Free Energy in Introductory Physics Reading Reflection need help to Write a reading reflection on paper “Free Energy in Introductory Physics” attached below It should address one or more of the reading reflection prompts. Make sure it is focused and coherent (there is a a unifying theme to your reflection). The reading reflection should be around 250 words long.Reading reflection prompts: what are the author’s purpose in writing this article? What can you take from it that can be applied in your own classroom? How does the article impact your own physics content and pedagogical knowledge? Free Energy in Introductory Physics

Jeffrey J. Prentis and Michael J. Obsniuk

Citation: The Physics Teacher 54, 91 (2016); doi: 10.1119/1.4940172

View online: http://dx.doi.org/10.1119/1.4940172

View Table of Contents: http://scitation.aip.org/content/aapt/journal/tpt/54/2?ver=pdfcov

Published by the American Association of Physics Teachers

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Free Energy in Introductory Physics

Jeffrey J. Prentis, University of Michigan-Dearborn, Dearborn, MI

Michael J. Obsniuk, Michigan State University, East Lansing, MI

E

nergy and entropy are two of the most important concepts in science. For all natural processes where a system exchanges energy with its environment, the energy

of the system tends to decrease and the entropy of the system

tends to increase. Free energy is the special concept that specifies how to balance the opposing tendencies to minimize

energy and maximize entropy. There are many pedagogical

articles on energy1 and entropy.2 Here we present a simple

model to illustrate the concept of free energy and the principle

of minimum free energy.

Free energy is not a standard topic in introductory physics.

The recent guidelines established by the National Experiment

in Undergraduate Science Education (NEXUS) call for the

addition of the topic of free energy and its molecular meaning to the introductory physics curriculum for life-science

students.3 Most students encounter free energy for the first

time in a beginning biology or chemistry course, usually as a

computational tool to judge the feasibility of a chemical reaction from tables of standard data. A beginning physics course

is the natural place to provide insight into the mechanical underpinnings of energy, entropy, and free energy.

There are two kinds of free energy commonly used in thermodynamics.4 The Helmholtz free energy F of a system is constructed from the energy E and the entropy S of the system:

F ? E TS.

(1)

The function F stores complete thermodynamic information

on a system in terms of the temperature T and volume V of the

system. The Gibbs free energy, G ? E TS + pV, is ubiquitous

in chemistry, where the natural thermodynamic variables are

temperature T and pressure p. Here we focus on F, which we

will simply refer to as the free energy.

The principle of minimum free energy states that the equilibrium condition for any system in thermal contact with its

environment (energy reservoir) is

F = minimum.

(2)

Two short proofs of this principle will be given in the last section of this paper. The condition of thermal contact implies

that the system can exchange energy (not volume or matter)

with the environment in the form of heat (not work). Whenever a system in thermal contact with an energy reservoir of

constant temperature undergoes a thermodynamic process

(spontaneous change of state), the free energy of the system

will decrease until a minimum is reached, at which point

further natural change is impossible (improbable), i.e., final

equilibrium is reached.

Just like a ball sits at rest at the bottom of a hill (potentialDOI: 10.1119/1.4940172

energy well), a gas in a container is in a state of stable equilibrium at the bottom of a free energy well. A thermodynamic

system falls downhill to minimize its free energy. J. W.

Gibbs, who formulated the general theory of equilibrium in

thermodynamics, regarded the criterion for equilibrium in

Eq. (2) as an extension of the criterion employed in ordinary

statics and emphasized that the transition from the systems

considered in ordinary mechanics to thermodynamic systems

is most naturally made by this formula.5 Gibbs referred to the

F in Eq. (1) as the force-function (potential energy) for constant temperature.5

E = min (falling) vs S = max (spreading)

Entropy S has been interpreted as a spreading function

for its role in quantifying the tendency for energy to spread

through space.2 In a similar way, we can view energy E as

a falling function because of the tendency for the energy

of a non-isolated system to fall in value.6 The two opposing

forces that drive natural processesthe fall and spread of

energyare automatically built into the free energy function.

The condition E TS = min represents the exact compromise

of nature to balance the conflicting tendencies to minimize

the amount E of energy within the system and maximize the

spread S of energy throughout the system. In the E versus S

competition, the temperature T determines the winner.

Lowering T tilts the balance toward E = min while raising T

favors S = max.

To illustrate the interplay of E and S in determining the

stable equilibrium situation, consider the gas-piston system in

Fig. 1. The temperature is kept constant by placing the system

in thermal contact with an energy reservoir. In a state of equilibrium, how does the

height of the piston depend on the temperature

of the gas?

The condition that

this system (gas, piston,

Earth) is in equilibrium

can be viewed as dynamLow T

High T

ic (net force on piston is

zero) or thermodynamic

(a)

(b)

(free energy of system

is minimum). For T = 0, Fig. 1. (a) For low T, ETS = min

is approximately E = min (falling

E TS = min reduces to

dominates), which is satisfied if the

E = min, which means

piston height is low. (b) For high

the piston will be low to

T, ETS = min is approximately TS

= min or S = max (spreading domiminimize the gravitational potential energy of nates), which is satisfied if the gas

molecules have access to a large

the piston-Earth. Therspace in which to wander around

modynamically speaking, and explore many microstates.

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the piston tends to fall to increase the entropy of the reservoir.6 For example, a fall of E in the system from 5 J to 3 J is

accompanied by a spread of 2 J in the reservoir since the total

energy must be conserved. For T ? 0, the mechanical condition E = min is not the whole story. Molecular impacts (thermal agitation at finite T) push the piston up. In statistical

mechanics, this molecular force is related to the probability

to acquire volume. Gaining volume increases the number of

spatial microstates accessible to the molecules of the gas. In

short, the piston tends to fall (decrease E) and the gas tends

to spread (increase S). When the free-energy function E TS

reaches its lowest value, the mechanical force of gravity is in

perfect balance with the thermodynamic force of entropy.

The bottom of the free-energy well uniquely determines the

equilibrium height of the piston.

Gas-piston model

We introduce a simple model that contains the essential

energetic and entropic elements of a real gas-piston system.

Consider a ball (molecule) bouncing up and down at high

speed between the ends of a vertical cylinder as shown

in Fig. 2. The top end is a movable weight (piston). The ball

M

v

m

H

Fig. 2. A gas consists of a ball of

mass m moving up and down at high

speed v in a cylinder fitted with a

movable piston of mass M. The rapid

series of impacts of the light ball

against the heavy piston hold the piston up at a height H. Adding energy by

heating causes the ball to move faster

and the piston to float higher.

translates up and down, but does not rotate. There is no

friction in the system. The ball and piston are rigid bodies

whose internal state does not change. The mass of the piston

is much greater than the mass of the ball. All collisions are

elastic. During each upward motion of the ball, the ball collides with the piston. The piston is free to fall but the collision

frequency is so high that the piston remains in a suspended

state of equilibrium.

A one-particle gas can be assigned a constant temperature T if the particle is allowed to exchange energy with its

surroundings of temperature T. For this purpose, we can

imagine placing the container in Fig. 2 in thermal contact

with an energy reservoir.7 This reservoir allows us to treat the

speed v of the ball as a parameter whose constant (root mean

square) value is simply related to the temperature of the reservoir.

The basic question is: How does the equilibrium height H

of the piston depend on the speed v of the ball? The principle

of minimum free energy provides a simple solution to the

general problem of finding H as a function of v, m, and M.

92

Calculating the free energy

The goal is to find how F depends on H and then minimize

the function F(H) with respect to H. The energy and entropy

functions of the system in Fig. (2) have simple forms:

E = MgH + E0

(3)

S = k ln H + S0 .

(4)

The additive terms E0 and S0 are independent of H. These

H-independent terms are not important in the free energy

method because when the derivative with respect to H is

taken to minimize F, these constant terms will vanish.

To understand the origin of Eqs. (3) and (4), imagine the

piston to be at the height H. Let the system consist of the ball,

the piston, and the Earth. The energy of the system is E =

MgH + mgH/2 + K, where MgH is the potential energy of the

piston-Earth, mgH/2 is the average potential energy of the

ball-Earth, and K is the average kinetic energy of the ball and

piston. For M>>m, the total energy reduces to E = MgH + K.

Since K only depends on speed (temperature), the H-dependent energy function has the form given by Eq. (3).

The statistical-mechanical meaning of entropy S for any

system is given by Boltzmanns principle:2,8

S = k ln W.

(5)

In this universal formula, k is Boltzmanns constant and

W is the multiplicity function that measures the number of

microstates with a certain energy that are accessible to the

system. To fix the notions of microstate and multiplicity

for the system in Fig. 2, we subdivide the space inside the cylinder into cells, each of length l, as shown in Fig. 3.

l

H

Fig. 3. The ball occupies locations in

the container (cells in state space)

like a checker occupies squares

on a checkerboard. Entropy is a

logarithmic measure of the number

of cells (H/l ) accessible to the ball.

A microstate of the ball is defined by specifying the position

(cell location) and the velocity of the ball. The number of

possible cells that the ball can occupy is H/l. For each of these

H/l position states, there are two velocity statesthe ball can

move up or down in a cell.9 Thus the total number of microstates is W = H/l 3 2. The entropy in Eq. (5) becomes S = k ln

H + k ln (2/l), which has the form stated in Eq. (4).10

Given Eqs. (1) and (5), all free energy functions have the

form F = E kT ln W. The factor kT is the natural unit of

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thermal energy. For a one-dimensional ideal gas of molecules at temperature T, the average translational kinetic energy of one molecule is kT/2.11 Therefore, the thermal parameter kT of the gas-piston model is associated with the kinetic

energy of the ball:

kT = mv2.

(6)

Given E, S, and T in Eqs. (3), (4), and (6), we can assemble

the free energy F ? E ? TS:

F = MgH mv2 ln H + F0 .

(7)

The H-independent constant is F0 ? E0 TS0.

The form of F in Eq. (7) reveals the competition between

lowering E and raising S. To make F in Eq. (7) as small as possible, we need to make E ~ H small or S ~ ln H large. A low

H will minimize the amount E of energy (piston falls down).

A high H will maximize the spread S of matter (ball explores

larger space). For large v, the entropy term (mv2 ln H) in Eq.

(7) wins and H will be large, i.e., a fast ball implies a high

piston. For small v, the energy term (MgH) in Eq. (7) dominates F, i.e., a slow ball implies a low piston. Since T ~ v2, this

energy-entropy analysis explains why the piston is low (high)

if the temperature is low (high).

F(H)

He

H

Fig. 4. The free energy well for the gas-piston model. The

equilibrium height He of the piston is the special value of H

determined by the condition F(He) = min.

Minimizing the free energy

The free energy function F(H) in Eq. (7) is sketched in Fig.

4. This function exhibits a minimum at H = He. To find this

equilibrium value of H, we take the derivative of F(H) in Eq.

(7) and set the result equal to zero:

v

(8)

Solving Eq. (8) for He gives

(9)

Note that He is independent of the length quantum l, which

was introduced solely to digitize space and count states. In

terms of the temperature T of the surroundings, the equilibrium height is He = kT/Mg.

Suppose the mass ratio is M/m = 1000, i.e., a Ping-Pong

ball molecule (5 g) colliding with a bowling-ball piston

(5 kg). For this mass ratio, Eq. (9) gives He = v2/1000g. If

v = 50 m/s, then He = 0.25 m. If energy is added until the ball

reaches a new speed v = 100 m/s, then the piston will rise to

He = 1.0 m. If v = 50 m/s and we assume this speed is constant,

then the ball hits the piston at the rate of v/2He = 100 collisions per second. During the time 0.01 s between collisions,

the piston falls from rest a distance of 0.00049 m, which is

0.2% of the equilibrium height. For v = 100 m/s, the value

of temperature in energy units, kT = mv2 = 50 J, provides a

meaningful measure of the average kinetic energy of the ball

or the vibration energy of the wall.7 Note that the corresponding value of temperature in kelvin units, T = 3.631024 K, has

no practical meaning as a thermometer reading when dealing

with a Ping-Pong ball gas.

Force and motion interpretation of F = min

A pedagogical feature of the gas-piston model is that it can

also be analyzed using force and motion without mention of

temperature or entropy. This force-motion analysis complements the energy-entropy analysis. The downward force of

gravity on the piston is Mg. The average upward force on the

piston due to the hail of impacts (elastic collisions) of the ball

against the piston is mv2/H. This force formula, familiar from

the kinetic theory of gases,12 derives from the product of two

terms: the momentum gained (2mv) by the piston per elastic

collision times the number of collisions (v/2H) per second:

2mv 3 v/2H = mv2/H. The balance-of-force condition, Mg =

mv2/H, is identical to the minimum-of-free-energy condition

in Eq. (8). For this gas-piston model, the mechanical (Newtonian) principle net force = zero is equivalent to the thermal

(Gibbsian) principle free energy = minimum.

We can gain additional dynamical insight into the nature

of thermodynamic equilibrium by analyzing the rise-and-fall

motion of the piston after colliding with the ball. If the collisions are timed just right, then the piston will hover at a

fixed height, simply undergoing small oscillations. To hover

at H, the piston must complete one rise-and-fall motion (after being hit by the ball at H) and return to H at exactly the

same time that the ball arrives back at H after completing

one down-and-up motion in the cylinder. Thus, the hovering condition is Vt ½ gt2 = 0 (net displacement of piston is

zero), where t = 2H/v is the round-trip time of the ball, and

V = mv/M (momentum conservation) is the speed of the

piston just after colliding with the ball. The solution to the

hovering condition, mv/M ½ g(2H/v) = 0, is H = mv2/Mg,

which is identical to the value of H that solves F = min.13 In

other words, thermodynamic equilibrium for this special case

is established if the ball-piston dynamics satisfy a resonance

condition: piston period (2V/g) = ball period (2H/v).

The Physics Teacher ? Vol. 54, February 2016

93

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Ideal gas of N molecules

We can generalize the free-energy analysis of the one-ball

gas to an analysis of a gas of N balls at temperature T in a

three-dimensional container with a piston. The H-dependent

energy function, MgH, in Eq. (3) is unchanged, while the Hdependent entropy function, k ln H, in Eq. (4) generalizes to

Nk ln H.14 The free energy principle, E TS = min, is

(10)

Using the ideal gas law, pV = NkT, where p is the pressure of

the gas on the piston of area A and V = AH is the volume of

the gas, Eq. (10) can be written as

Mg ? pA = 0.

(11)

Equation (11) says that the downward force of gravity on

the piston is equal to the upward force of the molecules on

the piston. Once again, we see how balancing opposite forces

(net force = 0) is equivalent to balancing opposing tendencies

(E TS = min).

Two proofs of F = min

Thermodynamic proof: For any system (sys) in thermal

contact with its environment (env), the second law of thermodynamics states

DSsys + DSenv ? 0.

(12)

The conservation of energy states DEenv + DEsys = 0. The

entropy change of the environment (reservoir) is equal to the

energy it absorbs from the system divided by the temperature

of the environment: DSenv = ?DEsys/T. The second law in Eq.

(12) can therefore be written in the form

DEsys TD Ssys ? 0,

(13)

or simply DFsys ? 0.

The remarkable thing about Eq. (13) is that it expresses

the total entropy change of the universe (sys + env) in terms

of a property of the system alone: T(DSsys+DSenv) = ? DFsys.

The increase in the entropy of the universe is equivalent to a

decrease in the free energy of the system. The following equilibrium conditions are therefore equivalent:

Fsys = min ó Ssys + Senv = max.

P(H) = a exp(MgH/kT) 3 H.

Note that P(H) is a product of two factors: [Boltzmann factor] 3 [multiplicity factor]. The constant a is independent of

H. The Boltzmann factor exp(?E/kT) for the energy level E =

MgH determines the probability to find the piston at H when

the ball is in one particular cell. Multiplying this single-cell

probability by the total number of possible cells, W ~ H, gives

the total probability P(H) for level H. Equation (15) can be

written in the form

P(H) = a exp[F(H)/kT],

References

1.

2.

3.

4.

5.

6.

Statistical mechanical proof. Consider the ball-piston

model. In statistical mechanics,4 the probability P(H) to find

the piston at height H is

94

(16)

where F(H) ? MgH ? kT ln H. The free energy function

F(H), which first appeared in Eq. (7), now emerges as the energy exponent in a Boltzmann factor.

The concept of free energy is usually applied to a macroscopic system. The statistical analysis presented here shows

how F = min applies to a one-particle system in thermal

contact with a reservoir. According to Eq. (16), the condition F(H) = min is equivalent to P…

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