chapter 5 simple applications of macroscopic thermodynamics
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Chapter 5 Simple Applications of Macroscopic Thermodynamics. Preliminary Discussion. Classical, Macroscopic, Thermodynamics Drop the statistical mechanics notation for average quantities. We know that All Variables are Averages Only ! - PowerPoint PPT PresentationTRANSCRIPT
Chapter 5Simple Applications of Macroscopic
Thermodynamics
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
• 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
• 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 Thermodynamics
1st Law: đQ = dE + pdV2nd Law: đQ = TdS
Combined 1st & 2nd LawsTdS = dE + pdV
Combined 1st & 2nd LawsTdS = dE + pdV
• Note that, in this relation, there are5 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.
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)
• 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.
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 yx x
• Most Ch. 5 applications use this with theCombined 1st & 2nd Laws of Thermodynamics
TdS = dE + pdV
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
xyF
yxF
yxxy FF
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!
Transposition
Inversion
Chain Rule Expansion
yx
fgyxgf
,,
,,
gfyxyx
gf
,,
1,,
yx
wzwzgf
yxgf
,,
,,
,,
Jacobian TransformationsJacobian TransformationsHave Several Useful PropertiesUseful Properties
• 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
dE = TdS – pdV (1)First, choose S & V as independent variables:
E E(S,V)
Properties of the Internal Energy E
dVVUdS
SUdU
SV
TSU
V
p
VU
S
Comparison of (1) & (2) clearly shows that
dE ∂E (2)
Applying the general result with 2nd cross derivatives gives:
VS Sp
VT
Maxwell RelationMaxwell Relation I! I!
and ∂E
∂E
∂E
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:
pS SV
pT
But, also:
and
Maxwell RelationMaxwell Relation II! II!
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!
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!
1. Internal Energy: E E(S,V)2. Enthalpy: H = H(S,p) E + pV3. Helmholtz Free Energy: F = F (T,V) E – TS4. 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
dyyzdx
xzdzNdyMdx
xy
yx xN
yM
pS SV
pT
VS Sp
VT
VT Tp
VS
pT TV
pS
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
Maxwell Relations: “The Magic Square”?
V F T
G
PHS
E
Each side is labeled with anEnergy (E, H, F, G).
The corners are labeled withThermodynamic 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.
SummaryThe 4 Most CommonMost CommonMaxwell Relations:Maxwell Relations:
PTPS
VTVS
TV
PS
SV
PT
TP
VS
SP
VT
Maxwell Relations: Table (E → U)
InternalEnergy
HelmholtzFree Energy
Enthalpy
Gibbs FreeEnergy
Maxwell RelationsMaxwell Relations from dE, dF, dH, & dG
Some Common Measureable PropertiesHeat Capacity at Constant Volume:
Heat Capacity at Constant Pressure:
∂E
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
Some Sometimes Useful RelationshipsSummary of Results
Derivations are in the text and/or are left to the student!
Entropy:
dTRTHdP
RTV
RTGd 2
Enthalpy:
Gibbs FreeEnergy:
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:
• Use a Maxwell Relation:
• Combining these expressions gives:
• Converting this result to a partial derivative gives:
• This can be rewritten as:
• The triple product rule is:
• Substituting gives:
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
• Using these in the previous expression finally gives the desired result:
• Using this result as a starting point, A GENERAL RELATIONSHIP
between theHeat Capacity at Constant Volume CV
& the Heat Capacity at Constant Pressure Cp
can be found as follows:
• Using the definitions of the isothermal compressibility κ and the volume expansion coefficient , this becomes
General Relationship between Cv & Cp
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
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:
Other, Sometimes Useful, Expressions
TCONSTANTdVVR
TPS
TCONSTANTdPPR
TVS
TCONSTANTdPTVTVH
P
P VTV
P
P PTP
P
P PTP
0.
0.
0.
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.
• 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 TTT 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:
• The Entropy Change ΔS for this process can also be easily calculated:
21
21
21
2
2121
2
21
2ln2
ln2lnln
lnln1 2
TTTTmcS
TTT
mcTT
Tmc
TTT
mc
TT
TT
mcTdT
TdTmcS
P
fP
fP
fP
ffP
T
T
T
TPf 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
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.
dPcdvv
cdPPTcdv
vTcTds
dPTvdTcdPTvTdTcTds
dvTdTcdvTPTdTcTds
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:
Internal Energy u(T,ν):
dvPvudTcTds
dvvudT
Tudu
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:
v
v
v
vvvvv
Pv
PT
Tc
Ps
PT
TsT
TPT
Ts
Ps
dvvsdP
Psds
vPss
1
),( Consider
P
P
P
pPPPP
Pv
vT
Tc
vs
vT
TsT
TvT
Ts
vs
dvvsdP
Psds
1
dvvTcdP
PTcTds
dvvT
TcdP
PT
Tcds
dvvsdP
Psds
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):