Goldfarb D. Biophysics DeMYSTiFied

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62 Biophysics De mystifieD

Comparing this with Eq. (4-2),

DU 5 q 2w

we see that Eqs. (4-8) and (4-2) are equivalent if

q = DH 2 V DP

This gives us some insight into what enthalpy is; it’s simply a way of dividing

up heat into two different terms: one term is the energy associated with a

change in pressure (V DP) and the other term is the change in enthalpy (DH).

Rearranging, we get

DH 5 q 1 V DP (4-9)

Enthalpy is just the heat flowing into the system plus the energy associated

with a change in pressure.

Fortunately, almost all biological processes occur at constant pressure. At con-

stant pressure, DP 5 0. This means that for a system at constant pressure, the

enthalpy change is exactly equal to the heat flowing into the system: q 5 DH.

still struggling

What is pressure times volume (pV)? how is V DP the energy associated with a

change in pressure? often we can understand a concept better by looking at the

units of the quantities involved. The units of P are force per area (n/m

). if we

multiply these units by meters per meter (which does not change the value at

all because meters per meter is always equal to 1) we see that the units of P are

nm/m

. But a nm is a joule, so the units of P are actually J/m

or energy per unit

volume. pressure is energy per unit volume, a kind of energy density. When we

multiple the pressure (J/m

) by the volume (m

), the units on PV are joules; PV is

the potential energy associated with pressure, and V DP is the change in this

energy due to a change in pressure.

PROBLEM 4-2

Choose example numbers to show that Eq. (4-6) gives correct results.

PROBLEM

Choose example numbers to show that Eq. (4-6) gives correct results.

PROBLEM

Choose example numbers to show that Eq. (4-6) gives correct results.

chapter 4 EnErgy and LifE 63

SOLUTION

The equation is D(PV) 5 P DV  V DP

The trick in solving this problem is knowing what value to use for the

constant pressure or constant volume on the right side of the equation.

Since both pressure and volume are changing, the appropriate value to use

for the constant pressure and constant volume is the average value of the

pressure or volume. Here we will make an assumption that the average

value is equal to the midway point between the initial and final values. This

assumption holds true under certain conditions, for example, if the pres-

sure and volume change at a constant rate.

If we write each of the changes (denoted with D, the Greek letter delta)

in terms of the difference between the final values and the initial values,

we have

D(PV) 5 P

2 P

P DV 5 P(V

2 V

)

V DP 5 V(P

2 P

)

Now, to show Eq. (4-6) gives correct results, let’s choose some example

numbers. After reading this solution, you should try it yourself with your

own numbers. Any numbers will do, but remember that, while the change

in pressure or volume can be negative (meaning the pressure or volume

decreased), each value of pressure or volume must be positive.

For our example, let’s say the pressure went from 10 pascals (N/m

) to 7

pascals, while the volume went from 13 m

to 19 m

Substituting these numbers into the left side of Eq. (4-6), we get

D(PV) 5 P

2 P

5 (7  19) 2 (10  13) 5 133 2 130 5 3 Nm 5 3 J

And substituting the same numbers into the right side, we get

P DV 5 P

avg

2 V

) 5 [(10  7)/2](19 2 13) 5 8.5(6) 5 51 J

V DP 5 V

avg

2 P

) 5 [(13  19)/2](7 2 10) 5 16(23) 5 248 J

Then

P DV  V DP 5 51 J 2 48 J 5 3 J

which is the same as D(PV).

SOLUTION

The equation is

✔

64 Biophysics DemystifieD

PROBLEM 4-3

Using the definition of pressure (force per unit area) and the definition of

work (force times distance), show that P DV from Eq. (4-7) is just the three-

dimensional work done by the system.

SOLUTION

Work, by definition, is force times distance moved:

w 5 F Dx

where Dx is the distance moved (change in position) along the same direc-

tion the force was applied.

Pressure, by definition, is force per unit area, or P 5 F/A.

To simplify the problem, consider a cylinder closed at one end, with a

piston moving up or down inside the cylinder from the other end; see

Fig. 4-1. The volume inside the cylinder (between the piston and the

PROBLEM

Using the definition of pressure (force per unit area) and the definition of

work (force times distance), show that

PROBLEM

Using the definition of pressure (force per unit area) and the definition of

SOLUTION

Work, by definition, is force times distance moved:

✔

FigurE 4-1 • A piston in a cylinder

allows us to simplify the case of work

associated with pressure and a vol-

ume change.

chapter 4 EnErgy and LifE 65

closed end) changes as the piston moves up and down. The surface area of

the piston is constant, so the volume change is just the area times the dis-

tance the piston is moved.

DV 5 A Dx

So for the cylinder, we have

P∆∆ ∆V

Ax55()

The A are canceled, leaving

P DV 5 F Dx 5 w

In the case where work is done in multiple directions, we can imagine the work

done in each direction as analogous to a piston in a cylinder aligned with that direc-

tion. Take, for example, the work done in compressing or expanding a balloon. We

can imagine that at each infinitesimally small part of the balloon’s surface there is

a cylinder of constant area and the expansion or contraction of the balloon is

equivalent to moving a piston up or down inside the cylinder. The total change in

volume for the balloon is then the sum of all of these individual volume changes

given by DV = A Dx. Similarly the pressure against the surface of the balloon can

be broken down, at each infinitesimally small part of the balloon’s surface, as a force

in a direction perpendicular to the balloon’s surface at that point, divided by the

area of the infinitesimal piston for which we are considering the volume change.

The volume change for any three-dimensional shape can, in this way, be broken

down into a sum of many simple volume changes, each of which can be expressed

as a constant area times a distance. And the pressure that causes that volume change

is just, at each point on the surface, the force divided by that small constant area.

So we see that a constant pressure times a change in volume P DV is just the

three-dimensional equivalent of force times distance, or work.

Ax Fx

DDD

==()

(4-10)

Entropy

The idea of entropy was developed by Rudolf Clausius in the mid-1800s. It was

well known at the time that heat engines of the day converted only a small

portion of their heat into useful work. Typically only about 1% of the heat was

66 Biophysics DemystifieD

transformed. The rest of the heat was lost, dissipated, merely warming up the

surroundings.

Clausius’s goal was to account quantitatively for this loss. He originally

called this lost energy the “equivalence value of all uncompensated transfor-

mations,” meaning that it was equivalent to the portion of the original heat

that could not be compensated for by transforming the resulting work back

into heat.

A few years later he coined the term entropy. Clausius himself said he

took the Greek word trope meaning “transformation” and put it in a form

intended to resemble the word energy. So, from trope and energy he got

entropy.

The change in entropy as the result of a process is defined as

DS 5 Q/T (4-11)

where S is the entropy and Q is the heat irreversibly lost to the surroundings at

absolute temperature T. Notice that the units on entropy are energy units

divided by temperature units, for example, joules per degree celsius.

If a process or transformation involves passing the heat through a working

body (for example, a heat engine) so that the heat is passing from a place at

temperature T

to a place at temperature T

, then the entropy change is

defined as

=−













(4-12)

Although Clausius explained how he came up with the term entropy, he

never indicated why he chose the letter S for entropy. Some have speculated

that it was in honor of another famous early researcher of thermodynamics,

Sadi Carnot. Figure 4-2 shows a diagram by Sadi Carnot illustrating the flow of

heat and work in a heat engine.

Working

body

HOT

COLD

FigurE 4-2 • Diagram of heat engine by Sadi Carnot from

the year 1824.

chapter 4 EnErgy and LifE 67

PROBLEM 4-4

A heat engine takes 400 J of heat and converts it into 3 J of work. Assuming

that reversing the process will convert all of the work back into heat, how

much of the original heat is not compensated for by reversing the process?

What is the entropy change for this process at a temperature of 298 K?

SOLUTION

The amount of heat not compensated for by reversing the process is

Uncompensated heat 5 original heat 2 work back into heat

The original heat is 400 J. Assuming reversing the process converts all

the work back into heat, the heat compensated for by converting the work

back into heat is 3 J. Therefore the uncompensated heat is

Uncompensated heat 5 400 J 2 3 J 5 397 J

The entropy change, by definition, is

DS 5 Q/T 5 397 J/298 K 5 1.33 J/K

PROBLEM 4-5

If the heat engine from Prob. 4-4 transfers heat from a body at 350 K to a

body at 300 K, what is the entropy change for this process?

SOLUTION

From Prob. 4-5, the amount of heat lost is 397 J. The entropy change in

going from 350 K to 300 K is then

S525397

300

350

0 189J

J/K

(

)













Entropy Is Related to the Number of Ways of Doing Something

It is worth mentioning here that entropy is related to order and disorder. You

may have already heard this concept. Although Clausius originally defined

entropy as a simple thermodynamic property (related to energy and work),

research over the years has shown that entropy is actually a measure of disorder

in a system.

PROBLEM

A heat engine takes 400 J of heat and converts it into 3 J of work. Assuming

that reversing the process will convert all of the work back into heat, how

PROBLEM

A heat engine takes 400 J of heat and converts it into 3 J of work. Assuming

SOLUTION

The amount of heat not compensated for by reversing the process is

✔

PROBLEM

If the heat engine from Prob. 4-4 transfers heat from a body at 350 K to a

body at 300 K, what is the entropy change for this process?

PROBLEM

If the heat engine from Prob. 4-4 transfers heat from a body at 350 K to a

SOLUTION

From Prob. 4-5, the amount of heat lost is 397 J. The entropy change in

✔

68 Biophysics DemystifieD

But what do we mean by disorder? For now, let’s just say that order and

disorder in physics are not, as common sense might imply, a matter of neat

versus messy. Rather, order and disorder relate directly to the number of different

ways of accomplishing one particular thing. The more ways there are for a system

to do one particular thing or to be in one particular state, the more disorder and

the more entropy the system has.

For example, when there are only a few ways for the molecules of a system to

arrange themselves to achieve a particular temperature, then we say that the system

is highly ordered. But when there are many different ways for the same molecules

to achieve a given temperature, then we say that the system is disordered.

gibbs Energy

About 10 years after Clausius coined the term entropy, Josiah Willard Gibbs

used Clausius’s concept of entropy to define available energy, energy available

to do useful work. Later Gibbs renamed it free energy, meaning the energy that

is free to do useful work as compared with energy that is bound to be lost

through dissipation as entropy. This latter way of thinking reflected the increas-

ing realization that some heat energy is inevitably lost and simply can never be

converted into work no matter how hard one tries. This idea that it is impos-

sible to convert all of the energy into work without some loss gave rise to the

second law of thermodynamics.

The free energy as Gibbs defined it is now called Gibbs free energy, or simply

Gibbs energy, or the Gibbs function. It is defined as

G 5 H 2 TS (4-13)

where H is enthalpy, T is the absolute temperature, and S is the entropy. You

can see clearly from this definition that Gibbs energy is what remains of the

enthalpy after subtracting out the energy loss due to entropy.

Gibbs Energy Tells Us Something about

the Spontaneity of a Process

The Gibbs energy change for a process is defined as

DG 5 DH 2 T DS (4-14)

It didn’t take long for scientists to realize that if we measure the Gibbs

energy change for a process that happens spontaneously, then the Gibbs energy

chapter 4 EnErgy and LifE 69

of the final state is always less than that of the initial state. That is, for a process

that happens spontaneously (i.e., on its own, with no outside push), Gibbs

energy always decreases; DG is negative.

This was in fact what Gibbs was striving for when he proposed a way to

measure the available energy. Gibbs knew well, from the branch of physics

called mechanics, that a force is what moves a physical object. He also knew

that there is a simple mathematical relation between force and movement

(F 5 ma). Physical chemists at that time were looking for a property analogous

to force which could be measured and which could be shown to be the force

that drives chemical reactions. Gibbs found it. Just as the force of gravity causes

a ball to roll down hill, so too, for any process that takes a system from state A

to state B, if state B has less Gibbs energy than state A, then the process will

proceed spontaneously, rolling down the Gibbs energy hill from A to B.

The Second Law of Thermodynamics

No review of thermodynamics would be complete without some discussion of

the second law of thermodynamics. In all likelihood you’ve heard of the second

law and learned something about it. And you probably know that it has

something to do with entropy increasing. Perhaps you even know there are

many ways to state the second law of thermodynamics, and as different as they

all may sound, they are all various views of the same thing. For example,

The universe is winding down.

•

Heat spontaneously flows from a place of higher temperature to a place •

of lower temperature.

When converting energy into work, some of the energy is always lost to

•

heat.

In any process, the entropy of a closed system will either increase or

•

remain constant.

In any process, the entropy of the open system and its surroundings, taken

•

together, will either increase or remain constant.

In any process, a system taken together with its surroundings will always

•

go from a more ordered state to a less ordered state.

Some statements of the second law are very precise. Others, such as “The uni-

verse is winding down,” can be ambiguous. Care must be taken not to misinterpret

70 Biophysics DemystifieD

their meaning. For example, it is not scientifically correct to make statements like

“His dorm room is proof of the second law of thermodynamics.”

When a living thing develops from an undifferentiated mass of cells to a fully

grown plant or animal, it is clear that it has gone from a less ordered state to a

more ordered state. Because of this fact and the fact that living things continue

to propagate their order by having offspring, scientists have speculated that

living things may somehow violate the second law. Or, at least, they considered

it a possibility there are physical laws we have yet to discover that are apparent

only in the case of living matter.

But living things are open systems. When taken together with their surround-

ings, the impact of the living is to increase the disorder of the environment to a

greater extent than they increase their own order. Overall the result is an increase

in disorder and the second law of thermodynamics is not violated.

The gibbs Function as the Driver of Biophysical Processes

For the most part (and this may strike you as odd) biophysics is not concerned

with the second law of thermodynamics, at least not as stated. What concerns

us more is the Gibbs function. The Gibbs function effectively has the second

law of thermodynamics built into it, so by focusing on the Gibbs function alone,

everything is taken care of.

In this context we can make the following statement, which is akin to other state-

ments of the second law: All else being equal, in any process, an increase in entropy

makes a favorable contribution to the Gibbs energy change, whereas a decrease in

entropy makes an unfavorable contribution to the Gibbs energy change. By favorable

and unfavorable we mean whether or not the process is driven forward.

So forget about the second law of thermodynamics. Well, don’t actually for-

get about it, but focus on biophysical processes in terms of the Gibbs function

being the force that drives the process.

DG 5 DH 2 T DS

We can see from the Gibbs function that there are three things that contrib-

ute to negative DG and thus to rolling down the Gibbs energy hill and to driv-

ing a process forward.

1. Releasing energy. When a system goes from a higher-energy state to a lower-

energy state, it releases energy; its enthalpy decreases. DH is negative, and

a negative DH contributes toward a decrease in Gibbs energy.

chapter 4 EnErgy and LifE 71

2. Increasing disorder. When a system goes from more ordered to less ordered,

its entropy increases, DS is positive, and so, all else being equal, its Gibbs

energy decreases.

3. Changing temperature. Note that the effect of temperature depends on

whether the entropy change is positive or negative. For a positive entropy

change, increasing temperature contributes to a favorable Gibbs energy

change.

Some processes are driven forward by simultaneously releasing energy and

increasing disorder. For other processes it may be the case that either the

enthalpy change or the entropy change are unfavorable, yet the process is still

driven forward because the overall Gibbs energy change is favorable. In such a

case if the process is driven forward primarily by releasing energy, then the

process is said to be enthalpy driven. On the other hand if the process is driven

forward primarily by increasing disorder, then the process is said to be entropy

driven.