant Pressure Cp
can be found as follows:
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• Using the definitions of the isothermal compressibility κ and the volume expansion coefficient , this becomes
General Relationship between Cv & Cp
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Simplest Possible Example: The Ideal Gas
P
RTPRT
vPRT
PRT
PvPv
v
T
RTR
vPR
PRT
TvTv
v
TT
PP
1
11
1
11
2
• For an Ideal Gas, it’s easily shown (Reif) that the Equation of State (relation between pressure P, volume V,
temperature T) is (in per mole units!): Pν = RT. ν = (V/n)• With this, it is simple to show that the volume expansion
coefficient β & the isothermal compressibility κ are:
and
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and
• So, for an Ideal Gas, the volume expansion coefficient & the isothermal compressibility have the simple forms:
• We just found in general that the heat capacities at constant volume & at constant pressure are related as
• So, for an Ideal Gas, the specific heats per mole have the very simple relationship:
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Other, Sometimes Useful, Expressions
TCONSTANTdVV
R
T
PS
TCONSTANTdPP
R
T
VS
TCONSTANTdPT
VTVH
P
P VTV
P
P PTP
P
P PTP
0
.
0
.
0
.
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More Applications: Using the Combined 1st & 2nd Laws (“The TdS Equations”)
Calorimetry Again! • Consider Two Identical Objects, each of mass m, &
specific heat per kilogram cP. See figure next page.
Object 1 is at initial temperature T1.Object 2 is at initial temperature T2.
Assume T2 > T1.• When placed in contact, by the 2nd Law, heat Q
flows from the hotter (Object 2) to the cooler (Object 1), until they come to a common temperature, Tf.
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• Two Identical Objects, of mass m, & specific heat per kilogram cP. Object 1 is at initial temperature T1. Object 2 is at initial temperature T2.
• T2 > T1. When placed in contact, by the 2nd Law, heat Q flows from the hotter (Object 2) to the cooler (Object 1), until they come to a common temperature, Tf.
Object 1Initially
at T1
Object 2Initially
at T2
Q Heat Flows
221 TT
T f
• After a long enough time, the two objects are at the same temperature Tf. Since the 2 objects are identical, for this case,
For some timeafter initialcontact:
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• The Entropy Change ΔS for this process can also be easily calculated:
21
21
21
2
2121
2
21
2ln2
ln2lnln
lnln1 2
TT
TTmcS
TT
Tmc
TT
Tmc
TT
Tmc
T
T
T
Tmc
T
dT
T
dTmcS
P
fP
fP
fP
ffP
T
T
T
TP
f f
• Of course, by the 2nd Law,the entropy change ΔS must be positive!! This requires that the temperatures satisfy: 0)(
02
42
2
221
212
22
1
21212
22
1
2121
TT
TTTT
TTTTTT
TTTT
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Some Useful “TdS Equations”• NOTE: In the following, various quantities are
written in per mole units! Work with theCombined 1st & 2nd Laws:
Definitions:• υ Number of moles of a substance. • ν (V/υ) Volume per mole.• u (U/υ) Internal energy per mole. • h (H/υ) Enthalpy per mole. • s (S/υ) Entropy per mole. • cv (Cv/υ) const. volume specific heat per
mole. • cP (CP/υ) const. pressure specific heat per mole.
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dPc
dvv
cdP
P
Tcdv
v
TcTds
dPTvdTcdPT
vTdTcTds
dvT
dTcdvT
PTdTcTds
vP
vv
PP
PP
P
vv
v
• Given these definitions, it can be shown that the Combined 1st & 2nd Laws (TdS) can be written in at least the following ways:
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Internal Energy u(T,ν):
dvPvu
dTcTds
dvvu
dTTu
du
Tv
Tv
Enthalpy h(T,P):
• Student exercise to show that, starting with the previous expressions & using the definitions (per mole) of internal energy u & enthalpy h gives:
![Page 41: Chapter 5 Simple Applications of Macroscopic Thermodynamics.](https://reader033.fdocuments.us/reader033/viewer/2022061510/56649d365503460f94a0edc2/html5/thumbnails/41.jpg)
v
v
v
vvvvv
Pv
P
T
T
c
P
s
P
T
T
sT
TP
T
T
s
P
s
dvv
sdP
P
sds
vPss
1
),( Consider
P
P
P
pPPPP
Pv
v
T
T
c
v
s
v
T
T
sT
Tv
T
T
s
v
s
dvv
sdP
P
sds
1
dvv
TcdP
P
TcTds
dvv
T
T
cdP
P
T
T
cds
dvv
sdP
P
sds
PP
vv
P
P
v
v
Pv
• Student exercise also to show that similar manipulations give at least the following different expressions for the molar entropy s: Entropy s(T,ν):