Thermodynamics - Review and Example Problems:...

Outline Review Example Problem 1

ThermodynamicsReview and Example Problems: Part-2

X Bai

SDSMT, Physics

Fall 2014

X Bai Thermodynamics


1 Review

2 Example Problem 1Exponents of phase transformation



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



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.



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 = Nf 12kT



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!



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



Interacting systems: The connection between themicroscopic and the macroscopic - cnt.

1. Interacting thermal systems: Staring point

Ωcombined = Ω1 × Ω2

Scombined = S1 + S2

At equilibrium Pa+b Pa Pb

Away from equilibrium P’a+b P’a P’b




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



Heat Engines and Refrigerators

Qh

Qc

W

The efficiency of an engine:ε ≤ 1− Tc

Th



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



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)




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




T

P



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 red line: ΔP1 = ΔP2 ΔT1 = ΔT2

dPdT = δS

δVdPdT = L

TδV



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− 3

v2

The vdW equation in terms of reduced variables.



Maxwell construction plots

P/Pc

V/Vc

1 2

3

4 5

6

7



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



Final Exam

Time:

Start at 9:00 AM MT/10:00 AM CT, Tuesday, October 14,2014.Due at 1:00 PM MT/2:00 PM CT (4 hours), Tuesday,October 14, 2014.

Location: Not required to be in class.



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 quadraticterm 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− 3

v2. When we use reduced quantities, v = V /Vc , the expansion is

in 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



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



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)|+ 32(9t − 8)

∣∣(v − 1)3∣∣

So, the −6(t − 1)(v − 1)− 32(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.



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.



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



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)



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.


Thermodynamics - Review and Example Problems:...

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