Thermal Physics: 425/625

• Acknowledgement: I acknowledge the excellent efforts by Rutgers Physics Professors Misha Gershenson and Weida Wu in preparing powerpoint course material for this textbook. The powerpoint presentations in this course are rooted in their work, with permission from Dr Wu, with modifications by me.

Physics 425/625     Thermal Physics      Fall 2011

Professor:    Pat Arnott

Office:         Leifson Physics RM 213

e-mail:          patarnott@gmail.com

phone:    775-784-6834

office hour:  Wed 1:00-3:00 PM and by appointment

Class:       Tuesday and Thursday 11:00 am to 12:15 pm

Prerequisites: Physics 182, 301.

Physics 425/625     Thermal Physics      Fall 2011

 Textbook: An Introduction to Thermal Physics,

D.V. Schroeder, Addison-Wesley-Longman, 2000 

Web Site Access: http://www.patarnott.com/phys625

Homework assignments, announcements and notes will be posted here. The link must be visited regularly.

Thermal Physics = Thermodynamics + Statistical Mechanics

- conceptually, the most important, enjoyable subject of the undergraduate physics program.

Thermodynamics provides a framework of relating the macroscopic properties of a system to one another. It is concerned only with macroscopic quantities and ignores the microscopic variables that characterize individual molecules (both strength and weakness).

Statistical Mechanics is the bridge between the microscopic and macroscopic worlds: it links the laws of thermodynamics to the statistical behavior of molecules.

Macroscopic Description is Qualitatively Different!

Why do we need to consider macroscopic bodies as a special class of physical objects?

For a single particle: all equations of classical mechanics, electromagnetism, and quantum mechanics are time-reversal invariant (Newton’s second law, F = dp/dt, looks the same if the time t is replaced by –t and the momentum p by –p).

For macroscopic objects: the processes are often irreversible (a time-reversed version of such a process never seems to occur). Examples: (a) living things grow old and die, but never get younger, (b) if we drop a basketball onto a floor, it will bounce several times and eventually come to rest - the arrow of time does exist.

“More is different”, Phil Anderson, Science, 177, 393 (1972)

Ancients struggled with heat: What is it?

http://www.infinite-energy.com/iemagazine/issue37/mysteries.html

“It is astonishing to realize that many modern conceptions (or “laws”) in the science of heat— thermodynamics— arose during the nineteenth century, a period of utter confusion about the fundamental nature of heat. How could it have been otherwise, given that the very existence of atoms was still in question! “

The Main Idea of the Course

Statistical descriptionof a large system

of identical (mostly, non-interacting) particles

Irreversibility of macro processes,

the 2nd Law of Thermodynamics

all microstates of an isolated system occur with the same probability, the concepts of multiplicity (configuration

space), Entropy

Equation of state for macrosystems

(how macroparameters of the system and the temperature

are interrelated)

Thermodynamic Systems, Macroscopic Parameters

Open systems can exchange both matter and energy with the environment.

Closed systems exchange energy but not matter with the environment.

Isolated systems can exchange neither energy nor matter with the environment.

Internal and external macroscopic parameters: temperature, volume, pressure, energy, electromagnetic fields, etc. (average values, fluctuations are ignored).

No matter what is the initial state of an isolated system, eventually it will reach the state of thermodynamic equilibrium (no macroscopic processes, only microscopic motion of molecules).

A very important macro-parameter: Temperature

Temperature is a property associated with random motion of many particles.

Introduction of the concept of temperature in thermodynamics is based on the the zeroth law of thermodynamics:

A well-defined quantity called temperature exists such that two systems will be in thermal equilibrium if both have the same temperature.

Temperature Measurement

Properties of a thermoscope (any device that quantifies temperature):

1. It should be based on an easily measured macroscopic quantity a (volume, resistance, etc.) of a common macroscopic system.

2. The function that relates the chosen parameter with temperature, T = f(a), should be monotonic.

3. The quantity should be measurable over as wide a range of T as possible.

The simplest case – linear dependence T = Aa (e.g., for the ideal gas thermometer, T = PV/NkB).

Thermometer a thermoscope calibrated to a standard temp. scale

T

a

the ideal gas thermometer, T = PV/NkB

the resistance thermometer with a semi- conductor sensor, TR exp

The Absolute (Kelvin) Temperature Scale

The absolute (Kelvin) temperature scale is based on fixing T of the triple point for water (a specific T =

273.16 K and P = 611.73 Pa where water can coexist in the solid, liquid, and gas phases in equilibrium).

TPP

PKT 16.273

- for an ideal gas constant-volume

thermoscope

absolute zero

T,K

PPTP

273.16

PTP – the pressure of the gas in a constant-volume gas thermoscope

at T = 273.16 K

0

Our first model of a many-particle system: the Ideal Gas

The ideal gas model - works well at low densities (diluted gases)

• all the molecules are identical, N is huge;

• the molecules are tiny compared to their average separation (point masses);

• the molecules do not interact with each other;

• the molecules obey Newton’s laws of motion, their motion is random;

• collisions between the molecules and the container walls are elastic.

Models of matter: gas models (random motion of particles)

lattice models (positions of particles are fixed)

Air at normal conditions:

~ 2.71019 molecules in 1 cm3 of air (Pr. 1.10)

Size of the molecules ~ (2-3)10-10 m, distance

between the molecules ~ 310-9 m

The average speed - 500 m/s

The mean free path - 10-7 m (0.1 micron)

The number of collisions in 1 second - 5 109

The Equation of State of Ideal Gases

P – pressure [Newtons/m2]

V – volume [m3]

n – number of moles of gas [mol]

T – the temperature in Kelvins [K]

R – a universal constant

nRTPV

8.315J

Rmol K

The ideal gas equation of state:

An equation of state - an equation that relates macroscopic variables (e.g., P, V, and T) for a given substance in thermodynamic equilibrium.

In equilibrium ( no macroscopic motion), just a few macroscopic parameters are required to describe the state of a system.

The Ideal Gas Law

isotherms

In terms of the total number of molecules, N = nNA

TNkPV Bthe Boltzmann constant kB = R/NA 1.3810-23 J/K

(introduced by Planck in 1899)

Avogadro’s Law: equal volumes of different gases at the same P and T contain the same amount of molecules.

The P-V diagram – the projection of the surface of the equation of state onto the P-V plane.

NA 6.022045×1023

Avogadro’s number

The equations of state cannot be derived within the frame of thermodynamics: they can be either considered as experimental observations, or “borrowed” from statistical mechanics.

Connection between KEtr and T for Ideal Gases

T of an ideal gas the kinetic energy of molecules

Pressure – the result of collisions between the molecules and walls of the container.

Strategy: Pressure = Force/Area = [p / t ]/Area

vx

Piston area A

L

Volume = LA

For each (elastic) collision: px = 2 m vx

Intervals between collisions: t = 2 L/vx

Momentum

Vmv

Vvp

AvL

mvP xxx

x

xi

111

/2

2 2

22x

N

ix vmNmvPV For N molecules -

?

no-relativisticmotion

Connection between KEtr and T for Ideal Gases (cont.)

22x

N

ix vmNmvPV

TNkPV B

22222

2

3

2

1

2

1xzyxtr vmvvvmvmKE

Tkvm Bx 2

TkKE Btr 2

3 - the temperature of a gas is a direct measure of the

average translational kinetic energy of its molecules!

Average kinetic energy of the translational motion of molecules:

TNkKEU Btr 2

3

The internal energy U of a monatomic ideal gas is independent of its volume, and depends only on T (U =0 for an isothermal process, T=const).

UPV3

2 - for an ideal gas of non-relativistic particles,

kin. energy (velocity)2 .

Comparison with Experiment

TNkU B2

3

- for a point mass with three degrees

of freedom

dU/dT(300K) (J/K·mole)

Monatomic

Helium 12.5

Argon 12.5

Neon 12.7

Krypton 12.3

Diatomic

H2 20.4

N2 20.8

O2 21.1

CO 21

Polyatomic

H20 27.0

CO2 28.5

Testable prediction: if we put a known dU into a sample of gas, and measure the resulting change dT, we expect to get

moleJ/K5.12

J/K1038.1mole1062

32

3

231-23

BNkdT

dU

Conclusion: diatomic and polyatomic gases can store thermal energy in forms other than the translational kinetic energy of the molecules.

Degrees of Freedom

The degrees of freedom of a system are a collection of independent variables required to characterize the system.

Diatomic molecules: 3 + 2 = 5 transl.+rotat. degrees of freedom

Polyatomic molecules: 6 (transl.+rotat.) degrees of freedom

Degrees of Freedom (cont.)

Plus all vibrational degrees of freedom. The one-dimensional vibrational motion counts as two degrees of freedom (kin. + pot. energies):

22

2

1

2

1xkxmxUK

For a diatomic molecule (e.g., H2), 5 transl.+rotat. degrees of freedom plus 2 vibrational degrees of freedom = total 7 degrees of freedom

Among 7 degrees of freedom, only 3 (translational) degrees correspond to a continuous energy spectrum (classical) , the other 4 – to a discrete energy

spectrum (Quantum).

x

U(x)

E1

E2

E3

E4

x

U(x)

-3

-2

-1

0

1

2

3

4

1.5 2.0 2.5 3.0 3.5 4.0

distance

Ene

rgy

Approx.

“Frozen” degrees of freedom

For an ideal gas

PV = NkBT

U = f/2 NkBT

U /kBT

3/2N

5/2N

7/2N

10 100 1000 T, K

Translation

Rotation

Vibration

one mole of H2

The rotational energy levels are ~15 meV apart, the difference between vibrational energy levels ~270 meV. Thus, the rotational degrees start contributing to U at T > 200 K, the vibrational degrees of freedom - at T > 3000 K.

Example of H2:

An energy available to a H2 molecule colliding with a wall at T=300 K: 3/2 kBT ~ 40 meV. If the difference between energy levels is >> kBT, then a typical collision cannot cause transitions to the higher (excited) states and thus cannot transfer energy to this degree of freedom: it is “frozen out”.

x

U(x)

E1

E2

E3

E4

kBT

Equipartition of Energy

vx

Piston – a mechanical system with one degree of freedom. Thus,

TkuMvm

Bx

2

1

22

22

M – the mass of a piston, u2 the average u2, where u is the piston’s speed along the x-axis.

Thus, the energy that corresponds to the one-dimensional translational motion of a macroscopic system is the same as for a molecule (in this respect, a macrosystem behaves as a giant “molecule”).

Equipartition Theorem: At temperature T, the average energy of any “quadratic” degree of freedom is 1/2kBT.

“Quadratic” degree of freedom – the corresponding energy = f(x2, vx2)

[ translational motion, (classical) rotational and vibrational motion, etc. ]

- holds only for a system of particles whose kinetic energy is a quadratic form of x2, vx

2 (e.g., the equipartition theorem does not work for photons, E = cp)

Questions to solve in class: Think about them beforehand.

a. What are the molecules in the air in this room? Give % estimates.

b. How many molecules are in the air in this room?

d. What is the total internal energy of the air in this room?

c. On average, how far apart are the molecules from each other?

Questions Continued

e. Assuming (bad assumption) that we could use all this energy to run a hair dryer (1 kW), how long could we run the air dryer on energy?

f. Assuming the walls, floor, and ceiling of the room are perfect blackbodies, how much power is emitted by them all?

Questions Continued

g. Use the idea of the deBroglie wavelength and the average kineticenergy to calculate the thermal deBroglie wavelength for the molecules in this room as a function of temperature. How does the thermal deBroglie wavelength compare with the inter particle spacingat room temperature? As T goes to zero? Interpret.

http://en.wikipedia.org/wiki/Thermal_de_Broglie_wavelength

http://en.wikipedia.org/wiki/Matter_wave

The root-mean-square speed

m

Tkvv B

rms

32 - not quite the average speed, but close...

For H2 molecules (m ~21.710-27 kg ) at 300K: vrms~ 1.84 103 m/s

For N2 – vrms (Pr. 1.18), for O2 – vrms= 461 m/s

v

D(v)

vrms

This speed is close to the speed of sound in the gas – the sound wave propagates due to the thermal motion of molecules.

Problem 1.16 The “exponential” atmosphere

Consider a horizontal slab of air whose thickness is dz. If this slab is at rest, the pressure holding it up from below must balance both the pressure from above and the weight of the slab. Use this fact to find an expression for the variation of pressure with altitude, in terms of the density of air, . Assume that the temperature of atmosphere is independent of height, the average molecular mass m.

z

z+dzP(z+dz)A

P(z)A Mg

area A AzPMgAdzzP )()(

A

MgzPdzzP )()(

AdzM gdz

dP

the density of air: Tk

Pm

Tk

PVN

V

Nm

V

M

BB

PkT

mg

dz

dP

kT

mgzPzP exp)0()(Assuming T is independent of z

The First Law of Thermodynamics (Ch.1)

Outline:

1. Internal Energy, Work, Heating

2. Energy Conservation – the First Law

3. Quasi-static processes

4. Enthalpy

5. Heat Capacity

Internal Energy

system U = kinetic + potential

system boundary

“environment”

The internal energy of a system of particles, U, is the sum of the kinetic energy in the reference frame in which the center of mass is at rest and the potential energy arising from the forces of the particles on each other.

Difference between the total energy and the internal energy?

The internal energy is a state function – it depends only on the values of macroparameters (the state of a system), not on the method of preparation of this state (the “path” in the macroparameter space is irrelevant).

U = U (V, T)In equilibrium [ f (P,V,T)=0 ] :

U depends on the kinetic energy of particles in a system and an average inter-particle distance (~ V-1/3) – interactions.

P

V T A

B

For an ideal gas (no interactions) : U = U (T) - “pure” kinetic

Internal Energy of an Ideal Gas

The internal energy of an ideal gas with f degrees of freedom: TNk

fU B2

f 3 (monatomic), 5 (diatomic), 6 (polyatomic)

How does the internal energy of air in this (not-air-tight) room change with T if the external P = const?

PVf

Tk

PVNTkN

fU

BroominBroomin 22

(here we consider only trans.+rotat. degrees of freedom, and neglect the vibrational ones that can be excited at very high temperatures)

- does not change at all, an increase of the kinetic energy of individual molecules with T is compensated by a decrease of their number.

Work and Heating (“Heat”)

We are often interested in U , not U. U is due to:

• Q - energy flow between a system and its environment due to T across a boundary and a finite thermal conductivity of the boundary

– heating (Q > 0) /cooling (Q < 0)(there is no such physical quantity as “heat”; to emphasize this fact, it is better to use the term “heating” rather than “heat”)

• W - any other kind of energy transfer across boundary - work

Heating/cooling processes:conduction: the energy transfer by molecular contact – fast-moving molecules transfer energy to slow-moving molecules by collisions;

convection: by macroscopic motion of gas or liquid

radiation: by emission/absorption of electromagnetic radiation.

HEATING

WORK

Work and Heating are both defined to describe energy transfer across a system boundary.

The First Law

For a cyclic process (Ui = Uf) Q = - W.

If, in addition, Q = 0 then W = 0

The first law of thermodynamics: the internal energy of a system can be

changed by doing work on it or by heating/cooling it.

U = Q + W conservation of energy.

P

V T

An equivalent formulation:

Perpetual motion machines of the first type do not exist.

Sign convention: we consider Q and W to be positive if energy flows into the system.

Quasi-Static Processes

Quasi-static (quasi-equilibrium) processes – sufficiently slow processes, any intermediate state can be considered as an equilibrium state (the macroparamers are well-defined for all intermediate states).

Examples of quasi-equilibrium processes:

isochoric: V = const

isobaric: P = const

isothermal: T = const

adiabatic: Q = 0

For quasi-equilibrium processes, P, V, T are well-defined – the “path” between two states is a continuous lines in the P, V, T space.

P

V T

1

2

Advantage: the state of a system that participates in a quasi-equilibrium process can be described with the same (small) number of macro parameters as for a system in equilibrium (e.g., for an ideal gas in quasi-equilibrium processes, this could be T and P). By contrast, for non-equilibrium processes (e.g. turbulent flow of gas), we need a huge number of macro parameters.

Work

The sign: if the volume is decreased, W is positive (by compressing gas, we increase its internal energy); if the volume is increased, W is negative (the gas decreases its internal energy by doing some work on the environment).

2

1

),(21

V

VdVVTPW

The work done by an external force on a gas enclosed within a cylinder fitted with a piston:

W = (PA) dx = P (Adx) = - PdV

x

P

W = - PdV - applies to any shape of system boundary

The work is not necessarily associated with the volume changes – e.g., in the Joule’s experiments on determining the “mechanical equivalent of heat”, the system (water) was heated by stirring.

dU = Q – PdV

A – the piston area

force

W and Q are not State Functions

P

V

P2

P1

V1 V2

A B

CD

- the work is negative for the “clockwise” cycle; if the cyclic process were carried out in the reverse order (counterclockwise), the net work done on the gas would be positive.

01212

211122

VVPP

VVPVVPWWW CDABnet

2

1

),(21

V

VdVVTPW

- we can bring the system from state 1 to state 2 along infinite # of paths, and for each path P(T,V) will be different.

U is a state function, W - is not thus, Q is not a state function either.

U = Q + W

Since the work done on a system depends not only on the initial and final states, but also on the intermediate states, it is not a state function.

PV diagram

P

V T

1

2

the difference between the values of some (state) function z(x,y) at these points:

Comment on State Functions

U, P, T, and V are the state functions, Q and W are not. Specifying an initial and final states of a system does not fix the values of Q and W, we need to know the whole process (the intermediate states). Analogy: in classical mechanics, if a force is not conservative (e.g., friction), the initial and final positions do not determine the work, the entire path must be specified.

x

y z(x1,y1)

z(x2,y2)

dyy

zdx

x

zzd

xy

, ,x ydz A x y dx A x y dy - it is an exact differential if it is

x

yxA

y

yxA yx

,,

, ,dz z x dx y dy z x y

A necessary and sufficient condition for this:

If this conditionholds:

y

yxzyxA

x

yxzyxA yx

,

,,

,

e.g., for an ideal gas:

dV

V

TdT

fNkPdVdUQ B 2

- cross derivativesare not equal

dVPSdTUd

U

V S- an exact differential

In math terms, Q and W are not exact differentials of some functions of macroparameters. To emphasize that W and Q are NOT the state functions, we will use sometimes the curled symbols (instead of d) for their increments (Q and W).

Problem

Imagine that an ideal monatomic gas is taken from its initial state A to state B by an isothermal process, from B to C by an isobaric process, and from C back to its initial state A by an isochoric process. Fill in the signs of Q, W, and U for each step.

V, m3

P, 105 Pa

A

BC

Step Q W U

A B

B C

C A

2

1

1 2

+ -- 0

-- + --

+ 0 +

T=const

TNkf

U B2 BPV Nk T

The Enthalpy

Isobaric processes (P = const):

dU = Q - PV = Q -(PV) Q = U + (PV)

The enthalpy is a state function, because U, P, and V are state functions. In isobaric processes, the energy received by a system by heating equals to the change in enthalpy.

Q = H

isochoric:

isobaric:

in both cases, Q does not depend on the path from 1 to 2.

Consequence: the energy released (absorbed) in chemical reactions at constant volume (pressure) depends only on the initial and final states of a system.

H U + PV - the enthalpy

The enthalpy of an ideal gas:(depends on T only)

TNkf

TNkTNkf

PVUH BBB

1

22

Q = U

Heat Capacity

T

QC

The heat capacity of a system - the amount of energy

transfer due to heating required to produce a unit temperature rise in that system

C is NOT a state function (since Q is not a state function) – it depends on the path between two states of a system

T

V

T1

T1+dT

i

f1 f2 f3

The specific heat capacitym

Cc

( isothermic – C = , adiabatic – C = 0 )

Quasistatic Processes in an Ideal Gas

isochoric ( V = const )

isobaric ( P = const )

021 W

TCTTNkQ VB 02

31221

0),( 12

2

1

21 VVPdVTVPW

TCTTNkQ PB 02

51221

21QdU

2121 QWdU

V

P

V1,2

PV= NkBT1

PV= NkBT21

2

V

P

V1

PV= NkBT1

PV= NkBT212

V2

(see the last slide)

Isothermal Process in an Ideal Gas

1

221 ln),(

2

1

2

1V

VTNk

V

dVTNkdVTVPW B

V

V

B

V

V

f

iBfi V

VTNkW ln

Wi-f > 0 if Vi >Vf (compression)

Wi-f < 0 if Vi <Vf (expansion)

isothermal ( T = const ) :

V

P

PV= NkBT

V1V2

W

2121 WQ

0dU

Adiabatic Process in an Ideal Gas

adiabatic (thermally isolated system)

PdVdTNkf

dUTNkf

U BB 22

( f – the # of “unfrozen” degrees of freedom )

dTNkVdPPdVTNkPV BB PVPdVf

VdPPdV 2

fP

dP

fV

dV 21,0

21

constVPPVP

P

V

V

111

1

lnln

The amount of work needed to change the state of a thermally isolated system depends only on the initial and final states and not on the intermediate states.

021 Q 21WdU

to calculate W1-2 , we need to know P (V,T)

for an adiabatic process

2

1

),(21

V

V

dVTVPW

011

P

P

V

V P

dP

V

dV

V

P

V1

PV= NkBT1

PV= NkBT21

2

V2

Adiabaticexponent

Adiabatic Process in an Ideal Gas (cont.)

V

P

V1

PV= NkBT1

PV= NkBT21

2

22 2

1 1 1

11 11 2 1 1

1 1 1 12 1

1( , )

1

1 1 1

1

VV V

V V V

PVW P V T dV dV PV V

V

PVV V

1+2/31.67 (monatomic), 1+2/5 =1.4 (diatomic), 1+2/6 1.33 (polyatomic)(again, neglecting the vibrational degrees of freedom)

constVPPV 11

An adiabata is “steeper” than an isotherma: in an adiabatic process, the work flowing out of the gas comes at the expense of its thermal energy its temperature will decrease.

V2

Prove 1 2 2 2 B

f fW PV Nk T U

Summary of quasi-static processes of ideal gas

Quasi-Static process

U Q WIdeal gas

law

isobaric (P=0)

isochoric (V=0)

0

isothermal (T=0)

0

adiabatic (Q=0)

0

fi

i f

VV

T T

2 2B

f fU Nk T P V

f iU U U

2

2

fP V

P V

2 2B

f fU Nk T P V

2

fP V

fi

i f

PP

T T

i i f fPV P Vln fB

i

VNk T

VW

i i f fPV P V 2 2B

f fU Nk T PV U

Problem

Imagine that we rapidly compress a sample of air whose initial pressure is 105 Pa and temperature is 220C (= 295 K) to a volume that is a quarter of its original volume (e.g., pumping bike’s tire). What is its final temperature?

2211

222

111

VPVP

TNkVP

TNkVP

B

B

1

2

1

2

12

1

1121

2

11

2

112 T

T

V

VT

T

VPTNk

V

VP

V

VPP B

constVTVT 122

111

KKKV

VTT 51474.12954295 4.0

1

2

112

For adiabatic processes:

Rapid compression – approx. adiabatic, no time for the energy exchange with the environment due to thermal conductivity

constTP /1also

- poor approx. for a bike pump, works better for diesel engines

Non-equilibrium Adiabatic Processes

- applies only to quasi-equilibrium processes !!! constTV 1

2. On the other hand, U = Q + W = 0

U ~ T T – unchanged

(agrees with experimental finding)

Contradiction – because approach

#1 cannot be justified – violent

expansion of gas is not a quasi-

static process. T must remain the

same.

constTV 11. V – increases

T – decreases (cooling)

Free expansion

CV and CP

dT

PdVdU

dT

QC

V = const

P = const

VV T

UC

the heat capacity at constant volume

the heat capacity at constant pressure

PP T

HC

To find CP and CV, we need f (P,V,T) = 0 and U = U (V,T)

nRf

Nkf

C BV 22

For an ideal gas TNkf

U B2

# of moles

12 B

fH Nk T

12P

fC nR

For one mole of a monatomic ideal gas:

RCRC PV 2

5

2

3

Another Problem

During the ascent of a meteorological helium-gas filled balloon, its volume increases from Vi = 1 m3 to Vf = 1.8 m3, and the pressure inside the balloon decreases from 1 bar (=105 N/m2) to 0.5 bar. Assume that the pressure changes linearly with volume between Vi and Vf.(a) If the initial T is 300K, what is the final T?(b) How much work is done by the gas in the balloon?(c) How much “heat” does the gas absorb, if any?

P

V

Pf

Pi

Vi Vf

K2701mbar1

1.8mbar5.0K300

3

3

ii

ffif

BB VP

VPTT

Nk

PVTTNkPV

(a)

(b) f

i

V

V

ON dVVPW )(

bar625.1bar/m625.0 3 VVP

(c) ONWQU

Jmbarmbarmbar 333 41066.04.05.08.05.0)( f

i

V

V

BY dVVPW

- work done on a system f

i

V

V

BY dVVPW )( - work done by a system

BYON WW

JJJ 445 105.41061.0105.112

3

2

3

BY

i

fiiONifBON W

T

TVPWTTNkWUQ