Physics 344 Lecture 2 Jan. 15th, 2007 1 Mon. 1/15 Wed. 1/17 Fri. 1/19

C 11.5-11.5.3, S 1.4-1.6 Heat & Work Calorimetry C 10.1-10.3.0, S 2.1- 2.3 Probabilities

HW2:S 28,32,34,36,38,41,45,47 HW3: S. 1, 5ab, 9, 10

Lab I G , HW1: S 4,12,16,18,23 Lab C.

Equipment

• Office Hours Survey • Reading Notes and Labs to return • Piston Demo

o Boiling flask with syringe or piston attached o Hot plate o Water o Beaker to put flask in

Administrative

• Juniors, look into REU’s Recap last time

o Particle Picture • Mean Free Path • Pressure

• How it relates to average energy and speed o Temperature o Ideal Gas Law

• How temperature relates to average energy and speed

• _____

..23

EKkT =

o Equipartition Theorem • A system with f accessible, quadratic degrees of freedom per

member has accessible energy kTf

NU2

=

State Variables

• The Ideal Gas Law is known as an equation of state because it relates all the variables that you tend to care about in thermodynamics – a gas’s pressure, volume, temperature, and particle count. These four variables give a complete thermodynamic description of an ideal gas, of its state of being. If any one of these changes, we say that the gas has ‘changed states.’

• There’s an analogous use of ‘state’ in kinematics. An object’s state of motion can be described by its three momentum components. If one of these changes, we say that the state of motion has changed.

This Time

• Introductory demo: Boiling flask with syringe. Put the flask in boiling water and the syringe moves.

• Question: In terms of the microscopic particle picture, how is energy getting into the ‘system’ of air in flask & syringe?

• Water molecules have a lot of kinetic energy, they bump into the atoms constituting the glass and impart some of their energy, who in turn bump into the contained air particles and transfer energy.

Physics 344 Lecture 2 Jan. 15th, 2007 2

• Question: What is changing mechanically? Syringe Plunger goes up.

While the Work-Energy relation is applicable, it’ll be convenient to rephrase it to discuss what we’re doing to this thermodynamic system.

• Energy Transfer classifications o Conservation of energy dictates that if your system’s energy content

increases or decreases, then it must have come from or gone to somewhere. But how does it cross the threshold? In thermodynamics, we broadly classify energy transfer mechanism as either work or heat.

o Heat (Q): “any spontaneous flow of energy from one object to another, caused by a difference in temperature between the objects.” § Ex. Hot water pored into a coffee cup transfers energy to the cup

via heat flow. o Work (W): Any other energy transfer mechanism.

§ Ex. Compressing a piston transfers energy to the enclosed gas via work.

o Scales: It’s often true that what we classify as “work” is performed by a macroscopically observable agent – a person pushes up hill, a paddle stirs water. In contrast, the ‘agent’ is usually not macroscopically observable for what we classify as “heat” – naked eye observation won’t tell you how energy gets from the hot coffee to the cold cup. While on the microscopic level, you can identify many individuals working; from a macroscopic level it’s convenient to classify the collective effect separately as heating.

o History: This somewhat artificial distinction can be traced to an earlier time when folks conjectured that “heat” was actually the transfer of a “caloric fluid” that flowed from high to low temperature just as electric charge flows from high to low voltage (it may be no coincidence that both heat and charge are symbolized with the same letter). Indeed, though our understanding of ene rgy transfer has evolved, we still use some of the antiquated language when we talk of heat “flowing.”

o Transfer only: Heat and Work describe the transfer of energy, they’re actions, virtually verbs. We don’t say that a system contains work or contains heat anymore than we say that a wallet contains earn.

o Units: § Joule: Energy and its transfer, both work and heat, have the same

SI units. § calorie: 1 cal = 4.186 J. The traditional unit of heat is the calorie.

Perhaps because food makers don’t want to scare us, they say “calorie” when they mean “kilocalorie.”

o Mechanisms (we’ll get more into these later) § Heating mechanism are generally placed in one of three classes.

• Conduction: transfer of energy by molecular contact: via molecular collisions, kinetic energy is generally transferred from the more energetic to the less energetic.

• Convection: transfer of energy by bulk motion: warm gas or liquid moves into the system, thus raising its average energy. Commonly, this motion is driven by the increased buoyancy of heated objects.

Physics 344 Lecture 2 Jan. 15th, 2007 3

• Radiation: light is an oscillation in electric and magnetic field, thus it mediates a force on a charged particle and imparts motion – energy.

The First Law of Thermodynamics: modified Work – Energy Relation

• Return to Demo. In terms of heat and work, how would we describe energy transfer with the air in the boiling flask?

o In terms of energy we’d say that energy is transferred in via heating, systemwaterboilingQ →. , and work is done by the system of enclosed air

plungersystemW → . The overall change in energy in the system is the difference

between what we put in and what it put out: →→ −=∆ systemsystemsystem WQU • Equation 1.24

o →→ −=∆ systemsystemsystem WQE , systemsystemsystem WQU →→ +=∆ o Like Newton’s second Law in that we have interaction terms on one side,

and resulting change in state on the left: transfer heat in, do work on the system, as a result internal energy changes. (apply net force, as a result momentum changes)

o Essentially the work-energy relation, but the interaction terms are split into two, one for the macroscopic work and one for the microscopic internal energy transfer (radiative, conductive, or convective).

o Sloppiness and Inconsistency Warnings • Different books follow different sign and naming conventions for

the work term (some call W the work done on the system and others call it the work done by the system).

• Especially given this confused state, it’s best to specify who’s doing the work on whom in some meaningful subscripts.

Ex. 2 Say in eating 0.85kg of ice, you do 800 J of work crushing it up and your body conducts 416000 J of internal energy into the ice / water. By how much has the internal energy of the ice/water changed?

JJJU

QWU

ice

iceiceyouice

416800416000800.int

.int

=+=∆

+=∆ →

1.5 Compression Work / 19.8 A Closer Look at Heat and Work / 19.10 Some

Special Cases of the First Law of Thermodynamics / 20.11 The Adiabatic Expansion of an Ideal Gas.

19.8 A Closer Look at Heat and Work

• Historically: Thermodynamics was the science of the industrial revolution. Gains in thermodynamics drove developments in industry and vice versa. With that in mind, it probably isn’t surprising that the ‘poster child’ system for talking about work and thermodynamic systems is a piston. Let’s look at one.

Physics 344 Lecture 2 Jan. 15th, 2007 4 Demo: put flask with syringe connected in boiling water, see it expand. • Q: In terms of heat flow, internal energy, and work, what happened?

o A: we heated the flask – energy flowed in. This raised the internal energy. The particles zipped around more violently and pushed on the walls harder & made the piston head move – did work on the piston head.

• Rephrasing work in terms of gas properties (P, V, N, T) o Work is Done: What we’ve just seen is a piston in action. The

relationship demonstrated here could be put to work in driving machinery – heat gas, piston head is pushed up, perhaps it pushes something else. Gas cools, and piston head comes back down. – Mechanical Work is being done.

o Work =?: § Question: What is the relationship between work, force and

displacement? We’ll start with a differentially small compression of a piston and consider the work done on it.

• ydFdW systemsystem

rr•= →→ , considering an interval of

constant force. • Note: dyFydF systemysystem →→ =• .||

rr

• Question: What is the force applied to the piston, in terms of pressure and piston-head area?

o PAFA

FP A

A =⇒≡ ||||

o If Ar

is chosen parallel to ydr

, then F||A=F||dy

o So PAdydyFydF systemysystem ==• →→ .||

rr

o cross-sectional area times height is Volume, Ay = V

o So PdVPAdy = • PdVdWsystem =→ • Summing over all differentially small compressions

(they may all be in the same direction or in different directions) from initial volume to final volume we have

o ∫=→

f

i

V

Vsystem PdVW

o Sign note: here I’ve specified work done by the system, thus the + sign. Schroeder gives no subscript but means work done on the system, and thus gives the opposite sign.

• Quasistatic o The piston’s motion must be slow compared to the

dispersion of its push through the gas, so that, at any given instant, the whole volume of gas can be characterized by a single pressure. If that is not the case, one must wonder what P represents; perhaps the average pressure?

yr

∆

Volume

Pres

sure

Volume

Pres

sure

Physics 344 Lecture 2 Jan. 15th, 2007 5 o This condition simply requires that the piston move

much slower than the speed of sound in the gas. o We say that a (relatively) slowly changing system is

quasistatic – it isn’t really static, it is changing, but it is always within internal equilibrium – energy is uniformly distributed, particles are uniformly distributed…

o Non-quasistatic § If the expansion is quite fast, and if P represents

the spatially averaged pressure, then it’s safe to assume that the pressure just behind the moving piston head is less, and thus so is the work

∫<→

f

i

V

Vsystem PdVW (non-quasistatic)

• Many Paths o Given initial state, defined by temperature, pressure,

and volume, and a final state, there are many possible routs between the two, just as there are many conceivable paths between two points on a map. However, some take more effort (greater average pressure) and some take less, so each path entails a different amount of work. (area under the Pressure – Volume curve.)

• Closed Paths -Cycles o If a thermodynamics system returns to its initial state,

we say that it has undergone a full cycle. CQ (Check point 5) Consider the four different paths between the same two initial and final states. Rank them (a) according to change in internal energy (b) work done by the gas, and (c) the magnitude of heat flow.

1.5.1 Compression of an Ideal Gas • We have a handful of state properties: V, P, T, and N for an ideal gas. We’ve

just seen how Work relates to pressure and volume. It is useful to go further in relating the work-energy relation (Law 1) to the state variables – see how they change as energy is transferred, or conversely, see how changing or not changing them effects a system’s energy. In broad strokes, we can classify processes by which state properties change and which remain the same.

sysSyssys QWU →→ +=∆

Volume

Pres

sure

f

i

V

P

1 2

3 4

Physics 344 Lecture 2 Jan. 15th, 2007 6 o Classifications

• Adiabatic: No Heating syssys WU →=∆

• Isochoric: No work / volume change sysisys QU →=∆

• Isothermal: No temperature / internal energy change syssys QW →→ +=0

• Recall kTf

NU2

=

• Isobaric: No pressure change VPPdVWWf

i

V

Vsyssys ∆−=−=−= ∫→→

We’ll look at some of these specific cases now. There are a few different results worth seeking: how much work is done, how much internal energy changes, how much heating happens, and how does pressure vary as a function of volume. Often Pressure Vs. Volume is plotted. This is useful in two ways. A) The area under the curve is the work and B) it helps us to visualize the evolution of the system.

Adiabatic Expansion of an Ideal Gas Intro: Heating can be slow. So in practice, adiabatic processes (those with negligible heating) are very fast. For example, a piston suddenly being compressed, or, as in your car quite suddenly being expanded.

• Adiabatic: Q = 0 PdVdWdU system −==⇒ →

• Ideal Gas: PV = NkTNkPV

T =⇒

• Equipartition Theorem: dTNkdU f ⋅= 2 where f is the number of degrees of freedom

• Find P vs. V

Physics 344 Lecture 2 Jan. 15th, 2007 7

o

( )( )

( )

22

22

22

22

lnln

2

2

21

2

2

2

2

2

2

++

+

+

=

=

=

+−=

+−=

+−=

+−=⋅

−=⋅+⋅⋅

−=⋅

−=

⋅=

−=⋅=

∫∫

ff

f

f

iiff

f

i

i

f

f

i

i

f

f

f

f

f

f

f

VPVP

VV

P

P

VV

P

P

ff

VdV

PdP

ff

VdV

PdP

ff

PdVPdVVdP

PdVdVPVdP

PdVPVd

PdVNkPV

dNkdU

PdVdTNkdU

o Taking the initial values as constants, then we can drop the final subscripts and solve for P as a function of V.

o 2

22

2 1+

+

= f

f

VVPP ii

o The exact shape of the plot naturally depends f, the number of degrees of freedom, but it generally looks something like this:

o Work:

• Graphically, it’s the area under the curve.

• Mathematically, 2

22

2 1+

+

= f

f

VVPP ii can be put into the work

integral which can then easily be done. o Heating: 0

Isothermal Expansion of Ideal Gas • Intro. In the ideal gas lab, for one exercise you’d slowly compressed the plunger

of a syringe and observed the corresponding change in pressure. If you

Adiabatic Expansion Pinit

Pfin

Vfin Vinit

Physics 344 Lecture 2 Jan. 15th, 2007 8 compressed it slowly enough, the air in the syringe had time to exchange with their surroundings the small bits of kinetic energy they gained by ricocheting off a moving plunger, so the distribution of internal kinetic energies remained constant and the temperature remained constant – room temperature.

• Isothermal: ∆T = 0

• Equipartition Theorem: syssys QWTkf

NU →→ =⇒=∆=∆ 02

o So, the system perfectly converts the heat input into work output, or vice versa.

• Ideal Gas: PV=NkT • P vs. V

o With both N and T constant, ( )V

NkTP1

=

o Graph • Note two things:

• The proportionality constant between P and 1/V has a factor of T, so the curve is higher or lower depending on the temperature.

• In comparison with Adiabatic processes, where P is proportional to 1/ (V(f+2)/2) and f is likely 3 or more, P proportional to 1/V is shallower.

o Question: Comparing the two curves, which process does more work for the same change in volume, isothermal or adiabatic expansion?

• Isotherm, it’s the higher curve, with greater area beneath it. Since the pressure stays higher (because temperature is not allowed to drop), the piston is pushed harder during the expansion – so more work is done.

• Work

o

( )

=−=

====

→

→ ∫∫∫

init

fininitfinsys

V

V

V

V

V

V

V

Vsys

V

VNkTVVNkTW

VNkTdVV

NkTdVV

NkTPdVW fin

init

fin

init

fin

init

fin

init

ln]ln[]ln[

]ln[1

• Heating:

o

== →→

init

finsyssys V

VNkTWQ ln

• Isobaric – Constant pressure

Isothermal Expansion Pinit

Pfin

Vfin Vinit Adiabat

Isotherm

Physics 344 Lecture 2 Jan. 15th, 2007 9 o Demo: Boiling flask in boiling water with a syringe (piston)

attached to the end. Watch the syringe plunger move. § Note: Imagine that there was no friction between the plunger

and the syringe walls and the we hold the syringe horizontal (so plunger weight doesn’t apply to the gas), then the pressure of the contained air (in the system) must be that of the room.

o Isobaric: VPPdVWP sys ∆==⇒=∆ ∫→0

o Ideal: PV=NkT

o Equipartition: Tkf

NU ∆=∆2

o P vs. V: § On a P vs. V plot, work is the area under the curve.

o Work: VPWsys ∆=→

• ClassWork Let’s run through some values. Say we took a sealed syringe from room temperature (24 C) and put it in boiling water (100 C). The plunger rises with a fairly uniform speed so that the volume changes from 10-3m3 to 1.2×10-3m3. Given the atmospheric pressure outside the flask (Patm = 105Pa) and the plunger’s weight 0.098 N and area 3×10-6m2, what is the work the gas in the syringe is doing on the plunger as the gas expands?

• VPW plungergas ∆=→ o P = Patm + Pplunger o Pplunger = wplunger / Aplunger,

• ( ) VAwPW plungplungatmplungergas ∆+=→ /

o ( )( ) JmmmNPaW plungergas 5.26100.1102.1103/098.010 3333265 =×−××+= −−−→

o Heating: Looking at the change in internal energy when pressure is held constant:

o sys

f

syssyssys

QVPTkN

QWU

→

→→

+∆−=∆

+=∆

2

§ P

TNkV

∆=∆ (by Ideal Gas Law, assuming N is constant,

knowing P is constant)

Pi=Pf

Vf V

P

0 Vi

Isobaric Expansion

Physics 344 Lecture 2 Jan. 15th, 2007 10

o

TkNQ

QTNkTkN

QWU

fsys

sysf

syssyssys

∆=

+∆−=∆

+=∆

+→

→

→→

22

2

o Question: What mechanism(s) of internal energy transfer accounts for Q in our boiling flask – syringe example? § Conduction. The hot water molecules strike the glass walls of

the boiling flask and excite those particles. They in tern hit more vigorously into the air molecules in the flask and heat them up.

CQ 2 If there is a 100 J heat flow into a gas (by radiation or conduction) and it expands but maintains a constant pressure. Can the internal energy of the gas be 200 J?

• No (not if that is the only energy transfer into the system.) As the volume expands, energy is carried out of the system – work is done. This reduces the energy left around as internal energy.

Isochoric – Constant Volume o In lab, you corked off a flask, put it in boiling water, and watched the

pressure rise, but didn’t allow for the volume to expand. o Question: Did the air in the flask do any macroscopic work?

§ No. Though it got hotter, and the particles zipped around faster and slammed into the walls harder – higher pressure, they didn’t make the walls go anywhere. Without displacement, no work is achieved, no matter how hard you push or pull.

• • This is an example of another broad category of thermodynamic processes:

Isochoric –Constant Volume. • Isochoric: ∫ ==⇒=∆ → 00 PdVWV sys

• Ideal: PV=NkT

• Equipartition: Tkf

NU ∆=∆2

• P vs. V: 0 • Work: 0 • Heating:

o

TkNQ

QTkN

QWU

fsys

sysf

syssyssys

∆=

+=∆

+=∆

→

→

→→

2

2 0

§ What heating is done to/by the system goes straight into/out of its store of internal energy.

• CQ 1. Push down on a bicycle tire pump with the nozzle closed off. Why would

it get warm? o You have done work on the system, decreased its volume (and increased

its pressure). As a result the internal energy rose. Some of that energy

Physics 344 Lecture 2 Jan. 15th, 2007 11 was transferred to the walls of the cylinder. This increased the wall temperature. Note: when the pump is decompressed, the gas energy decreases and it cools and some of the energy is transferred back from the cylinder walls, but not all – it is not a completely reversible process (2nd Law).

• CQ 5 (a) Is it possible for the temperature of a substance to rise without heat flowing into it?

• Yes. Just because energy doesn’t come in on the microscopic scale, doesn’t mean that it won’t end-up there. Ex. the afore mentioned bicycle pump and syringe. →−=∆ systemsystem WU

(b) Does the temperature of a substance necessarily have to change because heat flows into or out of it?

• No. Without imagining a specific system, you could think of an isothermal process, heat flows in, but work is done at the same time.

→→→→ =⇒+−==∆ systemsystemsystemsystem WQQWU 0int 1.6 Heat Capacity

• Intro: As I mentioned last Wednesday, Thermodaynamics was developed before folks had a clear understanding of what was going on on the fundamental, microscopic level. Consequently, folks empirically figured out the math that worked, but their way of conceptualizing it wasn’t always accurate. Heat Capacity is a great example of this. I strongly suspect that this concept was informed by the ideal of electrical capacity, i.e., capacitance. It’s defining equation looks quite similar:

o T

QC

∆≡ where C = heat capacity, Q = heat, and ∆T = change in

temperature.

o Compare that with V

QC

∆= where C is capacitance, Q is charge

and ∆V is voltage difference. o Folks once believed that, like electrical charge, there was a

substance, the “caloric fluid” that flowed from object to object in response to temperature differences. C was then some measure of how receptive an object was to storing up the fluid.

o Now we know better, now we know that it’s really energy getting passed from object to object; and we know that it can be passed either by heating or by working.

o Still, Heat Capacity is a useful tool. • Modern read

o Now a days, we simply say heat capacity is the amount of heating it takes to raise an object’s temperature by a degree. That amount varies from system to system, and even from process to process. Look back over the work we just did for the four types of expansions of an ideal gas. § Adiabatic: 00 =⇒= adiabaticCQ (not so useful) § Isothermal: ∞=⇒=∆ TCT 0 (not so useful)

Physics 344 Lecture 2 Jan. 15th, 2007 12

§ Isochoric: Nk

TU

TU

TQ

C

TNkQUV

f

VVV

f

2

20

=

∂∂

=

∆∆

=∆

=

∆==∆⇒=∆

§ Isobaric:

NkTV

PTU

TVPU

TQ

C

TNkQVPUP

f

PPPP

f

22

20

+=

∂∂

+

∂∂

=

∆∆+∆

=∆

=

∆=+∆−=∆⇒=∆

• Heat-capacity and variable degrees of freedom

o Look at our ideal-gas results, they contain f, the number of accessible degrees of freedom. As we talked about last time, this number is temperature dependent. Below about 100 K, only translational modes are accessible, f =3. Then you start being able to excite transitions in rotational states; for a dumb-bell shaped molecule like N2 or H2, that gives 2 more degrees of freedom, f = 5. Around 1000K, you start exciting vibrational transitions, that gives 2 more freedoms (changes in potential and kinetic energy) f = 7. Of course, at high enough temperatures the molecules just vibrate apart.

• Generalize -able o While you and I can determine the constant-pressure and constant-volume

heat capacities of ideal gases right now, the same general equations hold for all types of systems – gases, liquids, and solids; it’s just trickier to calculate them from first principles. We’ll actually do this for solids later. Still one can determine heat capacities experimentally, and they are tabulated for all kinds of materials (over different temperature ranges).

• (Mass) Specific Heat Capacity o Looking back at our ideal gas results, you can easily see that specific heat

scales with the amount of material in your system. Double the number of particles, and you double how much heating is necessary to raise its temperature. What’s often tabulated then is not the Heat Capacity but the mass Specific Heat Capacity, that is, the heat capacity per unit mass of a substance.

§ mC

c =

o Relating this back to heat and change in temperature:

§ TcmQT

Qcm ∆=⇒

∆=

• Simple Example: TmcQU ∆==∆ : o By how much does the internal energy of a 0.01kg Al spoon change when

it moves from room temperature (321 K) to boiling water (397 K)? Over this temperature range, Aluminum’s mass-specific heat capacity is about 900J/(kg K).

§ ( )JE

KKkgETcMEQ

KkgJ

684

32139701.0900

int

int

int

=∆

−⋅⋅=∆∆=∆≡

⋅

General Just Ideal Gas

General Just Ideal Gas

Physics 344 Lecture 2 Jan. 15th, 2007 13 CQ 15 Two identical mugs contain hot coffee from the same pot. One mug is full, while the other is only one quarter full. Sitting on the kitchen table, which mug stays warmer longer? Explain.

• The mug with more water has greater mass, and thus requires greater change in internal energy for the same change in temperature. One may suppose that it will thus take longer to cool the more massive cup of water.

Calorimetry: Transfer of internal energy

o Say you took two chunks of material, for example, an Al spoon, and a cup of hot tea (flavored water). And stuck them together.

o Question: What happens to the temperature of the spoon? § It rises.

o Question: what does this mean about the internal energy of the spoon? § It too rises: TcMEQ ∆=∆≡ int .

o Question: If the cup of water and the spoon are fairly well isolated from the rest of the world, i.e. they form an isolated system, what does conservation of energy say about the change in the spoon’s and the water’s energies?

§ Equal and opposite: waterspoon

waterspoon

EE

EEE

.int.int

.int.int 0

∆−=∆

=∆+∆=∆.

o Question: What does the equa tion look like that relates the spoon and water temperature changes?

§ waterwaterwaterspoonspoonspoon

waterspoon

TMcTMc

EE

∆−=∆

∆−=∆ .int.int

• In general, when item 1 and item 2 are brought into contact (assuming no phase change):

o 222111

2.int1.int

TMcTMcEE

∆−=∆∆−=∆

o How do the final temperatures of the two items relate? o Transfer of Kinetic Energy similar to a gas. Say you set a hot chunk of

metal touching a cold chunk. In general, the atoms along the surface of the warm chunk are jiggling a little more violently / energetically, than those along the surface of the cold chunk. This means that as the jiggle into each other, the hot ones tend to give-up energy to the cold ones. The hot chunk tends to loose energy to the cold one until they both have the same distribution of energetic particles. In this way, energy is slowly transferred from hot to cold and the two chunks are brought into thermal equilibrium, i.e., to have the same temperature.

222111

2.int1.int

TcmTcmEE

∆−=∆∆−=∆

Simple Example like 42: Say I poor 0.2 kg of boiling water into my room temperature 0.1 kg glass tea mug. Neglecting the effect of all the room temperature air around the cup, what would be the final temperature of the water and mug?

Physics 344 Lecture 2 Jan. 15th, 2007 14

)()( ....

.int.int

icfccciwfwww

cccwww

cw

TTcmTTcmTcmTcm

EE

−−=−∆−=∆

∆−=∆

Tw.f = Tc.f = Tf

CJ

kgkg

CkgCkgT

cmcmTcmTcm

T

TcmTcmTcmTcm

TTcmTTcmTcmTcm

EE

CJ

CkgJ

CkgJ

CkgJ

CkgJ

f

ccww

iwwwicccf

iwwwicccfccfww

icfcciwfww

cccwww

cw

°==⋅+⋅

°⋅+°⋅⋅=

++

=

+=+

−−=−∆−=∆

∆−=∆

°°⋅°

°⋅° 932.921

8573641862.08401.0

10041862.0248401.0

)()(

..

..

..

.int.int

19.7.3 Heats of Transformation

• Phase Change o When a solid melts or a liquid vaporizes, strong chemical bonds are

broken. When this happens, our approximation of linear expansion and our approximation that the potential energies vary linearly with particle separation breaks down in a big way. There is a huge energy difference between two atoms sitting 4 Å apart and covalently sharing an electron and two atoms sitting 4 Å apart and not sharing an electron. Gravitationally, it’s similar to the difference between having two planets side by side and having, or not having a third in between them.

o For a particular material, and for a particular transition (solid to liquid, liquid to gas) there is a calculable energy change which has nothing to do with separation, thus is not reflected in a change in temperature. § changebondEP ...∆ .

o Equation 19-16 For a whole chunk of material to change phases (break or form bonds), the total change in internal energy is

MLm

EPMEP

mM

EPNEpart

changebondchangebond

partchangebond =

∆=∆=∆=∆ .

..int

...... .

Where L is the ratio of the bond change energy to the mass of one particle. This value is very material dependent and depends on the transition. See table 12.3.

• Simple Example like 56: vapormelt mLQE ==∆ int How much energy does it take to boil a 0.2 kg of water?

JkgE

mLQE

kgJ

vapormelt

45int

int

1052.4106.222.0 ×=×⋅=∆

==∆

Physics 344 Lecture 2 Jan. 15th, 2007 15

• Example like 52: TmcmLQQU fusionwarmmelt ∆+=+=∆ How much energy is required to just melt 0.45 kg of Al at 130 °C? Tmelt = 660°C, Lmelt = 4.0×105J/kg.

• ( )JE

CCkgkgE

TmcmLQQE

CkgJ

kgJ

Alfusionwarmmelt

5int

5int

int

1095.3

13066090045.0100.445.0

×=∆

°−°⋅+×⋅=∆

∆+=+=∆

⋅°

Enthalpy = Energy in a system + energy of making room for the system, under constant pressure. I.e., it’s the energy that you expend in making it (some into the system, some into the environment.)

PVUH +≡ pdVdUdH += otherWpdVQdU +−=

together this spells otherother WQpdVWpdVQdH +=++−=

Work in terms of thermodynamic state variables

ydrExpand

Piston

System = gas in cylinder

dyFydF dysyssystem ||,→→ =•rrydFdW systemheadpistonsystemrr

•= →→ .

In terms of pressure:

PAFA

FP Asys

Asys =⇒= →→

||,||,

For A || dy:

A

PAdydyF

PAFF

dysys

Asysdysys

=⇓

==

→

→→

||,

||,||,

If A is constant while y changes:

PdVAyPdPAdy == )(So:

PdVydFdW syssys =•= →→rr

Summing over all changes in volume (needn’t be in the same direction):

∫=→

f

i

V

Vsys PdVW

Example: Consider the four different paths between the same two initial and final states. Rank them (greatest to least) (a)according to change in internal energy (b) work done by the gas, and (c) the magnitude of heat flow.

f

i

V

P

1

2

3

4

( )( )

( ) ( )

( )ff

ff

f

f

f

f

f

f

VdV

PdP

PdVPdVVdP

PdVdVPVdP

PdVPVd

PdVNkPV

dNkdU

PdVdTNkdU

2

2

2

2

2

2

2

2

1

+

+

−=

−=+

−=⋅

−=⋅+⋅⋅

−=⋅

−=

⋅=

−=⋅=Adiabatic (Q=0) P-V relation

NkPV

T =

rearrange

Group P’s and V’s

( )

( )( )

22

22

22

22

lnln

lnln 22

2

++

+

+

=

=

=

=

−=

+

+∫∫

ff

f

f

iiff

f

i

i

f

f

i

i

f

f

if

i

f

ff

VPVP

VV

P

P

VV

P

P

VV

P

PVdV

PdP

Integrate

Log Math: alnx =lnxa

Group finals and initials

Key Points from Ch 1

systemsystemsystem WQU →→ +=∆

1st Law of Thermo. (work-energy relation recast)

Expansion/Compression work(quasi-static)

∫−=→

f

i

V

Vsystem PdVW

Equipartition Theorem

kTf

NU2

=Ideal Gas Law

NkTPV =

Expansions/CompressionsAdiabatic: Isothermal:

Ideally: ff

ff

iiVPPV22 ++

=

Ideally:

=−= →→

i

fsystemsystem V

VNkTWQ ln

Isochoric:

Isobaric:

Ideally:

TNkWU

Qf

systemsystem

system

∆==∆

=

→

→

2

0

∫=−=

+=

=∆=∆⇒=∆

→→

→→

f

i

V

Vsystemsystem

systemsystem

fsystem

PdVWQ

QW

TNkUT

0

00 2

TNkUQ

PdVWV

fsystemsystem

V

Vsystem

f

i

∆=∆=

=−=⇒=∆

→

→ ∫

2

00

VPTNkQ

WUQ

VPWP

fsystem

systemsystemsystem

system

∆+∆=

−∆=

∆−=⇒=∆

→

→→

→

2

0

TNkTNkQ fsystem ∆+∆=→ 2