thermodynamics - review and example...
TRANSCRIPT
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Outline Review Example Problem 1 Example Problem 2
ThermodynamicsReview and Example Problems
X Bai
SDSMT, Physics
Fall 2013
X Bai Thermodynamics
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Outline Review Example Problem 1 Example Problem 2
1 Review
2 Example Problem 1Exponents of phase transformation
3 Example Problem 2Application of Thermodynamic Identity
X Bai Thermodynamics
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Outline Review Example Problem 1 Example Problem 2
Thermodynamics: contents
1 Basic Concepts: Temperature, Work, Energy, Thermal systems, Ideal Gas,etc.
2 The Second Law of Thermodynamics
3 Interacting systems: The connection between the microscopic and themacroscopic
4 Heat Engines and Refrigerators
5 Chemical Thermodynamics
X Bai Thermodynamics
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Outline Review Example Problem 1 Example Problem 2
1. Equilibrium
Equilibrium Exchanges
Thermal Thermal energyMechanical Volume
Diffusive ParticlesChemical Chemical reactions
H2O ↔ H+ + OH−H2SO4 ↔ 2H+ + SO2−4
2. Thermal systems
Isolated system: No interaction with environment, no matter andenergy exchange.Closed system: No matter exchange, has energy exchange.Open system: Has matter and energy exchange.
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Outline Review Example Problem 1 Example Problem 2
Ideal Gas, Carnot cycle, etc.
3. Ideal Gas
PV = nRTPV = NkTn: number of moles of gas.R constant R = 8.31 J
mol·K
4. Carnot cycle
Step 1: isothermal expansion, (V1,P1,Th)→ (V2,P2,Th)Step 2: adiabatic expansion, (V2,P2,Th)→ (V3,P3,Tc)Step 3: isothermal compression, (V3,P3,Tc)→ (V4,P4,Tc)Step 4: adiabatic compression, (V4,P4,Tc)→ (V1,P1,Th)
5. Equipartition theorem
For a system with N particles, each with f DoF, and there is NO othernon-quadratic temperature-dependent forms of energy, the total thermal energyin the system is
Uthermal = Nf12kT
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Outline Review Example Problem 1 Example Problem 2
The Second Law of Thermodynamics
1. Macrostate, Microstate, Multiplicity, Applications: Einstein Solid, Ideal Gas,
2. Entropy: S = klnΩThe famous Sackur-Tetrode Equation for Ideal Gas:
S ≈ kN[ln(
VN
(4πmU3Nh2
)3/2)+ 5
2
]3. The Second Law of Thermodynamics
Any large system in equilibrium will be found in the macrostate with thegreatest muitiplicity (aside from fluctuations that are normally too small tomeasure).
If we start from a non-equilibrium system, it means:
Multiplicity tends to increase as time goes.
Isolated systems evolve from a organized system to a more randomized system!
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Outline Review Example Problem 1 Example Problem 2
Interacting systems: The connection between themicroscopic and the macroscopic
Interacting thermal systems
To understand heat flow and the evolution of a thermal system (reversible orirreversible), we need to consider two systems that interact with each otherwith heat and/or matter exchanges.
Solid A NA, qA, UA, ΩA
Solid B NB, qB, ΩB, UB
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Outline Review Example Problem 1 Example Problem 2
Interacting systems: The connection between themicroscopic and the macroscopic - cnt.
1. Interacting thermal systems: Staring point
Ωcombined = Ω1 × Ω2Scombined = S1 + S2
At equilibrium Pa+b Pa Pb
Away from equilibrium P’a+b P’a P’b
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Outline Review Example Problem 1 Example Problem 2
Interacting systems: The connection between themicroscopic and the macroscopic - cnt.
2. What governs a non-isolated system? The typical method: Starting fromthe total entropy of the entire system.
Stotal = S + Senvironment
dStotal = dS + dSenvironment
Using conditions:
Te = T
dU + dUe = 0
One gets:
dStotal = dS −1
TdU = − 1
T(dU − TdS) = − 1
TdF
X Bai Thermodynamics
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Outline Review Example Problem 1 Example Problem 2
Interacting systems: The connection between themicroscopic and the macroscopic - cnt.
X Bai Thermodynamics
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Outline Review Example Problem 1 Example Problem 2
Heat Engines and Refrigerators
Qh
Qc
W
The efficiency of an engine:� ≤ 1− Tc
Th
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Outline Review Example Problem 1 Example Problem 2
Heat Engines and Refrigerators - cnt.
H
HH T
QS =Δ
C
CC T
QS =Δ
HQ
CQ
W
heat
work he
at
hot reservoir, TH
cold reservoir, TC
entr
opy
The coefficient of performance (COP):
COP ≤ 1Th/Tc−1
= TcTh−Tc
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Outline Review Example Problem 1 Example Problem 2
Chemical Thermodynamics: Thermal quantities andpotentials
Table: Quantities that govern the thermal processes
Process What governs the process
constant E and V Entropy Sconstant T and V Helmholtz Free Energy Fconstant T and P Gibbs Free Energy G
Table: Thermodynamic quantities
Extensive quantities Intensive quantities
Do change when the Do NOT change when theamount of matter changes amount of matter changes
V, N, S, U, H, F, G , mass T, P, ρ, µ (chemical potential)
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Outline Review Example Problem 1 Example Problem 2
Chemical Thermodynamics: Thermal quantities andpotentials
Potential variables identity
U(S ,V ,N) S ,V ,N dU = TdS − PdV + µdNH(S ,P,N) S ,P,N dH = TdS + VdP + µdNF (T ,V ,N) V ,T ,N dF = −SdT − PdV + µdNG(T ,P,N) P,T ,N dG = −SdT + VdP + µdN
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Outline Review Example Problem 1 Example Problem 2
Chemical Thermodynamics: Thermal quantities andpotentials
T
P
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Outline Review Example Problem 1 Example Problem 2
Clausius-Claypeyron Relation
What determines the phase boundary is the condition? - TheClausius-Clapeyron Relation.
(1799 – 1864) French engineer and physicist, one of the founders of thermodynamics.
(1822 – 1888) German physicist & mathemaBcian, one of the central founders of the science of thermodynamics.
ΔP1 ΔP2
ΔT1
ΔT2 On the lone: ΔP1 = ΔP2 ΔT1 = ΔT2
dPdT =
δSδV
dPdT =
LTδV
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Outline Review Example Problem 1 Example Problem 2
van der Waals Model
U(r)
short-distance repulsion
long-distance attraction
-3
-2
-1
0
1
2
3
4
1.5 2.0 2.5 3.0 3.5 4.0distance
Energy
r
Johannes van der Waals, 1873
Using critical variables Vc ,Tc ,Pc to normalize P,V ,T :t = T/Tc , p = P/Pc , v = V /Vc , the vdW equation becomesp = 8t
3v−1 −3v2
The vdW equation in terms of reduced variables.
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Outline Review Example Problem 1 Example Problem 2
Maxwell construction plots
P/Pc
V/Vc
1 2
3
4 5
6
7
P/Pc
V/Vc
1 2
3
4 5
6
7
SA>0
SB
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Outline Review Example Problem 1 Example Problem 2
Vapor pressure and critical point
Understand phase transition using van der Waals Model.(1) vapor pressure Pv for each T: at which the liquid-gas transformation takesplace(2) critical pressure Pc , critical temperature Tc , and critical volume Vc : thecritical point.
Pc
Tc Vc
Pv
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Outline Review Example Problem 1 Example Problem 2
Problem 1
Review Exercise 01:This series of problems are dedicated to the study of the behavior of van derWaals fluid near the critical point.
(1) Expand the vdW equation in a Taylor series in (V − Vc), keeping termsthrough order (V − Vc)3. Prove that, for T close enough to Tc , the termquadratic in (V − Vc) becomes negligible compared to the others and may bedropped.Answer:To be simple, we take the vdW equation in terms of reduced variablesp = 8t
3v−1 −3v2
. When we use reduced quantities, v = V /Vc , the expansion isin terms of (V /Vc − Vc/Vc) = (v − 1).Taylor expansion is: f (x0 + h) = f (x0) + hf
′(x0) +h2
2!f ′′(x0) + . . .
The derivatives we need:∂p∂v
= −24t(3v − 1)−2 + 6v−3∂2p∂v2
= 144t(3v − 1)−3 − 18v−4∂3p∂v3
= −1296t(3v − 1)−4 + 72v−5
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Outline Review Example Problem 1 Example Problem 2
Problem 1 (1) - cnt.
p(v) ≈ p(1) + ∂p∂v
∣∣1
(v − 1) + 12∂2p∂v2
∣∣∣1
(v − 1)2 + 16∂3p∂v3
∣∣∣1
(v − 1)3
p(v) ≈ (4t − 3)− 6(t − 1)(v − 1) + 9(t − 1)(v − 1)2 − 32(9t − 8)(v − 1)3
T being close enough to Tc means T − Tc ≈ 0, or in reduced variables,t − 1 ≈ 0, let’s note this as t − 1 ≈ O(0).
When v − 1 ' t − 1:p(v) ≈ (4t − 3)− 6O2(0) + 9O3(0)− 3
2(9t − 8)O3(0)
When (v − 1)2 ' t − 1:p(v) ≈ (4t − 3)− 6O(0)(v − 1) + 9O2(0)− 3
2(9t − 8)(v − 1)O(0)
So, 9(t − 1)(v − 1)2 is always smallest, which we can drop. Then we have:p(v) ≈ (4t − 3)− 6(t − 1)(v − 1)− 3
2(9t − 8)(v − 1)3
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Outline Review Example Problem 1 Example Problem 2
Problem 1 - cnt.
(2) The expression for P(V ) is antisymmetric about the point V = Vc . Usethis fact to find an approximate formula for the vapor pressure as a function oftemperature.
Answer: p(v) ≈ (4t − 3)− 6(t − 1)(v − 1)− 32(9t − 8)(v − 1)3
When (v − 1) > 0, p(v) ≈ (4t − 3)− 6(t − 1) |(v − 1)| − 32(9t − 8)
∣∣(v − 1)3∣∣When (v − 1) < 0, p(v) ≈ (4t − 3) + 6(t − 1) |(v − 1)|+ 3
2(9t − 8)
∣∣(v − 1)3∣∣So, the −6(t − 1)(v − 1)− 3
2(9t − 8)(v − 1)3 is antisymmetric. The pressure of
the phase transition is p(v) = (4t − 3) - This line split the enclosed area intotwo parts with the same area.
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Outline Review Example Problem 1 Example Problem 2
Problem 1 - cnt.
(3) Find an expression for the difference between volume between the gas andthe liquid phases at the vapor pressure. You should find(Vg − Vl) ∝ (Tc − T )β , where β is a critical exponent. Experiments show thatβ has a universal value of about 1/3, but vdW model predicts a larger number.
Answer:
The volumes of the liquid and gas at the transition pressure are those at thetransition pressure p = 4t − 3. That is,(4t − 3) = (4t − 3)− 6(t − 1)(v − 1)− 3
2(9t − 8)(v − 1)3
−6(t − 1)− 32(9t − 8)(v − 1)2 = 0
At T ' Tc , or t ' 1,−6(t − 1)− 3
2(v − 1)2 = 0
v − 1 = ∓2√
1− t or, v = 1∓ 2√
1− tvg − vl = 4
√1− t
So, β is 12.
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Outline Review Example Problem 1 Example Problem 2
Problem 1 - cnt.
(4) Use the previous result to calculate the latent heat of the transformation asa function of T , and sketch this function.
Answer:
The Clausius-Clapeyron relation: dPdT
= LTδV
So, the latent heat is
L = T (Vg − Vl) dPdT = tPcVc(vg − vl)dpdt
L = tPcVc · 4√
1− t · 4L = 16PcVct
√1− t
L = 16 38kNTct
√1− t
LkNTc
≈ 6t√
1− t
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Outline Review Example Problem 1 Example Problem 2
Problem 1 (4) - cnt.
LkNTc
≈ 6t√
1− t by red line.L
kNTc≈ 6√
1− t by black line.
t0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
cL/
kNT
1
2
3
4
5
6.0*sqrt(1-x)
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Outline Review Example Problem 1 Example Problem 2
Problem 1 - cnt.
(5) The shape of the T = Tc isotherm defines another critical exponent, calledδ : (P − Pc) ∝ (V − Vc)δ. calculate δ in vdW model.
Answer:
When t = 1, from the Taylor expansion of the vdW equation:p(v) ≈ (4t − 3)− 6(t − 1)(v − 1)− 3
2(9t − 8)(v − 1)3
p(v) ≈ 1− 32(v − 1)3
p(v)− 1 ≈ − 32(v − 1)3
P − Pc ≈ − 32 (V − Vc)3
So, δ = 3
Note: The experimental value of δ is typically around 4 and 5.
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Outline Review Example Problem 1 Example Problem 2
Problem 2
Review Exercise 02:Find the internal energy U of vdW gas as a function of T ,V ,Cv . Assume thenumber of particles are fixed.Answer:vdW equation:
(P +
(NV
)2a)
(V − Nb) = NkT
P(N,V ,T ) = NkTV−Nb −
(NV
)2a(
∂P∂T
)V
= NkV−Nb
T(∂P∂T
)V− P =
(NV
)2a
Use equation (see next slide)dU = CV dT +
(T(∂P∂T
)V− P
)dV
One hasdU = CV dT +
(NV
)2adV
Integrating both sides of the equation gives:
U = U0(V0,T0) + CV (T − T0)− N2a(
1V− 1
V0
)
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Outline Review Example Problem 1 Example Problem 2
Problem 2 - cnt.
Review Exercise 02:How do we get this equation: dU = CV dT +
(T(∂P∂T
)V− P
)dV ???
We can start fromdU =
(∂U∂T
)VdT +
(∂U∂V
)TdV
anddS(V ,T ,N) =
(∂S∂T
)VdT +
(∂S∂V
)TdV ......(a)
The derivative of S can be written asdS(V ,T ,N) = Q
T= dU+PdV
T, U = U(T ,V ,N)
dS(V ,T ,N) = 1TCV dT +
(1T
(∂U∂V
)T
+ PT
)dV ......(b)
By comparing these equation (a) and (b) , one get(∂S∂T
)V
= 1TCV =
1T
(∂U∂T
)V(
∂S∂V
)T
= 1T
(∂U∂V
)T
+ PT
Note 1: Problem 3.33: CV = T(∂S∂T
)V.
Note 2: Textbook (3.39): P = T(∂S∂V
)U,N
.
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Outline Review Example Problem 1 Example Problem 2
Problem 2 - cnt.
In the meantime, the derivative of S is∂2S∂V∂T
= ∂∂V
(1T∂U∂T
)V
= ∂2S
∂T∂V= ∂
∂T
(1T
(∂U∂V
)T
+ PT
)So, we have1T
∂2U∂V∂T
= − 1T 2
(∂U∂V
)T
+ 1T
∂2U∂T∂V
− PT 2
+ 1T
(∂P∂T
)V
Identical terms cancel, we get1T 2
(∂U∂V
)T
= − PT 2
+ 1T
(∂P∂T
)V
Or,(∂U∂V
)T
= −P + T(∂P∂T
)V
So,dU =
(∂U∂T
)VdT +
(∂U∂V
)TdV
dU =(∂U∂T
)VdT + (−P + T
(∂P∂T
)V
)dV
X Bai Thermodynamics
ReviewExample Problem 1Exponents of phase transformation
Example Problem 2Application of Thermodynamic Identity