phy304: statistical mechanics · 2021. 2. 15. · that is, a certain reversible work is required to...

PHY304: StatisticalMechanicsLecture 5,Wednesday, January 13, 2021

Dr. Anosh JosephIISER Mohali

Microstates and Macrostates

Equal probability of all compatible microstates is anassumption ...

... that can be justified by an experimentalexamination of the consequences.

As of today no one has invented an experiment whichdisproves this assumption.

Can say that for every macroscopic thermodynamicstate, there exists a large number of microscopicrealization possibilities.

PHY304: Statistical Mechanics Dr. Anosh Joseph, IISER Mohali


We may understand the effort of a system to reach amaximum in entropy ...

... as a transition into the most probable state.

That is, to a state with the largest number ofmicroscopic realization of possibilities.

S is thus a measure for the number of possiblemicrostates of a system ...

... when a macrostate is given.

Can formulate the second law of thermodynamics indifferent ways.



(a.) There is no perpetual motion machine of thesecond kind.

That is, an engine that does nothing but performwork while cooling a heat bath cannot exist.

An engine of this type, if it exists, will transformheat into work with 100% efficiency.

(b.) Each isolated macroscopic system wants toassume the most probable state, that is, the statewhich is characterized by the largest number ofmicroscopic realization possibilities.



Let us use both the laws to derive some consequences...

... for the state variables T , p,µ,φ, · · · in an isolatedsystem in equilibrium.

Imagine that the completely isolated system is dividedinto two parts.

The total system is characterized by the state variablesS, V , and N .



U1 U2S1, T1V1, p1N1, μ1

S2, T2V2, p2N2, μ2

⋯ ⋯

Figure: A system with two sub-systems separated by a wall.



Internal energy U is a function of these variables.

All these state variables are constants since the totalsystem is isolated.

However, the partial systems can exchange work andheat between themselves.



We have

U = U1 + U2 = constant, (1)

V = V1 + V2 = constant, (2)

S = S1 + S2 = constant, (3)

N = N1 + N2 = constant, (4)...



The first law for a reversible change of state for bothpartial systems tells us

dU1 = T1dS1 − p1dV1 + µ1dN1 + · · · , (5)

dU2 = T2dS2 − p2dV2 + µ2dN2 + · · · . (6)



Since U = constant we must have

dU1 + dU2 = 0. (7)

Now, if we add Eqs. (5) and (6) we get

0 = (T1 − T2)dS1 − (p1 − p2)dV1

+(µ1 − µ2)dN1 + · · · . (8)



In the above we used

dS1 = −dS2, dV1 = −dV2, · · · . (9)

Since change of variables S1, V1, N1, · · · in system 1underlies no restrictions, ...

... above equation is true if it separately holds that

T1 = T2, p1 = p2, µ1 = µ2, · · · . (10)



These are the necessary conditions for thermodynamicequilibrium.

Note that the imagined partition is arbitrarily chosen.

Thus we can conclude that ...

... if an isolated system is in equilibrium, then it willhave the same constant temperature, pressure,chemical potential, etc. everywhere.



Suppose there is a real wall instead of a fictitious wall,...

... and this prevents the change of volume or of theparticle number: dN1 = 0, dV1 = 0.

In that case, only the condition

T1 = T2 (11)

remains.

And the system is in thermal equilibrium.



If p1 = p2 we have mechanical equilibrium.

If µ1 = µ2, we have chemical equilibrium.


Global and Local Equilibrium

If a system is in thermodynamic equilibrium, that is,...

... if it has the same temperature, pressure, andchemical potential everywhere, then we have a globalequilibrium.

If the system can be divided into partial systems andthey themselves have their own values for thethermodynamic quantities ...

... then we have a system with local equilibrium.

Differences in thermodynamic quantities induce flowsin fluxes coming from differences in thermodynamicpotentials.PHY304: Statistical Mechanics Dr. Anosh Joseph, IISER Mohali


As an example we can look at heavy iron collisions(nucleus - nucleus collisions).

Drops of nuclear matter collide and different parts ofthe nuclei have, at a given time, very differenttemperatures, densities, particle numbers etc.

Here we can assume partial regions to beapproximately in (local) equilibrium.



Another example is the different parts of a star.

Studying systems with local potential differencesrequires the use of transport theories ...

... and it is beyond the scope of this course.


Thermodynamic Engines

We benefit from cyclic heat engines in our day to daylives.

Nuclear power plants and combustion engines areexamples of such engines.

Can invoke the first and second laws ofthermodynamics to arrive at far reaching conclusions...

... concerning the transformation of heat into work.



Figure: A steam engine is a heat engine that performsmechanical work using steam as its working fluid.



The experience from the second law tells us that ...

... the work performed in reversible processes issmallest and the heat largest

δWirr > δWrev = −pdV , (12)

δQirr < δQrev = T dS. (13)



If the ideal gas expands into vacuum withoutperforming work, we have δWirr = 0.

If the gas expands reversibly (always in equilibriumwith the external force) ...

... then it performs the work

δWrev = −pdV . (14)



If dV > 0, we have

δWirr = 0 > δWrev = −pdV . (15)

If the gas is reversibly compressed, we have

δWrev = −pdV > 0. (16)



That is, a certain reversible work is required tocompress the gas.

To compress the gas irreversibly, we require morework than in the reversible case.

We have

δWirr > δWrev = −pdV (dV < 0). (17)



In the irreversible expansion system performs lesswork than in the reversible case.

For irreversible compression we require more workthan in the reversible case.

If we have a cyclic engine, which leads the workingmaterial back to the initial state after one cycle, ...

... it holds according to the first law that∮dU = 0. (18)



Thus,0 = ∆Wrev + ∆Qrev = ∆Wirr + ∆Qirr. (19)

From the above considerations we see that among allpossible processes, ...

... reversible processes produce the largest amount ofutilizable work.

Let us compute the efficiency of a general reversiblecyclic process.



Heat Bath

Heat BathTc

Th

engine cycleprocessΔU = 0

ΔQh

ΔQc

ΔW

Figure: Working parts of a heat engine.



Each engine needs a heat reservoir (T = Th) fromwhich to extract the heat energy and ...

... a second reservoir (T = Tc) to absorb the waste heatof the process (that is, to cool the engine).

An engine which works with only one reservoir cannotperform utilizable work in a cyclic process.

According to the first law

0 = ∆W + ∆Qh + ∆Qc. (20)



In the earlier lecture we saw the definition of efficiencyη as the fraction

η =|∆W |

∆Qh. (21)

This tells us how much heat energy ∆Qh istransformed into work

(∆W < 0,∆Qh > 0,∆Qc < 0):

ηirr < ηrev = −∆W∆Qh

=∆Qh + ∆Qc

∆Qh. (22)



The engine must work reversibly.

So it must hold that

δQh = Th dS, (23)

δQc = −Tc dS. (24)



In above, dS is the change in entropy, in a smallpartial step of the cycle.

dS 6= 0 here, although only equilibrium states occur.

The reason is that the engine (the working material) isnot an isolated system.



The signs in Eq. (23) indicate the directions given inthe figure above.

Since ∆W < 0 (performed work) we have

η =|∆W |

∆Qh=

Th − Tc

Th. (25)



We also note thatη 6 1. (26)

Equation (25) always holds, ...

... independent of the working material and thetechnical realization of the engine.

If there were two reversible cycles with differentefficiencies, then ...

... we could construct a perpetual motion machine ofthe second kind.



Th

Tc

WA (WB − WA)QhA QhB

QcA QcB

A B

Figure: Perpetual motion machine of the second kind.



Suppose we have two heat engines A and B withefficiencies ηA and ηB.

Consider the case ηB > ηA.

We can connect the two engines as shown in the figureabove.



Engine A works here in the reverse direction (as arefrigerator).

It expels energy WA and the heat QcA from coldreservoir (T = Tc) as heat QhA into the hot reservoir(T = Th).

The energy WA is generated by the process B.



We assume that the process B has higher efficiencythan process A.

An amount or work WB − WA remains as useful work.

Choosing the signs according to the figure, we have

WA = ηAQhA, (27)

WB = ηBQhB, (28)

QcA = QhA − WA, (29)

QcB = QhB − WB. (30)



Now adjust the engine such that QhA = QhB = Qh .

Then there will be no change in the hot reservoir in along time scale since the amount of heat is taken offas pumped back.

Then,

QcA = (1 − ηA)Qh > QcB = (1 − ηB)Qh (31)

since ηB > ηA.



Thus the heat

∆Qc = QcA − QcB = (ηB − ηA)Qh (32)

is effectively drawn off from the cold reservoir.

Hence the engine performs the work

WB − WA = (ηB − ηA)Qh (33)

while cooling the cold reservoir.

This is nothing but the perpetual motion machine ofthe second kind.



It permanently performs work and merely cools a heatreservoir.

All the failed attempts to construct such a machine ...

... (one that respects the energy conservation law buyviolates the entropy law) ...

... resulted in the common understanding that

∆Qc = WB − WA = 0, (34)

orηA = ηB =

Th − Tc

Th(35)

for all reversible processes given at Th and Tc.


Work Diagrams of Engines

In the figure below we show the pV diagrams and TSdiagrams of the Carnot and Otto engines.

The work performed per cycle corresponds to theshaded area.

∆W = −

∮pdV =

∮T dS. (36)



P

T V

S

Otto engine Carnot process

Figure: Work diagrams of Otto and Carnot engines.



It is exactly as large as the difference of the heats

∆Qh = Th∆S (37)

and∆Qc = Tc∆S. (38)

Real processes (for example, the Otto engine) deviatemore or less from this diagram.


References

I W. Greiner, L. Neise, H. Stocker, and D. Rischke,Thermodynamics and Statistical Mechanics,Springer (2001).


End


phy304: statistical mechanics · 2021. 2. 15. · that is, a certain reversible work is required to...

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