The Fundamental Efficiency of Osmotic Engines

Karl Duderstadt

Advisor: Dan Styer

May 21, 2004

Contents

1 Introduction 4

1.1 The Nature of Theoretical Research . . . . . . . . . . . . . . . . . . . . . . . 4

1.2 Foundations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 The Fundamental Limitation of the Osmotic Engine 8

2.1 The Second Law of Thermodynamics

Restated . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.2 A Method for Defining Chemical Potential and the Fundamental Efficiency . 9

2.3 Properties of the Osmotic Engine1 . . . . . . . . . . . . . . . . . . . . . . . . 12

3 The Connection to Entropy 19

3.1 A State Function2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.2 The Connection to Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3.3 What is Chemical Potential? . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

4 The Osmotic Carnot Cycle 26

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4.2 The Osmotic Carnot Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

5 Biological Osmotic Engines 37

5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

5.2 Coupled Transport . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

5.3 Antibiotic Resistance Via Coupled Transport . . . . . . . . . . . . . . . . . . 42

A Chemical Potential in Dilute Solutions3 47

1

B Equal Chemical Potential Implies Equilibrium4 53

C Osmotic Pressure5 55

D Solvent Density is Constant for Isopotic Transitions 58

E An Experimental Method for Determining the Transport Rate of TetA 60

2

List of Figures

2.1 Abstract concentration engine diagram.6 . . . . . . . . . . . . . . . . . . . . 12

4.1 Osmotic Carnot engine. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.2 Representation of the piston position throughout a complete Carnot osmotic

engine cycle (ABCDA). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4.3 (µ, N) Diagram of the Carnot osmotic engine cycle (ABCDA). . . . . . . . . 30

4.4 The correct (p, V ) diagram. . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

5.1 Coupled anitport (left) and symport (right) via membrane transport proteins. 38

5.2 Abstract representation of a coupled particle transporter. . . . . . . . . . . . 39

C.1 Osmotic pressure. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

3

Chapter 1

Introduction

1.1 The Nature of Theoretical Research

“What has been said so far may have seemed to imply that normal science is

a single monolithic and unified enterprise that must stand or fall with any one

of its paradigms as well as with all of them together. But science is obviously

seldom or never like that. Often, viewing all fields together, it seems instead a

rather ramshackle structure with little coherence among its various parts.”

—Thomas S. Kuhn [3]

In The Structure of Scientific Revolutions, Kuhn reminds us that the scientific theories

which form the foundation of all modern research are nothing more then a set of disconnected

problems with elegant solutions. No one understands this point better than those who

attempt to adapt and extend these theories to new problems. Theoretical research is as

much an exploration of the “gaps and cracks” left in the structure and coherence of scientific

theories over the centuries as it is an exploration of a particular new problem or set of

problems. We must keep this in mind as we begin our exploration of osmotic engines and

the various theories that we will use to explain them.

The emergence of new fields of physics brings with it new problems that force us to con-

tinually go back and re-examine the theories that make-up the core of the discipline. During

these periods of re-examination the “gaps and cracks” that Kuhn refers to are discovered

and old theories must be readapted in order to interpret new results. The 20th Century has

seen what some call the golden age of biology with the rise of new technologies for examining

4

the microscopic cellular systems that make-up all living creatures, the discovery of DNA and

the following genetic revolution. These discoveries have given way to a wide range of new

problems which make up a new field known as biophysics. Chapter 5 examines a particular

biophysical system but the analysis is applicable to many different biological systems. The

problem considered in chapter 5 and the experiment in appendix D were the original moti-

vations for this thesis. So in many ways this thesis is the re-examination of an old theory in

hopes of a better adaptation to interpret new results.

This thesis attempts to answer two main questions: If we rewrite the statement of Clau-

sius in terms of particle diffusion, as is done in section 2.1, can we derive a rigorous definition

of entropy using the modified statement? How can we extend our theoretical understand-

ing of particle diffusion to two new applications, an osmotic Carnot engine and a biological

particle transporter?

In Thermodynamics [1], Fermi uses the Clausius statement of the second law, the pos-

tulate of Lord Kelvin, and Carnot’s model for a reversible heat engine to derive a rigorous

definition of entropy. This thesis starts at the same place as Fermi and attempts a similar

derivation using modified definitions and models for use with osmotic engines. These mod-

ified definitions outline the fundamental limitations placed on osmotic engines irrespective

of their mechanistic details. So in answering the first question, in chapters 2 and 3, we will

develop all of the theory needed to address the second question, in chapters 4 and 5. Answer-

ing the first question will also present us with an opportunity to unearth the foundations of

thermodynamics giving us a further appreciation of how scientific theories are constructed.

Chapter 4 is devoted to developing an osmotic engine similar to Carnot’s heat engine.

This model will help clarify exactly how reversible osmotic engines can perform mechanical

work and the characteristics of an osmotic engine cycle. Our reversible osmotic engine

performs mechanical work using osmotic pressure so we can plot its cycle on a (p, V ) diagram

to allow for strait forward comparison with Carnot’s heat engine.

Chapter 5 is devoted to a quite new and different application of the theory of osmotic

engines. In this chapter we examine two types of coupled transport that are both used by

a wide variety of membrane proteins found in biological cells. Using theory developed in

earlier sections we can find the fundamental efficiencies of these coupled osmotic engines and

other quantities of interest.

5

1.2 Foundations

The systems that make up the world around us, the macroscopic world we live in, are

composed of immensely large sets of molecules. The laws of statistical mechanics devel-

oped by Maxwell, Gibbs, and Boltzmann consider these enormous collections of particles

and provide us with tangible mechanistic models whose bulk properties are described by

thermodynamics. But thermodynamics emerged, long before the refinements presented by

Statistical Mechanics, out of the experiments of Carnot, Clausius, Kelvin, Joule and others.

This thesis is a tribute to Carnot’s research, which has remained influential over the last

two centuries. Carnot set out to answer the simple question of whether there is a definite

limit on the work a stream engine can perform and ended up discovering the fundamental

principles of thermodynamics. In 1824, he published his classic work Reflections on the

Motive Power of Heat [4] containing a theorem that the fundamental efficiency of all heat

engines depends only on the temperatures of the reservoirs connected to it and that the most

efficient heat engine operating between two particular temperature reservoirs is a reversible

one. Carnot determined this fundamental theorem in a time when heat was considered a

fluid, even before Joule had determined the first law of thermodynamics, in the late 1830’s

and 40’s. Even though our concept of heat has changed drastically since his time, Carnot’s

efficiency result has remained correct ever since its discovery.

Carnot’s results were expanded in the 1850’s by Kelvin into a statement of the limitation

on all conversions between heat and mechanical energy. This statement is now known as the

second law of thermodynamics. At the same time, working independently, Rudolf Clausius

had formulated his own statement of the second law of thermodynamics, which is equivalent

to Kelvin’s statement (proved by Fermi in [1]).

Although thermodynamics had its foundation constructed around Carnot’s heat engine

and the fundamental limitations of heat transfer, the same thermodynamic information is

observed in many different processes. This thesis is an exploration of what thermodynamic

information can be determined simply from observations of osmotic processes and their fun-

damental limitations. If in fact it wasn’t Carnot’s research that Kelvin based his statement

on but the research of someone else working on the limitations of osmotic processes would his

statement have encapsulated all of the information in the second law of thermodynamics?

That someone else could have been Abbe Nollet, who was the first to record an observation

of the property of osmotic pressure in 1784 [2]. Nollet placed a bladder filled with wine in a

6

vat of water and to his surprise the bladder burst. The only plausible explanation was that

water had seeped through the bladder raising the internal pressure to such an extent that the

bladder broke. We will start our exploration of osmotic processes and the thermodynamic

information encapsulated in them with Nollet’s observation starting in section 2.2.

7

Chapter 2

The Fundamental Limitation of the

Osmotic Engine

2.1 The Second Law of Thermodynamics

Restated

The first law of thermodynamics rules out the possibility of constructing a machine

that can spontaneously create energy. The first law, however, places no restrictions on the

possibility of transferring energy from one form into another. So according to the first law of

thermodynamics alone we may transfer energy from work into heat or from heat into work

as the heat engine does. In this chapter we will prove that it is also possible to transfer work

into osmotic energy or osmotic energy into work.

Utilizing the first law alone, we could imagine all sorts of wild energy transfers with few

limitations. For example, we could construct a machine capable of performing work simply

by absorbing heat from its surroundings. According to the first law alone this machine could

tap the enormous reservoirs of thermal energy that surround us performing extraordinary

amounts of work. Similarly, nothing in the first law forbids us from constructing a ma-

chine capable of collecting the minute amount of gold found in sea water and accumulating

enormous wealth. Unfortunately, as we know from personal experience, no such perpetual

motion machines exist in nature.

Since as of yet there have been no perpetual motions machines discovered in nature we

must conclude that the limitations imposed by the first law of thermodynamics alone are not

8

sufficient to explain the world we live in. The further limitation which was first discovered

by Carnot and later clearly stated by Clausius and Kelvin is known as the Second Law of

Thermodynamics. Clausius stated this limitation explicitly in 1850 [2] and it is now know

as the postulate of Clausius (Statement 1):

If heat flows by conduction from a body A to another body B, then a transforma-

tion whose only final result is to transfer heat from B to A is impossible.

This statement of the second law can be used to determine the exact relationships that

govern heat engines. Since we are interested not in the conduction of heat but the diffusion

of particles we will restate the postulate of Clausius as follows in terms of particle conduction

rather than heat conduction (Statement 2):

If a particular type of particle flows by diffusion from a body A to another body

B, then a transformation whose only final result is to transfer this same type of

particle from B to A is impossible.

At this point we don’t know if statement 2 will produce the same results as statement 1, but

we will use statement 2 as an axiom in constructing our understanding of osmotic engines.

In chapter 3 we will determine whether a rigorous definition of entropy can be derived from

statement 2.

2.2 A Method for Defining Chemical Potential and the

Fundamental Efficiency

Since thermodynamics was originally a theory based exclusively on experimental obser-

vation it is very possible that thermodynamics could have been based on statement 2 instead

of statement 1. If Nollet had taken his initial observation of osmotic pressure a little more

seriously then perhaps he would have continued to experiment. He would have found that

for a given amount of wine the bladder would always burst. If he added the same amount

of diluted wine he would have discovered that the bladder increased in volume but didn’t

burst. In all of his experimenting he may have concluded that if we consider A to be one

body and B to be another body for a given set of initial conditions (wine concentration,

volume as well as reservoir volume etc..) the outcome is always the same. This conclusion

9

would have led him directly to statement 2. If water flows into the bladder for a given set

of initial conditions then it is impossible for water to suddenly flow out of the bladder for

those initial conditions and visa versa.

In reality Nollet didn’t fulfill the role of Carnot in the history of thermodynamics, but

this line of reasoning was simply to show how statement 2 could have come about on its own

instead of as a restatement of the postulate of Clausius. In any case now that we have it we

can move on and analyze its implications.

At this point we need to find some way of defining the property of bodies A and B that

causes particles to diffuse in only one direction between them. Nollet did not make the next

theoretical leap all on his own, but all of the information needed for that leap is wrapped up

in his observation. The expansion of the bladder was a form of mechanical work performed

by the water particles diffusing in. According to the first law of thermodynamics the total

energy of the two body system must remain constant throughout the transformation. Since

energy is required to perform mechanical work we must conclude that some energy was lost

in expanding the wall of the bladder. Since the bladder didn’t break until it was added to

the reservoir and water diffused in, we must also conclude that the energy needed to perform

the mechanical work came from the water particles of the reservoir.

Let’s define the energy lost by the reservoir to be Mr. Since only particles are exchanged

during this transformation the energy lost by the reservoir, Mr, must be proportional to the

change in particle number, dNr, of the reservoir. If we define µr to be the proportionality

constant then Mr becomes

Mr = µr dNr. (2.1)

According to statement 2 the particles must remain in the bladder after the transformation

and in fact the bladder must absorb at least some of the energy lost by the reservoir as we

will prove in section 2.3. For now we proceed by defining the energy gained by the bladder

as

Mb = µb dNb (2.2)

where µb is the proportionality constant that relates the change in number to the energy

gained. In this expression we are not necessarily assuming that the bladder absorbs any

energy because we could simply set µb = 0. So this will give us a very general expression

that even allows for the possibility of the bladder absorbing no energy. Now we can solve

for the total mechanical work, L, which is equal to the energy absorbed by the bladder

10

subtracted from the energy lost by the reservoir.

L = Mr −Mb (2.3)

Since we know that L > 0 for this transformation, because mechanical work is performed

and the bladder breaks, it follows from equation (2.3) that µr > µb. So we can conclude

that the relationship between µr and µb defines the direction of diffusing particles in the two

body system. If, for example, µr < µb then by (2.3) L < 0 and instead of the wall of the

bladder expanding it would contract as particles diffused out of the bladder.

In Nollet’s system the mechanical work performed is PV work. Without having any

knowledge of the laws of thermodynamics we could use the experimentally derived ideal gas

law to interpret exactly what happens for different relationships between µr and µb. Without

justifying the above transformations with the ideal gas law we can’t clearly interpret the

meaning of equation (2.3). For example, if µr = µb then we know that L = 0 but from the

definitions above that doesn’t imply that no particles are flowing from the reservoir to the

bladder. If we apply the ideal gas law we see that with constant temperature and constant

volume an increase in number would cause a change in pressure so number must be constant

for this situation in order for there to be no PV work performed.

One further property of this transformation that will become more important in chapter

2 and 3 is its efficiency. Lets define the efficiency as the ratio,

η =benefit

cost=

L

Mr

=Mr −Mb

Mr

= 1− Mb

Mr

, (2.4)

of the work performed by the transformation to the energy change of the reservoir. For

Nollet’s system dNr = dNb because particles that leave the reservoir must enter the bladder,

there is no where else for them to go. With this additional fact we can simplify the efficiency

expression.

η = 1− µb

µr

. (2.5)

For this irreversible transformation if µr = µb then η = 0 which makes sense because no work

is performed in that case. When µb = 0, η = 1 which is the condition when all of the energy

from the reservoir is used for PV work. You may be asking yourself whether it is possible

for the particles that diffuse into the bladder to give up all of their energy to PV work —

certainly they still have some kinetic energy. In fact we will find that this isn’t possible for

either a reversible engine or an irreversible engine.

11

2.3 Properties of the Osmotic Engine1

Figure 2.1: Abstract concentration engine diagram.2

In this section we determine the efficiency limitations of an arbitrary concentration

engine utilizing all of the expressions derived in section 2.2; but for the sake of generality

all r (reservoir) subscripts will become h’s (high) and all b (bladder) subscripts will become

l’s (low). The fundamental properties of a concentration engine proven in this section are

independent of all mechanistic details. The only condition we impose on this arbitrary

engine, for the sake of our argument, is that it be cyclic: at the end of the process it must

return to its initial state.

The engine discussed in section 2.1 wasn’t much of an engine at all because it wasn’t cyclic.

Particles diffused into the bladder but once the inside of the bladder reached equilibrium

1This section follows arguments for the efficiency of heat engines by Fermi in chapter III from [1].2This diagram was inspired by a similar diagram on p. 123 of Thermal Physics by Schroeder [5].

12

with the reservoir nothing changed. In cases where the bladder broke, the experiment was

destroyed in only one transformation. Since we are interested in plausible engines that can

continually perform work for long periods of time we must consider cyclic engines.

Figure 2.1 on page 12 is a representation of an abstract osmotic engine operating be-

tween two reservoirs of chemical potential µh (high chemical potential) and µl (low chemical

potential) where µh > µl. Rewriting expressions (2.1) and (2.2) from section (2.2) in terms

of our new definitions we obtain

Mh = µh dNh and Ml = µl dNl. (2.6)

If we now plug these into expression (2.3) from section (2.2) we obtain,

L = Mh −Ml, (2.7)

the total work performed during one cycle of the abstract osmotic engine. In figure 2.1 on

page 12, dNh particles diffusing from the µh reservoir to the µl reservoir carry with them an

amount of osmotic energy Mh, which is split between the osmotic energy absorbed by the µl

reservoir and the osmotic energy that is transformed into work, L. Here the term osmotic

energy for the osmotic engine parallels the term heat for the heat engine.

In order to complete our exploration of osmotic engine efficiencies we must first deter-

mine the relationship between work and osmotic energy. The modified Clausius statement,

statement 2 from (2.1), determines the natural flow of particles in a system but doesn’t de-

scribe the relationship between work and osmotic energy. To accomplish this task we must

modify the postulate of Kelvin so that it applies to osmotic engines. Our modified postulate

of Kelvin (statement 3):

A transformation whose only final result is to transform into work osmotic energy

extracted from a source which is at the same chemical potential throughout is

impossible.

Eventhough statement 2 doesn’t describe the relationship between work and osmotic energy

it is equivalent to statement 3, as we will now prove.

If statement 3 were not true then we could perform a transformation whose only final

result was to transform into work a quantity of osmotic energy taken from a single source at

chemical potential µl. We could then transform this work into an addition of osmotic energy

to a source at chemical potential µh. If µh > µl then our only final result would have been to

13

transfer osmotic energy from one body (the source at chemical potential µl) to another body

at higher chemical potential, µh. This is in violation of statement 2 from section (2.1). So

we have proven that if statement 3 is invalid then statement 2 must be invalid. To complete

the proof of equivalence of the two statements we must prove that if statement 2 is invalid

then statement 3 must be invalid.

Let us assume, in contradiction to statement 2, that it were possible to transfer a certain

amount of osmotic energy Mh from a reservoir at chemical potential µl to a reservoir at

higher chemical potential µh in such a way that no other change in the state of the system

occurred. With the aid of Nollet’s bladder we could then transform this amount of mixing

energy into a corresponding amount of work L (mechanical PV work). The reservoir at

chemical potential µh receives and gives up the same amount of osmotic energy so it suffers

no final change. Thus, the process just described would have as its only final result the

transformation into work, osmotic energy extracted from a reservoir which is at the same

chemical potential µl throughout. This result is contrary to statement 3, thus, we have

demonstrated that statement 2 implies statement 3.

Now we prove that if the total amount of work done by the engine is greater than zero

than the two osmotic energies must be greater than zero, that is if L > 0 then Mh, Ml > 0.

This will shore up the dilemma we had in section (2.1) when considering whether the bladder

actually absorbed any of the energy taken from the reservoir. If we assume that Ml ≤ 0 it

follows that the engine absorbed an amount of osmotic energy −Ml from the µl reservoir

during the cycle (it might help to look at figure (2.1) on page 12). If we then allowed particles

to flow by diffusion between the two reservoirs transferring −Ml of osmotic energy from the

µh reservoir to the µl reservoir the µl reservoir would be returned to its initial state. The

only result of this cycle would then be the transformation of −Ml of osmotic energy into an

amount of work L. This result is in contradiction to our original postulate so it must be the

case that Ml ≥ 0. Now that we know that Ml ≥ 0 it follows directly that Mh ≥ 0 due to the

fact that L ≥ 0 by assumption.

Lets now consider a second concentration engine operating between the same chemical

potential reservoirs, µh and µl, as the first engine with the quantities L′, M ′l and M ′

h as

the total work and the two osmotic energies, respectively. We will prove the following

fundamental theorem:

14

a. If the first engine is a reversible one, then,

Mh

Ml

≥ M ′h

M ′l

. (2.8)

b. If the second engine is also a reversible one, then,

Mh

Ml

=M ′

h

M ′l

. (2.9)

If we recall the efficiency expression (5.9) we determined on page 41 we see that the theorem

stated above implies that reversible osmotic engines are more efficient than all other osmotic

engines.

In follows from equation (2.7) on page 13 that the work performed by the two engines is

L = Mh −Ml and L′ = M ′h −M ′

l , (2.10)

for the first engine and the second engine, respectively. The ratio Mh

M ′h

can be approximated

by a rational number to as high an accuracy as we may wish. So we can set our ratio equal

to a ratio of two positive integers N and N ′,

Mh

M ′h

=N ′

N. (2.11)

We now consider a process consisting of N ′ cycles of the second engine and N reverse cycles

of the first engine. This is allowed because we assumed that the first engine was reversible.

When operated in reverse the first engine absorbs an amount of work L during each cycle,

giving up an amount of osmotic energy Mh to the source µh and absorbing an amount of

osmotic energy Ml from the source µl. The total work performed by the two engines during

the complex process described above is:

Ltotal = N ′L′ −NL. (2.12)

The total amount of osmotic energy absorbed from the reservoir µh is:

Mhtotal= N ′M ′

h −NMh. (2.13)

And the total amount of osmotic energy given up to the source µl is:

Mltotal= N ′M ′

l −NMl. (2.14)

15

Now using our original definitions of the total work each engine performs in a cycle from

expression (2.10) we obtain the result,

Ltotal = Mhtotal−Mltotal

. (2.15)

But substituting expression (2.11), which relates the ratio of osmotic energies to the ratio of

cycles performed, into expression (2.13) we see that:

Mhtotal= 0. (2.16)

Hence,

Ltotal = −Mltotal. (2.17)

Equation (2.16) states that the complete process produces no exchange of particles at the

high chemical potential reservoir µh, and equation (2.17) states that the osmotic energy

absorbed from the source µl is transformed into the work Ltotal. At the completion of this

process both engines will come back to their initial states. During this cycle we know that

Ltotal cannot be positive; for if it were positive, then the only result of this process would

be the transformation into work, Ltotal, of osmotic energy, −Mltotal, absorbed from a source

at chemical potential µl throughout. But this would contradict our equivalent of the Kelvin

postulate (statement 3). Hence, the two following inequalities must be satisfied:

Ltotal ≤ 0 and Mltotal≥ 0. (2.18)

Now using expression (2.14) for Mltotal, we obtain:

N ′M ′l ≥ NMl. (2.19)

If we eliminate N ′ and N from this expression with the aid of equation (2.11), we get, since

all the quantities in (2.11) are positive,

MhM′l ≥ M ′

hMl orMh

Ml

≥ M ′h

M ′l

, (2.20)

which is identical to part (a) of the fundamental theorem we set out to prove.

If we take the second engine to be reversible, we have, on interchanging the two engines

and applying the inequality (2.20), from part (a) of our theorem, to the new arrangement,

M ′h

M ′l

≥ Mh

Ml

. (2.21)

16

Both this inequality and the reverse (2.20) must hold for this case because both engines are

reversible. Since both inequalities (2.20) and (2.21) must hold if both osmotic engines are

reversible in follows thatMh

Ml

=M ′

h

M ′l

, (2.22)

which is identical to part (b) of the fundamental theorem we set out to prove.

We can restate the theorem just proven as follows (statement 4):

If there are several cyclic osmotic engines, some of which are reversible, operating

around cycles between the same chemical potentials µl and µh, all the reversible

engines have the same efficiency, while the nonreversible engines have efficien-

cies, which can never exceed the efficiency of the reversible engines.

According to statement 4 the efficiencies of two reversible osmotic engines must be equiv-

alent. This is a direct consequence of expression (2.9) on page 15. Two reversible osmotic

engines with osmotic energies Mh, Ml, M ′h and M ′

l must observe the relationship:

Mh

Ml

=M ′

h

M ′l

1− Mh

Ml

= 1− M ′h

M ′l

η = η′. (2.23)

According to statement 4 the efficiency of a reversible osmotic engine must always be

greater than the efficiency of an irreversible osmotic engine. This is a direct consequence

of expression (2.8) on page 15. A reversible osmotic engine with osmotic energies Ml and

Mh and an irreversible osmotic engine with osmotic energies M ′l and M ′

h must observe the

relationship:

Mh

Ml

≤ M ′h

M ′l

1− Mh

Ml

≥ 1− M ′h

M ′l

η ≥ η′. (2.24)

At this point all we know is that the efficiencies of osmotic engines are related to the

ratio of the osmotic energy, Mh extracted from the high chemical potential reservoir and the

osmotic energy, Ml absorbed by the low chemical potential reservoir. In the case of reversible

17

osmotic engines there is one further refinement we can make to the efficiency expressions. For

reversible engines the initial state of the engine must be equivalent to the final state of the

engine. This means that the engine can neither gain nor lose particles after the completion of

one cycle. Now if we recall the definitions of osmotic energy given in (2.6) and set dNh = dNl

the fundamental efficiency of a reversible osmotic engine becomes

η = 1− Mh

Ml

= 1− µh

µl

. (2.25)

This is an extremely surprising result which tells us that the efficiencies of all reversible os-

motic engines, irrespective of their mechanistic details, are limited by the chemical potentials

of the reservoirs they are connected to. In conjunction with statement 4 we must conclude

that the efficiency of any osmotic engine, ηe, is limited by the chemical potentials of the

reservoirs it is connected to in the following manner,

ηe ≤ 1− µh

µl

. (2.26)

18

Chapter 3

The Connection to Entropy

3.1 A State Function1

Now let’s attempt to develop a rigorous definition of entropy by considering a cyclic

system A. We suppose that during the cycle the system receives osmotic energy from or

surrenders osmotic energy to a set of sources having the chemical potentials µ1, µ2, . . . , µn.

Let the amounts of osmotic energy exchanged between the system and these sources be M1,

M2, . . . , Mn, respectively; we take the Ms positive if they represent osmotic energy received

by the system and negative in the other case. We shall now prove that:

n∑i=1

Mi

µi

≤ 0, (3.1)

and that the equality sign holds if the cycle is reversible.

In order to prove the theorem above we must introduce, besides the n sources listed

above, another source of osmotic energy with an arbitrary chemical potential µ0, and also n

reversible cyclic engines (which we will call, C1, C2, . . . ,Cn) operating between the chemical

potentials µ1, µ2, . . . , µn, respectively, and the chemical potential µ0. We shall choose the

ith Carnot cycle, Ci, which operates between the chemical potentials µi and µ0, to be of

such a size that it surrenders at the chemical potential µi the quantity of osmotic energy

Mi, that is, an amount equal to that absorbed by the system A at the chemical potential µi.

The amount of osmotic energy absorbed by system A could be positive or negative as stated

1This section is similar to Chapter 4 in [1].

19

in the first paragraph of this section so the corresponding osmotic energy transferred, Mi,

by the Cith Carnot cycle could also be positive or negative.

According to (2.25), the amount of osmotic energy absorbed by Ci from the source µ0 is

Mi,0 =µ0

µi

Mi. (3.2)

We now consider a complex cycle consisting of one cycle of the system A and one cycle of

each of the Carnot cycles C1, C2, . . . , Cn. The net exchange of osmotic energy at each of the

sources µ1, µ2, . . . , µn during the complex cycle is zero; the source µi surrenders an amount

of osmotic energy Mi to the system A, but it receives the same amount of osmotic energy

from the cycle Ci. The source µ0, on the other hand, loses an amount of osmotic energy equal

to the sum of the amounts (given by the above equation) absorbed by the Carnot cycles C1,

C2, . . . ,Cn. Thus, the source µ0 surrenders altogether an amount of osmotic energy equal to

M0 =n∑

i=1

Mi,0 = µ0

n∑i=1

Mi

µi

. (3.3)

Hence, the net result of our complex cycle is that the system composed of A and C1, C2,

. . . , Cn receives an amount of osmotic energy M0 from the source µ0. But we have already

seen that in a cyclic transformation the work performed is equal to the total osmotic energy

received by the system. Thus, since A, C1, C2, . . . , Cn return to their initial states at the

end of the complex cycle, the only final result of the complex cycle is to transform into work

an amount of osmotic energy received from a source at a uniform chemical potential µ0. If

M0 were positive, this result would be in contrandition to statement 3. It therefore follows

that M0 ≤ 0, or, from expression (3.3),

n∑i=1

Mi

µi

≤ 0, (3.4)

which is identical to our first definition.

If the cycle performed by A is reversible, we can describe it in the opposite direction, in

which case all the Mi will change sign. Applying our first definition to the reverse cycle, we

obtain:n∑

i=1

−Mi

µi

≤ 0 (3.5)

orn∑

i=1

Mi

µi

≥ 0. (3.6)

20

Thus, if the cycle is reversible, this inequality as well as (3.1), must be satisfied. This is

possible only if the equality sign holds. For a reversible cycle, therefore, we must have:n∑

i=1

Mi

µi

= 0. (3.7)

This completes the proof of our theorem.

In establishing our definition we assumed that the system exchanges osmotic energy with

a finite number of sources µ1, µ2, . . . , µn. It is important, however, to consider a continuous

distribution of sources. In that case, the sums in our definitions must be replaced by integrals

extended over the entire cycle.

Denoting by circle integral the integral extended over a cycle and by dM the infinitesimal

amount of osmotic energy received by the system from a source at the chemical potential µ,

we have: ∮dM

µ≤ 0, (3.8)

which is valid for all cycles, and ∮dM

µ= 0, (3.9)

which is valid only for reversible cycles.

Expression (3.9) tells us that the ratio of the change in osmotic energy, dM , over the

chemical potential, µ, is a state function. So if we were to pick any two equilibrium states,

B and C, of the system A the integral ∫ C

B

dM

µ(3.10)

is constant for all reversible transformations between B and C2. In order to prove this re-

lationship we must show that for any two arbitrary reversible transformations, 1 and 2,

between the equilibrium states B and C the ratios of the change in osmotic energy over the

chemical potential are equivalent,( ∫ C

B

dM

µ

)1

=

( ∫ C

B

dM

µ

)2

. (3.11)

If we consider the cyclic transformation from B to C along path 1 then back from C to B

along path 23 we have a complete reversible cycle since it is made of two reversible transitions.

2All of the states the system passes through between B and C must be equilibrium states because we areonly considering reversible cycles.

3In this transition we are using the fact that path 2 is a reversible transformation to run it in reverse.

21

Since we have a reversible cycle we can apply the property determined in expression (3.9) as

follows ∫B

1→C

2→B

dM

µ= 0. (3.12)

We can split this integral into a sum of two integrals,( ∫ C

B

dM

µ

)1

+

( ∫ B

C

dM

µ

)2

= 0( ∫ C

B

dM

µ

)1

=

( ∫ C

B

dM

µ

)2

. (3.13)

After reversing the end points of the second integral it becomes the negative of our original

path 2 integral and we have an expression equivalent to (3.11) which proves path indepen-

dence.

3.2 The Connection to Entropy

In these first chapters we set out to determine whether rewriting the Clausius statement

in terms of particle flow would give us a rigorous definition of entropy. Perhaps this wasn’t

exactly the right question to ask, because we set out to discover entropy by initially twisting

its definition around and then attempted to get back to the original definition. Of course the

equivalence of the Kelvin statement and the Clausius statement (which we proved in section

2.3) wasn’t a difficult matter but both definitions concerned heat transfers. Our statement

doesn’t even mention heat. Obviously if an equivalence did exist between our statement and

the original statement it would not be quite so simple as the one that exists between the

Kelvin statement and the Clausius statement.

In order for us to determine whether we actually have come up with a rigorous definition

of entropy we need some other way of defining what we are looking for. In Statistical

Mechanics, entropy is more commonly defined as a measure of the number of configurations

that are possible for a particular macroscopic state. This definition allows us to see that

entropy must be a state function because if a system returns to its initial state after a series

of transformations it should have the same number of configurations in that state as it had,

initially otherwise the two states wouldn’t be equivalent. In section 3.1 we discovered that

the ratio of the osmotic energy over the chemical potential is a state function. So now the

question becomes: is this the entropy state function?

22

In section 2.3 we defined the osmotic energy as the chemical potential times the change

in number so the ratio of the osmotic energy over the chemical potential is simply equal to

the change in number, ∫dM

µ= dN, (3.14)

and our state function is defined as

N(B)−N(A) =

∫ B

A

dM

µ, (3.15)

where we are considering a transformation between the states A and B each with total

number N(A) and N(B). We know that number is a state function but certainly it isn’t

equivalent to the entropy state function, so the only conclusion is that our statement of

Clausius in terms of particle flow doesn’t define entropy.

But let’s back up a little bit. We defined entropy simply as a measure of the total

configurations possible in a particular state. It is certainly the case that if you increase the

number of a system there are more possible configurations and if you decease the number of

a system there are fewer possible configurations. Our number state function certainly seems

like a measure of the number of configurations. If in fact our statement of Clausius was the

first the early scientists may have concluded that in fact the number state function was the

entropy state function.

At this point in the thesis we have answered the first of the two questions posed in the

introduction. We cannot derive a rigorous definition of entropy with the Clausius statement

rewritten in terms of particle flow. Now it is time to address the second question, but before

we apply the osmotic engine theory we have developed to some applications, we must connect

what we have determined to more commonly known thermodynamic relations.

3.3 What is Chemical Potential?

In chapter 2 we defined chemical potential, µ, simply as the proportionality constant

which converted the change in number to the amount of osmotic energy that could be

extracted (expressions (2.1) and (2.2)). Now we must define chemical potential in a more

general fashion and obtain a more intuitive understanding of what chemical potential means

in different systems. First let’s examine the entropy master function,

dS =1

TdE +

p

TdV − µ

TdN. (3.16)

23

The term which contains the chemical potential tells us that if two systems with the same

temperature and pressure are brought together so that they can exchange particles, then in

the system with high chemical potential the number will decrease while in the system with

low chemical potential the number will increase. With this interpretation in mind chemists

often call the chemical potential the “escaping tendency”, because particles tend to flow

from regions of high chemical potential to regions of low chemical potential [7].

The most important applications of chemical potential in this thesis are for the situation

where two compartments contain more than one type of particle but only exchange one of

the types of particles they contain. If we make the further assumption that we have dilute

solution conditions, which in fact we do make in many applications, then it follows that we

can use expressions (A.24) and (A.25) derived in appendix A for the chemical potential of

the solutes and the solvent, respectively. The chemical potential of the solvent (A.25) is

µ0(T, p,N0, · · · , Ng) = g0(T, p)− kBT

g∑i=1

(Ni

N0

). (3.17)

In this expression the first term is the amount of free-energy per particle of the solvent

particles and the second term, which is in fact a sum of terms, is the amount of free-energy

per particle due to the mixing of different types of particles. The second term exists simply

due to the fact that if you have two compartments with the same total number of particles but

one of the compartments possesses two different types of particles and the other compartment

only possesses one type, the compartment with two types of particles consistent with more

possible configurations. More configurations is equivalent to more total entropy.

Because of the second term in the solvent chemical potential, compartments composed

of larger amounts of solute particles have lower chemical potentials then those at the same

temperature and pressure with lower amounts of solutes. The various solute chemical po-

tentials,

µi(T, p,N0, · · · , Ng) = gi(T, p) + kBT log

(Ni

N0

)+ kBT, (3.18)

possess roughly the same two terms as the solvent. This first term is the amount of free-

energy per particle the solute particles in question possess alone. The second term is the

amount of free-energy per particle due to the mixing of solvent and solute particles.

For ideal solutions, all in the same phase, if the chemical potential of two compartments

is equal then the densities are also equal. Although equal chemical potential implies equal

density for this case, this isn’t always true. For example, at the freezing point of water, when

24

ice and liquid water are in equilibrium, the chemical potential is the same for both the solid

and the liquid, but the densities are clearly not equal [7].

25

Chapter 4

The Osmotic Carnot Cycle

4.1 Introduction

In chapers 1-3 we examined the limitations of an arbitrary osmotic engine. Now it is

time to apply the theorems we derived to a specific type of osmotic engine. Since the sim-

plest and most often discussed type of heat engine is the Carnot heat engine the osmotic

engine discussed in this section is similar to allow for comparison. Carnot’s heat engine was

originally considered as an attempt to construct the most efficient steam engine possible.

In section 2.3 we proved that reversible osmotic engines are always more efficient than ir-

reversible osmotic engines. The same is true of heat engines. Since Carnot was attempting

to construct the most efficient heat engine his engine was reversible. The efficiency of his

reversible engine was

ηr = 1− Tc

Th

. (4.1)

This famous result is very surprising because it is independent of the mechanistic details of

any particular heat engine and places a fundamental limitation on the efficiency of all heat

engines. Our result,

ηr = 1− µl

µh

, (4.2)

has the same implications for all osmotic engines but we haven’t yet put forth an actual

model of a specific osmotic engine. This section is dedicated to that task.

26

4.2 The Osmotic Carnot Cycle

Figure 4.1: Osmotic Carnot engine.

Let us consider the engine in figure 4.1 which consists of two large particle reservoirs

and a piston connected to both by semipermeable walls. The piston and reservoirs all

contain two distinct types of particles, N0 and N1, but the walls connecting the piston to

the reservoirs are only permeable to N0 type particles. We can control the permeability of

the walls such that the piston can be connected to neither of the reservoirs or either one

individually. Let us assume that we have dilute solution conditions, N0 � N1 and that the

reservoirs are sufficiently large that the engine is incapable of depleting them or changing

their concentrations. If we define the N0 particle type chemical potential of each reservoir

such that µh > µl1 then by connecting the piston to each of the reservoirs individually we

can obtain mechanical work from the system using osmotic pressure.

Osmotic pressure occurs when two compartments contain more than one type of particle

but exchange only one of the particle types they contain (this is described in more detail in

1The subscripts are h for high chemical potential and l for low chemical potential. Theses refer to the N0

particle type because it is the type that can diffuse through the permeable walls connected to the piston.

27

appendix C). In our engine the two compartments are the piston and whichever reservoir the

piston is connected to. Since both compartments contain two types of particles, N0 and N1,

but are only capable of exchanging N0 type particles, according to expression (C.11) from

appendix C if the densities of N1 type particles differ between the two compartments the

pressures of the compartments must differ. This difference in pressure allows us to extract

mechanical energy from the system.

One cycle of this osmotic engine consists of the following steps (see figure 4.2). The piston

is connected to the high chemical potential reservoir but it has a higher pressure than the

reservoir because it has a greater density of particle N1 then the reservoir. As a consequence

the piston’s volume will increase until the two chambers have the same pressure. At this

point the piston will be disconnected from the high chemical potential reservoir. When the

piston is connected to the low chemical potential reservoir it will have a lower pressure due to

a lower density of particle N1 than the reservoir. As a consequence the piston’s volume will

decrease until the two chambers have the same pressures. This process produces mechanical

work but from this description alone it isn’t reversible.

Since we wish to construct a reversible engine we will need to ensure that throughout

all transitions the system remains at equilibrium. According to appendix B if two com-

partments can exchange a particular type of particle then the two compartment system is

only at equilibrium when the chemical potential of that particle is equal between the two

compartments. So if the piston is connected to the high chemical potential reservoir then

for the two compartment system to be at equilibrium it must be the case that µh = µp.

The same relationship exists between the piston and the low chemical potential reservoir,

µl = µp, when they are connected.

For each of the two legs of our cycle where the piston is connected to one of the reservoirs

we know that the chemical potential of the piston is constant. We will call these isopotic

legs referring to constant chemical potential. To complete the cycle we need to get between

the two isopotic legs. We can’t simply disconnect the piston from one reservoir and then

reconnect it to the other one. If we did then upon reconnection the chemical potential of

the piston wouldn’t be equal to the chemical potential of the reservoir because µh > µl by

definition. If the two chemical potentials were different then we would have a non-equilibrium

state and the engine wouldn’t be reversible.

In order to solve this problem we must consider expression (A.25) for the solvent chemical

potential derived in appendix A. This expression applies because we have dilute solution

28

Figure 4.2: Representation of the piston position throughout a complete Carnot osmotic

engine cycle (ABCDA).

conditions and are considering ideal solutions. If we replace g0(T, p) so that it applies to

changes in pressure with respect to atmospheric pressure we obtain

µ0(T, p,N0, N1) = kBT log

(p

patm

)− kBT

(N1

N0

)(4.3)

as the solvent chemical potential as a function of temperature, T , pressure, p, and the number

of atoms N0 and N1. Throughout the entire cycle we will assume that the temperature is

constant, otherwise our engine would be both a heat engine and an osmotic engine (whether

constant temperature is a valid assumption is explored at the end of this section). Now we

can return to the question of how to get between the isopotic legs. Once disconnected from

one of the reservoirs we must somehow change the chemical potential of the piston to match

29

that of the other reservoir before we connect it. Expression (4.3) tells us that the solvent

chemical potential is a function of temperature, pressure, amount of solvent and of solute.

We know that the temperature is constant throughout the entire cycle and we know that the

amount of solvent and amount of solute must be constant when the piston isn’t connected

to a reservoir so we can only change the pressure in order to equalize the chemical potential.

Since µh > µl according to expression (4.3) when going from high to low chemical potential

we must decrease the pressure and when going from low to high chemical potential we must

increase the pressure.

Figure 4.3: (µ, N) Diagram of the Carnot osmotic engine cycle (ABCDA).

Our Carnot osmotic engine cycle ABCDA is represented in figure 4.3 in terms of changes

in number, N , and changes in chemical potential, µ, where the arrows represent the direction

the engine travels around the cycle. For isopotic transitions AB and CD the piston is at

constant chemical potentials µh and µl, respectively. We will call the other two transitions

BC and DA isonumbric referring to constant number.

Our osmotic engine from figure 4.1 operates using the same osmotic energy that we

discussed earlier in chapter 2. So according to expression (2.7) the total work performed by

the engine in one cycle must be equal to

L = Mh −Ml = µh dNh − µl dNl, (4.4)

where dNh is the number of N0 particles that enter the piston when it is connected to the

high chemical potential reservoir and dNl is the number of N0 particles that leave the piston

30

when it is connected to the low chemical potential reservoir. For a reversible cycle dNh = dNl

because the initial state of the piston must be equal to the final state of the piston. Figure

2.1 is an abstract representation of this situation. In the (N , µ) diagram from figure 4.3 the

area enclosed is equal to total mechanical work performed, L, and the difference between

NDA and NBC is equal to dNh and dNl.

Now that we have a clear understanding of the (N , µ) diagram it is time to consider the

(p, V ) diagram for this cycle. To find the (p, V ) diagram for this cycle, we need expressions

for the pressure of the piston as a function of its volume for both isopotic legs, AB and CD,

and for both isonumbric legs, BC and DA. Once we have these expressions we must ensure

that they are all equal at each of the four points A, B, C and D, where the transformation

we are considering changes from isopotic to isonumbric or isonumbric to isopotic.

There are two possible methods for modeling the isopotic legs, AB and CD. The first

method is to simply use the osmotic pressure expression (C.11) derived in appendix C with

atmospheric pressure added in, assuming that is the pressure of the reservoirs. For the AB

transition the pressure as a function of volume then becomes

P (V ) = kBT

(N1

V− N1B

VB

)+ Patm (4.5)

where we are assuming that the µh reservoir’s N1 particle density is equal to that of the

piston at B, N1B

VB. In expression (4.5) the osmotic pressure term is positive throughout the

transition until it becomes zero at B and the pressure of the piston is simply atmospheric

pressure. The corresponding expression for the pressure as a function of volume for the CD

transition is

P (V ) = kBT

(N1

V− N1D

VD

)+ Patm (4.6)

where we are assuming that the µl reservoir’s N1 particle density is equal to that of the

piston at D, N1D

VD. In expression (4.6) the osmotic pressure term is negative thoughout the

transition until it becomes zero at D and the pressure of the piston is simply atmospheric

pressure.

The pressure as a function of volume for the isonumbric legs can be modeled simply by

assuming that PV = constant because the number and temperature aren’t changing for the

transitions, BC and DA. So the two isonumbric functions are

PBC(V ) =(N1B + N0B) kBT

Vand PDA(V ) =

(N1D + N0D) kBT

V, (4.7)

31

where N1B is the number of N1 particles in the piston at B, N0B is the number of N0 particles

at B, N1D is the number of N1 particles in the piston at D and N0D is the number of N0

particles in the piston at D. We could have chosen N1C for the BC transition and N1A for

the DA transition because N1B = N1C and N1D = N1A. So now we can use expressions

(4.7), (4.5), and (4.6) to graph a (p, V ) diagram (figure 4.4) as long as we ensure that the

expressions are consistent and are equal to one another at the points A, B, C and D.

Figure 4.4: The correct (p, V ) diagram.

As we can see in figure 4.4, throughout the AB transition the pressure of the piston

is greater than the pressure of the µh reservoir because N0 particles are flowing from the

reservoir to the piston, which is expanding. Throughout the CD transition the pressure of

the piston is less than the pressure of the µl reservoir because N1 particles are flowing from

the piston to the reservoir and the piston is contracting. Assuming both reservoirs are at

atmospheric pressure, which is most likely the situation for any two natural reservoirs, the

piston’s internal pressure must be above atmospheric pressure for the AB transition and

below atmospheric pressure for the CD transition. The two isonumbric transitions, BD and

DA, in figure 4.4 simply connect the isopotic legs which wouldn’t intersect unless they were

at the same chemical potential.

We could alternatively use expression (4.3) to determine the pressure as a function of

volume for the isoptic transitions. For the AB transition we can start by setting expression

(4.3) equal to µh and multiplying N1

N0by 1/V

1/Vwhere V is the volume of the piston. After these

32

changes we obtain the expression

µh = kBT log

(p

patm

)− kBT

( N1

VN0

V

). (4.8)

If we then solve for the pressure, p and replace N0

Vwith ρ0 we obtain

P (V ) = Patm eµh

kBT+

N1V ρ0 , (4.9)

which is the pressure throughout the AB transition as a function of the chemical potential of

the reservoir, µh, the temperature, T , the number of N1 particles in the piston, the volume

of the piston, V , and the density of N0 particles, ρ0, which is constant throughout the AB

transition. The density of the N0 particles in the piston is constant throughout both isopotic

transitions, AB and DA, as a consequence of the fact that the osmotic pressure, expression

(C.11) from appendix C, is independent of the density of N0 particles (this property of

constant N0 density for osmotic systems which can freely exchange N0 particles is proven in

appendix D).

We can also easily obtain the equivalent to expression (4.9) for the CD transition when

the piston is connected to the low chemical potential reservoir. All we would have to do

is set expression (4.3) equal to µl and repeat the derivation above. In this expression we

also have constant N0 density as proven in appendix D. For the two isonumbric legs it

doesn’t make sense to use expressions for the pressure including chemical potential because

chemical potential isn’t constant for the transitions, BC and DA. If we wished to know how

the chemical potential changes during the isonumbric transitions as a function of pressure

we could use expression (4.3). If we wanted to know how chemical potential changes as a

function volume we could replace pressure, p, in expression (4.3) with one of the isonumbric

expressions (4.7). There are quite a few additional graphs that we could produce with very

interesting information using these relationships but that is material for another thesis.

Earlier we assumed that the temperature is constant throughout all transitions but didn’t

justify or explore the implications of that assumption. In Carnot’s original cycle the two

so-called adiabatic legs took the piston between two isothermal legs, which were at different

temperatures but in our engine the pistons volume changes but we keep the temperature

constant. During the two isopotic transitions particles are constantly exchanged between the

reservoirs and the piston so the transitions are isothermal because the temperature of the

reservoir is equal to the temperature of the piston throughout. During the two isonumbric

33

transitions, however, we must add heat, for the BD transition, and subtract heat, for the

DA transition, in order to keep the temperature constant. Since we are adding heat during

the BD leg and subtracting heat during the DA leg as long as the two transfers of heat are

equal there will be no net heat loss or gain for the system.

A method for making these heat transfers could simply be to run two Carnot osmotic

engines side by side except half a cycle off. That way as one is performing the BD transition

it can absorb heat from the other one performing the DA transition and visa versa for the

other isonumbric transitions.

4.3 Applications

The use of osmotic pressure in producing clean renewable energy is not a new idea but re-

cently the technology has become available to construct very large pressure retarded osmotic

systems for producing large amounts of energy. One of the major projects that has been

proposed is the construction of an osmotic power generation system between the Mediter-

ranean ocean and the Dead Sea (see [10]). Other projects have been proposed between fresh

water rivers and the ocean (see [11]).

Without knowing anything about the specifics of the osmotic systems involved in these

projects we can place a fundamental limit on the amount of energy that could be gained based

on the salinity content of the reservoirs the osmotic engine is connected to. As demonstrated

in section 4.2, reversible osmotic pressure engines have a fundamental efficiency, expression

(4.2), related to the chemical potential of the solvents in both of the reservoirs connected to

the osmotic engine. According to statement 4 this is the maximum possible efficiency for any

engine between those two reservoirs. In order to solve this problem we need an expression

for the total energy as a function of the salinity of the two reservoirs the osmotic engine

is operating between. Expression (4.4) on page 30 is the total work an reversible osmotic

engine can perform as a function of the solvent chemical potentials,

L = Mh −Ml = µh dNh − µl dNl. (4.10)

For a reversible engine dNh = dNl and for our purposes we can simply set these equal to

one. Substituting for µh and µl with the solvent chemical potential expression (A.25) from

34

appendix A leaves us with

L = g0h(T, p)− kBT

g∑i=1

(Nih

N0h

)− g0l(T, p) + kBT

g∑i=1

(Nil

N0l

). (4.11)

If we assume that both reservoirs have the same temperature, pressure and solvent properties

then we can cancel out the two Gibbs free-energy terms leaving us with

L = kBT

( g∑i=1

Nil

N0l

−g∑

i=1

Nih

N0h

). (4.12)

We can use expression (4.12) to calculate the total amount of energy that any type of osmotic

engine could hope to produce between two reservoirs if we know the ratio of solute to solvent

for all solutes in each reservoir. In the above expression we are subtracting the sum of the

solute-solvent ratios of the high chemical potential reservoir from the sum of the solute-

solvent ratios of the low chemical potential reservoir. As we learned in our discussion of

chemical potential in section 3.3, solvent reservoirs with high salinity have a lower chemical

potential then reservoirs with low salinity so expression (4.12) must always be positive or

zero because the l subscripts stand for low chemical potential and the h subscripts stand for

high chemical potential.

Using table 4.1, which contains the composition of salt in the ocean, and table 4.2, which

contains the composition of salt in the Dead Sea, we can now calculate the maximum amount

of energy per liter transferred that could be obtained from an osmotic engine connected

between them. Using the sums of all the ocean water and Dead Sea water salt mole fractions

we can substitute for the two summations in expression (4.12) to obtain a value of 24.363 kJL

for the maximum energy produced. So for every 150 L transferred we could obtain 1 kWh

of energy.

35

Ion Concentration ( gkg

) % salt weight mole fraction ( moles solutemoles solvent

)

Chloride (Cl−) 18.98 55.04 9.794× 10−3

Sodium (Na+) 10.56 30.61 8.403× 10−3

Sulphate (S02−4 ) 2.65 7.68 5.047× 10−4

Magnesium (Mg2+) 1.27 3.69 9.557× 10−4

Calcium (Ca2+) 0.40 1.16 1.826× 10−4

Potassium (K+) 0.38 1.10 1.778× 10−4

Bicarbonate (HCO−3 ) 0.14 0.41 4.197× 10−5

Bromide (Br−) 0.07 0.19 1.603× 10−5

Borate (mainly H3BO3) 0.03 0.07 2.088× 10−6

Strontium (Sr2+) 0.01 0.04 2.088× 10−6

Sum 2.009× 10−2

Table 4.1: Composition of salt in the ocean (concentrations ( gkg

) taken from [8]).

Ion Concentration ( gkg

) % salt weight mole fraction ( moles solutemoles solvent

)

Chloride (Cl−) 224.2 66.46 0.1180

Sodium (Na+) 41.3 12.24 3.354× 10−2

Sulphate (S02−4 ) 0.001 29.64× 10−5 1.943× 10−7

Magnesium (Mg2+) 42.12 12.48 3.235× 10−2

Calcium (Ca2+) 17.60 5.21 8.198× 10−3

Potassium (K+) 7.60 2.25 3.629× 10−3

Bicarbonate (HCO−3 ) < 0.001 0 0

Bromide (Br−) 4.5 79.9 1.051× 10−3

Sum 0.196

Table 4.2: Composition of salt at the surface of the dead sea (halite concentrations ( gkg

)

taken from [9]).

36

Chapter 5

Biological Osmotic Engines

5.1 Introduction

One important application of the osmotic engine theory we have developed is in biologi-

cal systems. All biological systems are made up of cells whose membranes compartmentalize

various cellular functions in order to ensure proper environmental regulation and continuous

operation of metabolic reactions. All of the various compartments within the cell and within

the intracellular medium have different specific required solute concentrations including dif-

ferent specific pH requirements. In the human body these requirements become even more

complex with the need for rapid selective movement of nutrients, waste products and other

molecules and proteins into and out of specific cells at specific times.

These complex operations continually performed by both single and multicellular systems

remained a mystery until the second half of the 19th century when the knowledge and tech-

nology finally became available to perform more precise measurements of cellular structure

and membrane transport. One of the first scientists to propose that osmotic energy could be

used to perform cellular functions was Peter Mitchell, who, in 1961, published a paper [12]

titled Coupling of Phosphorylation to Electron and Hydrogen Transfer by a Chemi-osmotic

Type of Mechanism in which he postulated that a proton gradient is used as the energy source

for ATP synthesis in mitochondria. Peter Mitchell won the Nobel Prize for this discovery

and it became the foundation for chemiosmotic theory.

Now it is known that there exist thousands of different types of membrane proteins

designed to transport many different kinds of molecules into their proper compartments using

37

osmotic energy from concentration gradients. Without knowing the mechanistic details of

these transport proteins we can apply all of the osmotic engine theory developed in previous

sections of this thesis in order to understand the limitations of these mechanisms. In fact the

mechanisms of these transport proteins aren’t very well understood by the leading scientists

in the field.

In this section we will focus on two main classes of cellular osmotic engines, antiporters

and symporters (see figure 5.1). Both of these types operate by coupling two osmotic engines

using the work from one engine to run the other engine. Antiporters transport molecules in

opposite directions, one type of molecule is transported into the cell and the other type of

molecule is transported out of the cell. Symporters transport molecules the same direction,

both types of molecules are transported either out of the cell or into the cell.

Figure 5.1: Coupled anitport (left) and symport (right) via membrane transport proteins.

5.2 Coupled Transport

Antiport and symport membrane proteins utilize the free-energy gained by allowing one

type of particle to flow from high chemical potential to low chemical potential (along its

natural gradient), to pump a different type of particle from low chemical potential to high

chemical potential (against its natural gradient). This situation is represented for antiporters

38

Figure 5.2: Abstract representation of a coupled particle transporter.

in figure 5.2, where engine 1 gains free-energy through transporting particle type N1 which,

is then used by engine 2 to transport particle type N2. Applying equation (2.7) on page 13,

the work performed by engine 1 is

L1 = M ′1 −M1 = µ′1 dN1 − µ1 dN1 (5.1)

where M ′1 and M1 are the osmotic energies exchanged with each reservoir for type N1 par-

ticles, µ′1 and µ1 are the chemical potentials of the two reservoirs for type N1 particles and

dN1 is the number of particles transferred. We can solve for the work performed by engine

2 as we did for engine 1 to obtain the expression

L2 = M2 −M ′2 = µ2 dN2 − µ′2 dN2, (5.2)

where all quantities are defined as above except for N2 type particles.

In the coupled antiport model represented in figure 5.2 the reservoirs satisfy the conditions

N ′1 > N1 and N ′

2 > N2 (5.3)

39

which, if we assume a dilute solutions approximation, implies

µ′1 > µ1 and µ′2 > µ2. (5.4)

For these conditions engine 1 will perform a positive amount of work (L1 > 0 from equation

(5.1)) and engine 2 will perform a negative amount of work (L2 < 0 from equation (5.2)).

Defining the total work of the two engine system to be the sum of the work performed by

each individual engine we have

Ltotal = L1 + L2. (5.5)

Depending on the values of µ′1, µ1, µ′2 and µ2 the total work could be greater than zero,

Ltotal > 0, equal to zero, Ltotal = 0, or less than zero, Ltotal < 0. If Ltotal < 0 the coupled

engines cannot run without an input of work from the outside. The situation where Ltotal <

0 occurs for coupled transport where ATP is phosphorylated, providing the extra energy

required for the engine to run. This is not the situation we wish to consider but there are a

large number of proteins that operate in this manner. We are most interested in the cases

where Ltotal ≥ 0.

Although the above argument and diagram apply to antiporters, with only one minor

change we can interpret the situation for symporters. Because symporters transport both

types of particles in the same direction, and if we run engine 2 from figure 5.2 in the same

direction as engine 1, then we have the situation for a symporter. The work performed by

engine 2 then becomes

L2 = M ′2 −M2 = µ′2 dN2 − µ2 dN2, (5.6)

where are all quantities are defined identically to those from expression (5.2).

The total work performed is still simply the sum of the work performed by each engine

but, in the symport model the reservoir must satisfy the conditions

N ′1 > N1 and N2 > N ′

2 (5.7)

which, if we assume a dilute solution approximation (N1 � N0 and N2 � N0), implies

µ′1 > µ1 and µ2 > µ′2. (5.8)

In this model it is still the case that engine 1 performs a positive amount of work and engine

2 performs a negative amount of work. It follows that the engine will only run in the absence

of ATP if Ltotal ≥ 0 as is the case for antiporter transport proteins.

40

Now that we understand the basic mechanics of the antiport and symport osmotic engines

we can consider their efficiency. We shall define the efficiency as

η =benefit

cost=−L2

L1

. (5.9)

We define the efficiency of the engine as above because we wish to know how much of the

available work produced by engine 1 is used to run engine 2. If most of the osmotic work

produced by engine 1 isn’t used then the whole osmotic engine is inefficient. On the other

hand, if all of the osmotic work produced by engine 1 is used by engine 2, then L1 = −L2;

the engine is 100% efficient. So how does this efficiency relate to the fundamental efficiencies

we found in earlier sections that depended solely on the chemical potentials of the reservoirs?

In section 2.3 we proved that reversible osmotic engines are the most efficient osmotic

engines of all those connected to a particular set of reservoirs (statement 4). In chapter 4

we applied this theorem and determined that the efficiency of a reversible osmotic piston

engine is dependent only on the chemical potentials of the reservoirs it operates between,

expression (2.25). To obtain this result we used the fact that a reversible piston engine must

have the same number of particles before and after one cycle.

In order to apply our theorem to antiporters and symporters we must first determine

whether they are reversible. Our criteria for determining reversibility has to be slightly

adjusted from our analysis of the osmotic piston engine because the coupled osmotic engine

is always in the same state. There is no chamber inside this engine so there is no way of

defining volume, number, temperature and pressure. In order to get around this problem we

must consider the two reservoirs to be part of the engine. The engine is only reversible if it

can run in the reverse direction and return both reservoirs to their initial state. It follows

from this definition that a coupled osmotic engine is reversible only if L1 = −L2 (Ltotal = 0)

because, if L1 > −L2, then it is not possible to run the coupled osmotic engine in reverse.

Now using expressions (5.1), (5.2) and (5.6) we can substitute for L1 and −L2 to find

the fundamental efficiency of both antiporters and symporters. It follows that reversible

antiporters have the efficiency,

η =(µ′2 − µ2) dN2

(µ′1 − µ1) dN1

, (5.10)

and reversible symporters have the efficiency,

η =(µ2 − µ′2) dN2

(µ′1 − µ1) dN1

. (5.11)

41

If the number of each type of particle transported is the same for either engine then we can

remove the dN1’s and dN2’s from the expressions above. We have demonstrated that the

limitations of antiporters and symporters can be understood using statement 4 and that their

fundamental efficiencies are dependent only upon the chemical potentials of the reservoirs

they are connect to and the number of particles exchanged.

5.3 Antibiotic Resistance Via Coupled Transport

One particular class of antiport membrane proteins, known as the Multidrug Facilitator

Superfamily (MFS), confer antibiotic resistance to many strains of bacteria [13]. This par-

ticular set of proteins is of interest to those in the medical profession because humans are

susceptible to infection from several of these antibiotic resistant strains.

Antibiotics are administered to a wide range of patients as a method for selectively

attacking bacterial infections which invade the body. Human cells are eukaryotic, possessing

a nucleus and highly evolved mechanisms for antibiotic resistance, so they aren’t susceptible

to antibiotics given for treatment. Bacterial cells are prokaryotic, lacking both a nucleus

and the complex mechanisms to resist antibiotic attacks. Antibiotics diffuse through the

membranes of bacterial cells and bind to the protein-making machinery, halting cell growth,

thus killing off the invading cells.

In recent decades the use of antibiotics in treatment has sky-rocketed. This has helped

keep us disease-free, but it has also caused the world to become one large incubation chamber

where only the most antibiotic resistant strains of bacteria survive and multiply. Several

strains of bacteria have developed coupled transport mechanisms that allow them to pump

antibiotics out at rates equivalent to or greater than antibiotics diffuse in. As a consequence

antibiotics are becoming ineffective at killing off bacterial infections.

The osmotic engine theory that we have developed allows us to make some predictions

about the limitations of these antibiotic resistant bacterial strains. Many members of the

MFS are antiport membrane proteins, which allow one particular type of molecule to en-

ter the cell (traveling along its natural gradient) in exchange for transporting an antibiotic

molecule out of the cell (against its natural gradient). In section 5.2 we determined that

coupled osmotic engines will only run if Ltotal ≥ 0 or equivalently if L1 ≥ −L2. Without

considering the specifics of any particular antiport protein, we can determine which combina-

tions of reservoir concentrations the protein will function in. This situation is an illustration

42

of the power of statement 4 and the unexpected result in section 5.2 that the efficiencies

of all reversible antiport and symport osmotic engines are independent of their mechanistic

details.

We can now solve for the combination of reservoir concentrations, at which any particular

antiport protein will stop functioning, by setting Ltotal equal to zero,

Ltotal = 0 = L1 + L2

−L2 = L1

−µ2 dN2 + µ′2 dN2 = µ′1 dN1 − µ1 dN1. (5.12)

For most antiport proteins every cycle involves transporting one molecule in and one molecule

out so expression (5.12) simplifies to

µ′2 − µ2 = µ′1 − µ1. (5.13)

Since both engines exchange solute molecules, L1 using N1 type molecules and L2 using

N2 type molecules (see figure 5.2), we can apply expression (A.24) to determine the solute

chemical potentials. Before we can substitute in the chemical potential expressions we must

add a term to account for the membrane potential difference, which occurs across all cell

membranes.

There are many transport proteins that pump ions across cell membranes continually as

a part of normal cellular function. These proteins maintain a constant potential difference

between the inside and the outside of the cell. The net potential difference between the

inside and outside of the cell can be determined using the ion concentrations on either side

of the membrane and the Nernst equation (equation (1.10) in [14]). In order to adjust the

solute chemical potential expressions we must add in a term to account for the free-energy

gained or lost due to the potential difference. The modified general solute chemical potential

expression is

µi(T, p,N0, · · · , Ng) = gi(T, p) + kBT log

(Ni

N0

)+ kBT + zi e V, (5.14)

where z is the charge of the Nith molecule, e is the elementary charge and V is the potential

of the chamber the chemical potential refers to.

Now substituting in the solute chemical potentials using the revised expression (5.14) for

43

ideal dilute solution conditions, expression (5.12) becomes

g′2(T, p) + kBT log

(N ′

2

N ′0

)+ kBT + z2 e V ′

−g2(T, p)− kBT log

(N2

N0

)− kBT − z2 e V =

g′1(T, p) + kBT log

(N ′

1

N ′0

)+ kBT + z1 e V ′

−g1(T, p)− kBT log

(N1

N0

)− kBT − z1 e V. (5.15)

If we assume that the Gibbs free-energy functions per particle, independent of mixing factors,

are equal for particles of the same type on either side of the membrane, g′2(T, p) = g2(T, p)

and g′1(T, p) = g1(T, p), which is a fare assumption because the temperature and pressure

inside and outside the cell should be nearly equal, then we obtain

kBT log

(N ′

2

N ′0

N0

N2

)+ z2 e Vm = kBT log

(N ′

1

N ′0

N0

N1

)+ z1 e Vm. (5.16)

We can simplify expression (5.16) if we consider the quantities of molecules to be con-

centrations instead of total quantities inside the cell and outside the cell. The concentration

of solvent, which is water in this case, is equal on either side of the membrane so our final

expression is

kBT log

([N ′

2]

[N2]

)+ z2 e Vm = kBT log

([N ′

1]

[N1]

)+ z1 e Vm, (5.17)

where Vm is the potential difference across the membrane. Expression (5.17) is the free-

energy for transporting one N2 molecule across the membrane (on the left) set equal to the

free-energy for transporting one N1 molecule across the membrane (on the right).

Surprisingly we can simplify this expression even further for situations where z1 = z2 to

obtain[N ′

2]

[N2]=

[N ′1]

[N1]. (5.18)

So if the charges on the two types of molecules exchanged are equivalent then the antiport

protein stops functioning when the ratios of their concentrations inside and outside the cell

are equal. In fact for all concentration ratios where the quantity on the left is less than or

equal to the quantity on the right, in expression (5.18), the protein still functions.

44

The combination of concentrations, at which any arbitrary symporter stops functioning,

can be derived similarly to that of antiporters. If we set Ltotal for a symporter equal to zero

and substitute in our revised solute chemical potential expression (5.14) where appropriate

we obtain

kBT log

([N2]

[N ′2]

)− z2 e Vm = kBT log

([N ′

1]

[N1]

)+ z1 e Vm. (5.19)

Once again we are essentially setting the free-energies of the two gradients equal to each

other. There is no way to simplify this expression further. Even if the two molecules

transported have equal charge, z1 = z2, we can’t cancel out the free-energies due to the

potential difference across the membrane. Upon closer examination this difference between

antiporters and symporters makes sense. For antiporters when z1 = z2 the free-energy due to

the potential difference across the membrane is always equal and opposite between the two

gradients so it cancels out. For symporters when z1 = z2 the free-energy due to the potential

difference across the membrane is always equal between the two gradients and doesn’t cancel

out.

Using the revised solute chemical potential expression (5.14) we can substitute for the

chemical potentials in the efficiency expressions for reversible antiporters and symporters,

expressions (5.10) and (5.11). We can then use these efficiency expressions to determine the

relative efficiency of the various symporters and antiporters.

The ability to calculate the relative efficiencies of various coupled transport proteins is

very exciting because encapsulated in these efficiencies is a great deal of information. As an

example lets consider the TetA tetracycline/proton antiporter that is a member of the MFS,

which confers high levels of tetracycline resistance to many bacterial strains. Using expression

(5.17) we can calculate the exact combination of concentrations where the protein stops

functioning. In reality this result places an upper limit on ratio of concentrations where the

protein stops functioning because all real coupled transport proteins operate with efficiencies

lower then reversible coupled transport proteins. This difference between the efficiency of real

coupled transport proteins and theoretical reversible coupled transport proteins can tell us a

great deal about a particular protein. Solving for Ltotal for the point when the protein stops

functioning can tell us roughly how much free-energy is being consumed by the mechanisms

of the protein. Armed with this information it is possible to start to address what is exactly

going on inside these proteins (what sets of conformational changes are possible with the

available free-energy Ltotal). Appendix E outlines an experimental procedure for measuring

the actual combination of concentrations at which TetA stops functioning (This measurement

45

was the original motivation for this thesis).

46

Appendix A

Chemical Potential in Dilute

Solutions1

In this appendix a useful expression is derived for the chemical potential of the various

constituents in a dilute solution as a function of temperature, pressure, moles of solvent and

moles of solutes. The expression derived enables us to compare the chemical potentials of

particles under different conditions.

Lets us consider a solution composed of N0 moles of solvent and N1, N2, . . . , Ng moles

of various solutes A1, A2, . . . , Ag, respectively. If our solution is very dilute we can assume:

N1 � N0 ; N2 � N0 ; · · · ; Ng � N0. (A.1)

We must first solve for the energy of our solution, U , which we will need in order to apply

the thermodynamic equations. Let u be the energy of a fraction of the solution containing

one mole of solvent. This fraction of solution also contains N1

N0moles of solute A1,

N2

N0moles

of solute A2, . . . , Ng

N0moles of solute Ag. The energy of the fraction will be a function of T ,

p and the ratios N1

N0, N2

N0, . . . , Ng

N0.

u = u

(T, p,

N1

N0

,N2

N0

, · · · ,Ng

N0

). (A.2)

The total energy is then simply the energy of the one mole fraction (A.2) times the number

1The beginning of this section follows arguments about the thermodynamics of dilute solutions by Fermiin chapter VII from [1].

47

of moles of solvent in our solution, N0.

U = N0 u

(T, p,

N1

N0

,N2

N0

, · · · ,Ng

N0

). (A.3)

Since our solution is dilute (A.1) we can assume that the solute to solvent ratios, N1

N0, N2

N0,

. . . , Ng

N0are very small. Since the ratios are small we can develop the function in powers of

the ratios and neglect all powers above the first. The Taylor Series approximation is

u = u0

(T, p, 0, 0, · · · , 0

)+

N1

N0

u1

(T, p, 0,

N2

N0

, · · · ,Ng

N0

)+

N2

N0

u2

(T, p,

N1

N0

, 0,N3

N0

, · · ·)

.

(A.4)

If we now substitute this expression in for u in (A.3) we obtain

U = N0 u0(T, p) + N1 u1(T, p) + · · ·+ Ng ug(T, p) =

g∑i=0

Ni ui(T, p). (A.5)

We can extend the reasoning of the approximation for the energy to the volume. We can

then derive the volume with equal accuracy.

V = N0 v0(T, p) + N1 v1(T, p) + · · ·+ Ng vg(T, p) =

g∑i=0

Ni vi(T, p). (A.6)

Now we can solve for the entropy of our solution. To accomplish this task we must

consider an infinitesimal reversible transformation in which T and p change by the amounts

dT and dp, while the number of moles of solvent and solutes remain constant. For this

transformation the change in entropy is

dS =dQ

T=

1

T(dU + p dV ) =

g∑i=0

Nidui + p dvi

T. (A.7)

Since dS is a perfect differential for all values of the N ’s, the coefficient of each N in

(A.7) must be a perfect differential. Since we know that the individual coefficients are perfect

differentials we can define a set of functions s0(T, p), s1(T.p), . . . , sg(T, p) where

dsi(T, p) =dui + p dvi

T. (A.8)

If we put this set of functions into expression (A.7) and integrate the total entropy of the

solution becomes

S =

g∑i=0

Ni si(T, p) + C(N0, N1, · · · , Ng). (A.9)

48

The constant of integration, C, which depends only on the number of moles of solvent and

solutes, can be solved for explicitly.

In order to solve for the constant of integration we must consider an alternate expression

for the total entropy of our solution. The expression for the change in heat, dQ, in terms of

the specific heat at constant volume, Cv, is

dQ = Cv dT + p dV. (A.10)

If we then divide expression (A.10) by temperature, T , remembering that the change in

entropy, dS, is equal to the change in heat over the temperature we obtain

dS =dQ

T=

Cv

TdT +

p

TdV. (A.11)

Using the equation of state, assuming we have one mole, we can replace the pressure in

expression (A.11) and integrate to obtain

S = Cv log(T ) + kB log(V ) + a, (A.12)

the alternate expression for the total entropy we were seeking where a is the constant of

integration. We can rewrite expression (A.12) by once again making a substitution using

the equation of state but this time for V . If we also use the relation Cp = Cv + R we obtain

the expression

S = Cp log(T )− kB log(p) + a + kB log(kB). (A.13)

Now in order to rewrite expression (A.13) including the number of molecules of solvent and

solutes we must replace the pressure, p, with the partial pressure (the total pressure times

the ratio of number of molecules of the substance we want the partial pressure of over the

total number of molecules of solution) leaving us with entropy for the particular substance in

question. We must then multiply this quantity by the number of molecules of that substance

we have. After making these changes we obtain

S =

g∑i=0

Ni

(Cpi log(T )− kB log

(p

Ni

N0 + · · ·+ Ng

)+ ai + kB log(kB)

)=

g∑i=0

Ni(Cpi log(T )− kB log(p) + ai + kB log(kB))

− kB

g∑i=0

Ni log

(Ni

N0 + · · ·+ Ng

). (A.14)

49

After examining expression (A.14), the alternate total entropy, closely we find that there

are two distinct pieces,

si = Cpi log(T )− kB log(p) + ai + kB log(kB),

and

C(N0, N1, · · · , Ng) = −kB

g∑i=0

Ni log

(Ni

N0 + · · ·+ Ng

), (A.15)

that match up with pieces from our previous expression (A.9). We can see that our previous

constant of integration, C, is not a function of T or p as we expected it wouldn’t be.

We must remember that in finding our alternate expression for the entropy we assumed

that the total entropy is equal to the sum of the partial entropies of the various substances.

Luckily we can avoid concern about whether this is a fare assumption for our solution because

the constant of integration, C, which is the only piece we are interested in, doesn’t depend

on temperature or pressure. If it did depend on temperature and pressure then we would

have to recognize the fact that the assumption only really holds for gases and may not hold

at lower temperatures for solutions.

Now if we substitute for the constant of integration, C, in expression (A.9) we obtain

S =

g∑i=0

Ni si(T, p)− kB

g∑i=0

Ni log

(Ni

N0 + · · ·+ Ng

), (A.16)

the total entropy of our solution. We can further simplify expression (A.16) by taking into

account the dilute solution approximation we assumed in the beginning with the inequalities

(A.1). By neglecting terms of an order higher than the first in the small quantities N1, N2,

. . . , Ng, the first term in the second summation becomes

N0 log

(N0

N0 + · · ·+ Ng

)= N0 log

(1

1 + N1

N0+ · · ·+ Ng

N0

)= N0

(− N1

N0

− N2

N0

− · · · − Ng

N0

)= −N1 −N2 − · · · −Ng. (A.17)

We can also further simplify all of the other terms in the second summation to obtain

Ni log

(Ni

N0 + · · ·+ Ng

)= Ni log

(Ni

N0

)(for i ≥ 1) (A.18)

50

which gives us

S = N0 s0(T, p) +

g∑i=1

Ni (si(T, p) + kB)− kB

g∑i=1

Ni log

(Ni

N0

).

Introducing the new functions

σ0(T, p) = s0(T, p)

σ1(T, p) = s1(T, p) + kB

...

σg(T, p) = sg(T, p) + kB (A.19)

we can rewrite the entropy of our solution as

S =

g∑i=0

Ni σi(T, p)− kB

g∑i=1

Ni log

(Ni

N0

). (A.20)

Now we can solve for the Gibbs function, G, of our system using the total entropy (A.20),

the total volume (A.6) and the total energy (A.5).

G = U − T S + p V

G =

g∑i=0

Ni[ui(T, p)− T σi(T, p) + p vi(T, p)] + kB T

g∑i=1

Ni log

(Ni

N0

). (A.21)

let the function gi be defined as

gi(T, p) = ui(T, p)− T σi(T, p) + p vi(T, p). (A.22)

The Gibbs function then becomes

G =

g∑i=0

Ni gi(T, p) + R T

g∑i=1

Ni log

(Ni

N0

). (A.23)

Using the Gibbs function we can find the chemical potentials, µi, of the solutes simply by

taking the derivative with respect to the particular constituent, Ni we are interested in. The

chemical potential of the ith constituent of our solution is

µi(T, p,N0, · · · , Ng) =dG

dNi

= gi(T, p) + kBT log

(Ni

N0

)+ kBT. (A.24)

51

In order to find the chemical potential, µ0, of the solvent in our system we must take the

derivative of the Gibbs function with respect to N0. The chemical potential of the solvent is

µ0(T, p,N0, · · · , Ng) =dG

dN0

= g0(T, p)− kBT

g∑i=1

(Ni

N0

). (A.25)

For an explanation of the significance of the various terms in the solvent and solute expression

above see section 3.3.

52

Appendix B

Equal Chemical Potential Implies

Equilibrium1

In this appendix it is demonstrated that if a particular type of particle can flow freely

between two chambers and the two chambers are at equilibrium then the chemical potentials

for that particle type are equal between the two chambers. The equality of chemical potential

is independent of pressure and temperature differences between the two chambers as long as

these differences aren’t imposed from outside the system. This appendix proves that the AB

and CD isothermal transitions of the osmotic engine in chapter 4 are also isopotic, which

means that the piston remains at constant solvent chemical potential equal to that of the

reservoir it is connected to throughout each of those transtions.

First let us consider an arbitrary two chamber system, which exchanges solvent particles

but doesn’t exchange solute particles and is always at equilibrium for all changes in pressure

and temperature. Since the system is always at equilibrium the total change in Gibbs free

energy must be equal to zero,

∆Gµh+ ∆Gpiston = 0 (B.1)

is equal to

[Vµh∆pµh

− Sµh∆Tµh

+ µhsolvent∆nhsolvent + µhsolute∆nhsolute]

−[Vµp∆pµp − Sµp∆Tµp + µpsolvent∆npsolvent + µpsolute∆npsolute] = 0 (B.2)

now since the two chambers are freely exchanging solvent particles we can assume that the

1This proof is similar to that of Reiss in [6] p.152-153.

53

temperature of the piston is equivalent to the temperature of the reservoir throughout the

transition thus eliminating the terms containing entropy. We can also eliminate the terms

containing changes in pressure because we are not going to do work by changing the pressure

externally. After these eliminations we are left with only terms containing changes in particle

number,

µhsolvent∆nhsolvent + µhsolute∆nhsolute + µpsolvent∆npsolvent + µpsolute∆npsolute = 0. (B.3)

Now remembering the restriction that solute molecules cannot pass freely between the piston

and reservoir we can eliminate the two terms with solute particle changes. We also know

that the positive change in the particle number of the piston is accompanied by an equal

negative change in the particle number of the reservoir,

−∆npsolvent = ∆npsolvent. (B.4)

Substituting this into equation (B.3) leaves us with,

[µpsolvent − µhsolvent]∆npsolvent = 0 (B.5)

or

µpsolvent = µhsolvent. (B.6)

This result tells us that if two chambers can freely exchange solvent particles then the solvent

chemical potentials must be equal between the two chambers. It follows from this result that

throughout the transitions AB and CD the chemical potential of the solvent in the piston,

from chapter 4, is equal to the chemical potential of the solvent in the reservoir.

54

Appendix C

Osmotic Pressure1

If we have two chambers that both contain two different types of particles but can only

exchange one of those types of particles, we wish to know what the osmotic pressure is if

one chamber has a different concentration then the other. This situation is represented in

figure C.1. Chamber A has a volume of V and contains N1 and N0 particles of the type that

cannot be exchanged and the type that can be exchanged, respectively. Chamber B has a

volume of V ′ and contains N ′1 and N ′

0 particles of the type that cannot be exchanged and

the type that can be exchanged, respectively.

Figure C.1: Osmotic pressure.

Now before we proceed any further we will make some definitions that will simplify our

1The beginning of this section follows arguments about the osmotic pressure by Fermi in chapter VIIfrom [1].

55

calculation. First we will want to define the volumes of the two chambers, V and V ′, in

terms of the volume per number quantities v0(T, p) and v1(T, p) as follows,

V = N0 v0(T, pa) + N1 v1(T, pa) (C.1)

and

V ′ = N ′0 v0(T, pb) + N ′

1 v1(T, pb), (C.2)

where pa represents the pressure of chamber A and pb represents the pressure of chamber B.

We will also want to define the change in volume and the total work for a given transformation

in the following manner,

dV = v0 dN0 (C.3)

and

W = P dV = P v0 dN0. (C.4)

Finally we will want to define the total free-energy of both chamber A and B and then add

them together to obtain the total free-energy of the two chamber system. The independent

free energies for chamber A and chamber B under dilute conditions as defined by Fermi are:

FA = N0 f0 + N1 f1 + kBT

(N1 log

(N1

N0

))(C.5)

and

FA = N ′0 f0 + N ′

1 f1 + kBT

(N ′

1 log

(N ′

1

N0

))(C.6)

and now adding them together,

Ftotal = (N0 + N ′0) f0 + (N1 + N ′

1) f1 + kBT

(N ′

1 log

(N ′

1

N0

)+ N1 log

(N1

N0

)), (C.7)

where f0 and f1 are the free energies per particle for the two types of particles and are a

function of temperature.

Let us assume that N1

N0>

N ′1

N ′0

so that particles will spontaneously diffuse from chamber

A to chamber B. Now lets consider a transformation where N ′0 and N0 change by amounts

dN0 and −dN0 when particles flow from chamber A to chamber B. The variation in Ftotal

for this transformation is

dF =∂F

∂N ′0

dN0 −∂F

∂N0

dN0

dF = −kBT

(f0 −

kBT

N0

N1

)dN0 −

(f0 −

kBT

N ′0

N ′1

)dN0

dF = −kBT

(N1

N0

− N ′1

N ′0

)dN0. (C.8)

56

Since we know that this transformation is reversible the negative of this quantity must be

equal to the work. Thus,

W = P v0 dN0 = kBT

(N1

N0

− N ′1

N ′0

)dN0 (C.9)

and

Posmotic = kBT

(N1

N0v0

− N ′1

N ′0v0

)(C.10)

and assuming dilute solution conditions, N1 � N0, it follows that N0 v0 = V and corre-

spondingly, N ′0 v0 = V ′, which gives us our final result,

Posmotic = kBT

(N1

V− N ′

1

V

)= KBT (ρN1A − ρN1B). (C.11)

This result tells us that the osmotic pressure of our two chamber system is equal to the

difference in density of the solute molecules between the two chambers A and B.

57

Appendix D

Solvent Density is Constant for

Isopotic Transitions

Let P1 and P2 be the pressures of the reservoir and the piston respectively for the tran-

sitions AB and CD. Since we are considering these pressures for the two isopotic transitions

we can conclude that their difference will be equal to the osmotic pressure, expression (C.11),

for any arbitrary point along the transitions,

P1 − P2 = Posmotic. (D.1)

Now if we employ the ideal gas law solving for pressure we recognize the fact that the pressure

is proportional to density at constant temperature. The total density proportional to P1 and

P2 in this case is the sum of the solvent and solute densities for each chamber,

P1 − P2 = kBT

((Nsolvent

V

)P1

+

(Nsolute

V

)P1

)−kBT

((Nsolvent

V

)P2

+

(Nsolute

V

)P2

)= kBT

((Nsolvent

V

)P1

−(

Nsolvent

V

)P2

)+kBT

((Nsolute

V

)P1

−(

Nsolute

V

)P2

). (D.2)

Since we know that the above quantity is equal to the osmotic pressure we can clearly see

that the density of solvent in the piston must equal the density of the solvent in the reservoir

58

because the osmotic pressure is simply equal to the difference of the solute densities. So for

the AB and CD transitions we have the relationship(Nsolvent

V

)P1

=

(Nsolvent

V

)P2

. (D.3)

59

Appendix E

An Experimental Method for

Determining the Transport Rate of

TetA

The TetA (class C) Tetracycline/proton antiporter expressed from pBR322 is part of the

MFS (Multidrug Facilitator Superfamily) of antibiotic efflux pumps [13]. TetA harnesses the

free-energy available from the proton and voltage gradients across the membrane of bacterial

cells to transport the antibiotic tetracycline-Mg2+ complex out of these cells. This enables

the cells to keep the cytoplasmic concentrations of Tetracycline below 10 µM that is the

predicted inhibitory concentration, at which the number of tetracycline molecules bound

to ribosomes inside the cell is sufficient to slow the protein making-machinery [15]. There

has been a great deal of work on TetA in the last decade but as of yet there is still no

crystallographic structure for the TetA transporter. There has however been a great deal

of structural predictions and it is generally accepted that the transporter is made up of 12

transmembrane α-helixes that assemble into two 6-helix bundles that are connected by a

cytoplasmic loop [16]. There is even a prediction of the orientation of these 12 helixes from

a group that solved the structure of the oxalate transporter, OxlT, a representative member

of the MFS of transporters [17].

There has also been a great deal of work done attempting to get very accurate mea-

surements of the transport rates of these MFS transporters including the TetA tetracy-

cline/proton antiporter. One common method for measuring the transport rate of the TetA

60

efflux pump is by preparation of everted membrane vesicles containing the transporter in

their membrane. These vesicles can be placed into a solution of radioactively labeled tetra-

cycline (containing hydrogen isotopes), which is then continually pumped into the vesicles,

as in [18], [19] and [20]. This solution can then be diluted and filtered at given time intervals

to enable measurement of the filters radioactive content and thus measurement of how much

tetracycline was taken up by the vesicles in the given time interval.

The traditional method for measuring the transport rate of TetA, using radioactively

tagged tetracycline described above, has the disadvantage that you can only measure the

transport rate at particular time intervals. A more accurate measurement of the transport

rate of TetA could be achieved through fluorescent studies. It just so happens that the TetR

repressor protein when complexed with tetracycline has a distinctly different fluorescent

spectrum then when alone [21]. The difference arises because the TetR and tetracycline

complex form a FRET system (Fluorescence Resonance Energy Transfer). In this system

the tetracycline quenches a nearby tryptophan residue within TetR and emits its absorbed

energy at 520 nm [22]. This distinct property has been used to perform measurements of

the diffusion rates of tetracycline through liposome membranes [22] but never to measure

active transport.

If everted membrane vesicles were prepared with the TetR repressor contained inside

as tetracycline was transported into the vesicles it would complex with the TetR repressor

producing the characteristic emission at 520 nm. This emission could be measured as a

function of time and as a function of tetracycline concentration. These measurement could

be used to precisely calculate all of the ratios of concentrations where the protein stops

functioning. With these ratios the efficiency of the coupled transport protein could be

determined and the amount of free-energy that is consumed by the protein during one cycle.

61

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64