lecture 8: maximum and minimum work, thermodynamic...

Lecture 8:Maximum and Minimum Work, Thermodynamic

Inequalities

Chapter II. Thermodynamic Quantities

A.G. Petukhov, PHYS 743

October 4, 2017

Chapter II. Thermodynamic Quantities A.G. Petukhov, PHYS 743Lecture 8: Maximum and Minimum Work, Thermodynamic InequalitiesOctober 4, 2017 1 / 12

Maximum WorkIf a thermally isolated system is in non-equilibrium state it may dowork on some external bodies while equilibrium is being established.The total work done depends on the way leading to the equilibrium.Therefore the final state will also be different. In any event, sincesystem is thermally isolated the work done by the system:

|A| = E0 − E(S),

where E0 is the initial energy and E(S) is final (equilibrium) one. Leus consider the case when Vinit = Vfinal but can change during theprocess.

∂ |A|∂S

= −(∂E

∂S

)V

= −Tfinal < 0

The entropy cannot decrease. Then it means that the greater is thechange of the entropy the smaller is work done by the system Themaximum work done by the system corresponds to the reversibleprocess when

∆S = Sfinal − Sinitial = 0


Clausius Theorem

dSR < 0

TBS > TB

A B δQA > 0

TA > TAS

dSR > 0

δQB < 0

The system following a closed path

A: System receives heat from a hot reservoir. Temperature of thethermostat is slightly larger then the system temperature

B: System dumps heat to a cold reservoir. Temperature of the systemis slightly larger then that of the thermostat


Clausius Theorem Cont’d

dSR < 0

TBS > TB

A B δQA > 0

TA > TAS

dSR > 0

δQB < 0

Entropy balance:

−dSAR =

δQA

TA≤ δQA

TSA

= dSAsys

dSBR = −δQB

TB≥ −δQB

TSB

= −dSBsys

− dSR = −dSAR − dSB

R =

∮δQ

T≤ dSA

sys + dSBsys = ∆Ssys = 0∮

δQ

T≤ 0

Here δQ is the heat absorbed by the system and T is the of thethermostat.


Carnot Engine

AAA A

Carnot Cycle:

Isothermal expansion at TH

Adiabatic expansion with cooling TH → TL

Isothermal compression at TL

Adiabatic compression with heating TL → TH


Carnot’s Engine Cont’d

Since Carnot cycle is reversible the Clausius equality holds:

0 =

∫dQ

T=QH

TH− QL

TL

because for two adiabatic processes Q1 = Q2 = 0. Efficiency:

η =A

QH=QH −QL

QH= 1− QL

QH= 1− TL

TH

The work extracted from the Carnot engine is the maximum work becausefro any other engine ∫

dQ

T≤ 0⇒ η ≤ 1− TL

TH


Minimum work done by an external medium

1

2

Work done by the system A12 < 0

Work done on the system A12 > 0

If 1→ 2→ 1 is a reversible process then A12 +A21 = 0. Therefore

(A12)max = − (A21)min

Let us consider a system in a thermostat, which is so large that anychanges in the system will not affect the macroscopic state of thethermostat (T0, p0).

Question: What is the minimum work that should be done by anexternal source to transform the body into the state (T, p) that is notin equilibrium with the medium?


Minimum Work cont’d

The work done on the body:

A = ∆E + ∆E0 ⇒ ∆E = A−∆E0 = A− T0∆S0 + p0∆V0

Here ∆E and ∆E0 are changes in the energy of the body and thethermostat respectively. As ∆(S + S0) ≥ 0 and ∆(V + V0) = 0 then

A = ∆E + T0∆S0 − p0∆V0 ≥ ∆E − T0∆S + p0∆V

Therefore, for a reversible process when ”=” :

Amin = ∆(E − T0S + p0V )

For any irreversible process A > Amin.

If a body is in equilibrium at every moment of time thendE = TdS − pdV and

dAmin = (T − T0)dS − (p− p0)dV


Minimum Work cont’d

Particular cases:I V = const and T = T0 Then

Amin = ∆(E − TS) = ∆F

Minimal work equals to the change of the Helmholtz Free energyI T = T0, p = p0 Then

Amin = ∆(E − TS + pV ) = ∆Φ

Minimal work equals to the change of the Gibbs Free energy


Thermodynamic inequalities

Again, consider a system in a thermostat with T = T0 and p = p0.Under these conditions the Gibbs free energy of the body has aminimum. By virtue of the previous consideration the minimum workrequired to bring the body from an equilibrium to a non equilibriumstate equals to ∆Φ > 0. Thus

δE − T0δS + p0δV > 0

But

δE(S, V ) =

(∂E

∂S

)0

δS +

(∂E

∂V

)0

δV+

1

2

[(∂2E

∂S2

)0

(δS)2 + 2

(∂2E

∂S∂V

)0

δSδV +

(∂2E

∂V 2

)0

(δV )2

]> T0δS − p0δV


Thermodynamic Inequalities cont’d

Using(∂E∂S

)0

= T0 and(∂E∂V

)0

= −p0 we obtain(∂2E

∂S2

)0

(δS)2 + 2

(∂2E

∂S∂V

)0

δSδV +

(∂2E

∂V 2

)0

(δV )2 > 0 (1)

Inequality (1) must be valid for arbitrary δS and δV . Setting δV = 0we get (

∂2E

∂S2

)0

> 0

In addition we can calculate the discriminant of the quadraticform (1) which yields(

∂2E

∂S2

)0

·(∂2E

∂V 2

)0

−(∂2E

∂S∂V

)2

0

> 0


Thermodynamic Inequalities cont’dThe meaning of these thermodynamic inequalities can be established asfollows:

∂2E

∂S2=

[∂

∂S

(∂E

∂S

)V

]V

=

(∂T

∂S

)V

= T1

T

(∂T

∂S

)V

=T

cV> 0

ThuscV > 0

By using Jacobians (see L&L) it is possible to show that the secondcondition is equivalent to (

∂p

∂V

)T

< 0

Finally one can show that

cp > cV > 0


lecture 8: maximum and minimum work, thermodynamic...

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