ThermodynamicsThermodynamics

•The First Law of Thermodynamics

•Thermodynamic Processes (isobaric, isochoric, isothermal, adiabatic)Thermodynamic Processes (isobaric, isochoric, isothermal, adiabatic)

•Reversible and Irreversible Processes

•Heat Engines

•Refrigerators and Heat Pumps

•The Carnot Cycle

•Entropy (The Second Law of Thermodynamics)

•The Third Law of Thermodynamics

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The Zeroth Law of ThermodynamicsThe Zeroth Law of Thermodynamics

If A is in thermal equilibrium with C and B in th l ilib i ith Cthermal equilibrium with C then A and B have to be in thermal equilibrium. No qheat flows!

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Internal Energy

In thermodynamics the internal energy of a

From http://en.wikipedia.org/wiki/Internal_energy.........

In thermodynamics, the internal energy of a thermodynamic system, or a body with well-defined boundaries, denoted by U, or sometimes E, is the total of the kinetic energy due to the motion of molecules(translational, rotational, vibrational) and the potential energy associated with the vibrational and electric energyenergy associated with the vibrational and electric energy of atoms within molecules or crystals. It includes the energy in all the chemical bonds, and the energy of the free conduction electrons in metalsfree, conduction electrons in metals.

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The First Law of ThermodynamicsThe First Law of Thermodynamics 

The first law of thermodynamics says the change in internal energy of a system is equal to the heat flow into theinternal energy of a system is equal to the heat flow into the system plus the work done on the system.

WQU −=Δ

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First Law of ThermodynamicsThe change in a systems internal energy is related to the heat and the work

ΔU= Uf ‐ Ui = Q ‐W

heat and the work.

Where:

Uf = internal energy of system @ endU i l f @Ui = internal energy of system @ start

Q = net thermal energy flowing into system during processy g p

Positive when system gains heatNegative when system loses heat

W net work done by the systemW = net work done by the systemPositive when work done by the systemNegative when work done on the system

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Thermodynamic ProcessesThermodynamic ProcessesA state variable describes the state of a system at time t, but it does not reveal how the system was put into thatbut it does not reveal how the system was put into that state. Examples of state variables: pressure, temperature, volume, number of moles, and internal energy.

Thermal processes can change the state of a system.

We assume that thermal processes have no friction or other dissipative forces. In other words: All processes are reversibleIn other words: All processes are reversible(Reversible means that it is possible to return system and surroundings to the initial states)REALITY: irreversible

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REALITY: irreversible

“Humpty Dumpty sat on a wall“Humpty Dumpty sat on a wall.Humpty Dumpty had a great fallAll the king’s horses and all the king’s menCouldn’t put Humpty Dumpty together again”

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Couldn t put Humpty Dumpty together again* Martin Schullinger-Krause (PH202 Winter 2008)

A PV diagram can be used to represent the state changes g p gof a system, provided the system is always near equilibrium.

The area under a PV curve gives the magnitude of the work done on a system. W<0 for compression and pW>0 for expansion.

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To go from the state (Vi, Pi) by the path (a) to the state (Vf, i i fPf) requires a different amount of work then by path (b). To return to the initial point (1) requires the work to be nonzero.

The work done on a system depends on the path taken in the PV diagram. The work done on a system during a closed cycle can be nonzero

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closed cycle can be nonzero.

An isothermal process implies that both P and pV of the gas change (PV∝T).

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Specific Heats under constant pressure and constant volume

Specific heatQ = m c ΔTQ

For a gas we use

Molar specific heatQ = n C ΔT

Constant Volume: CVConstant Pressure : CPP

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Thermodynamic Processes for an Ideal GasThermodynamic Processes for an Ideal Gas

No work is done on a system No work is done on a system when its volume remains constant (isochoric process). For an ideal gas (provided the

QWQU −=−=Δ 0

For an ideal gas (provided the number of moles remains constant), the change in internal

TUQ Δ=Δ= Cn Venergy is

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For a constant pressure (isobaric) process, the change in o a co s a p essu e ( soba c) p ocess, e c a geinternal energy is

WQU =Δ WQU −=Δ.TnCQ PΔ=TnRVPW Δ=Δ=where and

CP is the molar specific heat atspecific heat at constant pressure. For an ideal gas CP= CV+R.

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For a constant temperature (isothermal) process, ΔU = 0 and the work done on an ideal gas is

⎞⎛⎞⎛VVnRT

VVNkTW

ii⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛= .lnln ff

WQUii

=⇒=Δ⎠⎝⎠⎝

0

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We have found for a monoatomic gasΔU = 3/2 n R ΔTΔU = 3/2 n R ΔT

Constant volume: ΔU= Q

3/2 n R ΔT = n CV ΔT CV= 3/2 R

Constant pressure: Q = ΔU + W

n CP ΔT = 3/2 n R ΔT + n R ΔT

C 5/2 RCP= 5/2 R

CV – CP = R (always valid for any ideal gas)

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Adiabatic (“not passable”) processes(no heat is gained or lost by the system Q=0 i e system(no heat is gained or lost by the system Q 0, i.e. system perfectly isolated )

Q=0 and so ΔU= -W

P V = constant (isothermal)P Vγ = constant (adiabatic)

γ = CP/CV

For a monoatomic gas therefore γ = 5/3therefore γ 5/3

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Example: An ideal gas is in contact with a heat reservoir so that it remains at constant temperature of 300 0 K The gasthat it remains at constant temperature of 300.0 K. The gas is compressed from a volume of 24.0 L to a volume of 14.0 L. During the process, the mechanical device pushing the i t t th i f d t d 5 00 kJ fpiston to compress the gas is found to expend 5.00 kJ of

energy. How many moles of the ideal gas are in the system? How much heat flows between the heat reservoir and the gas, and in what direction does the heat flow occur?

lJWVRTW f 735000)ln( −

→ mol

VV

RTn

VnRTW

i

fi

f 7.3)24/14ln(31.8)ln(

)ln( =⋅

==→=

This is an isothermal process, so ΔU = Q - W = 0 (for an ideal gas) and W = Q = - 5 00 kJ Heat flows from

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an ideal gas) and W Q 5.00 kJ. Heat flows from the gas to the reservoir.

An ice cube placed on a countertop in a warm room will melt The reverse process cannot occur: an ice cube willmelt. The reverse process cannot occur: an ice cube will not form out of the puddle of water on the countertop in a warm room.

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Any process that involves dissipation of energy is notAny process that involves dissipation of energy is not reversible.

Any process that involves heat transfer from a hotter object to a colder object is not reversible.

The second law of thermodynamicsThe second law of thermodynamics(Clausius Statement): Heat never flows spontaneously from a colder body to a hotterspontaneously from a colder body to a hotter body.

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Heat EnginesHeat Engines

A heat engine is aA heat engine is a device designed to convert disordered

i d denergy into ordered energy. The net work done by an engine y gduring one cycle is equal to the net heat flow into the engine

QW

flow into the engine during the cycle (ΔU= 0).

netnet QW =

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The efficiency of an engine is defined as

HQW

e net

input heatenginetheby done worknet

==

Note: Q = Q Q(e.g. a efficiency of e=0.8 means 80% of the heat is Note: Qnet = Qin - Qoutmeans 80% of the heat is converted to mechanical work) outputnet work netW

1

inputheat ou pue wo

CCH

H

net

QQQQ

e

=−

=

==

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.1HH QQ

−==

Refrigerators and Heat PumpsRefrigerators and Heat PumpsHere, heat flows from cold to hot but withcold to hot but with work as the input.

Pump

Refrigerator

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K = Coefficient of performance

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Reversible Engines and Heat PumpsReversible Engines and Heat Pumps

A reversible engine can be used as an engine (heat g (input from a hot reservoir and exhausted to a cold reservoir) or as a heatreservoir) or as a heat pump (heat is taken from cold reservoir and

h t d t h texhausted to a hot reservoir).

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From the second law of thermodynamics, no engine can have an efficiency greater than that of an ideal reversible y gengine “Carnot engine” that uses the same two reservoirs. The efficiency of this ideal reversible engine is

1 CTe −= .1H

r Te −=

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Details of the Carnot CycleDetails of the Carnot CycleThe ideal engine of the previous section is known as a Carnot engineCarnot engine.

The Carnot cycle has four steps:

1. Isothermal expansion: takes in heat from hot reservoir; keeping the gas temperature at TH.

2. Adiabatic expansion: the gas does work without heat flow into the gas; gas temperature decreases to TC.

3 I h l i H Q i h d3. Isothermal compression: Heat QC is exhausted; gas temperature remains at TC.

4 Adiabatic compression: raises the temperature back to

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4. Adiabatic compression: raises the temperature back to TH.

The Carnot cycleThe Carnot The Carnot cycle illustrated

The Carnot engine model was graphically expanded upon by Benoit Paul Émile Clapeyronp yin 1834 and mathematically elaborated uponelaborated upon by Rudolf Clausius in the 18 0 d 601850s and 60sfrom which the concept of

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pentropy emerged

The Otto cycle

Its power cycle consists of adiabaticcompression heat addition at constantcompression, heat addition at constant volume, adiabatic expansion and rejection of heat at constant volume and characterized by four strokes, or reciprocating movements of a piston in a cylinder:cylinder:

intake/induction strokecompression stroke power strokepower stroke exhaust stroke

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EntropyEntropy 

Heat flows from objects of high temperature to objects at low temperature because this process increases the

f f ’disorder of the system. Entropy is a measure of a system’s disorder. Entropy is a state variable.

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If an amount of heat Q flows into a system at constant temperature, then the change in entropy is

.QS =Δ .T

SΔ

E i ibl i th t t l t f thEvery irreversible process increases the total entropy of the universe. Reversible processes do not increase the total entropy of the universe.

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Th d l f th d iThe second law of thermodynamics(Entropy Statement): The entropy of the universe never decreases.

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Example: An ice cube at 0.0 °C is slowly melting. What is the change in the ice cube’s entropy for each 1.00 g of ice that melts?

To melt ice requires Q = mLf joules of heat. To melt one gram of ice requires 333 7 J of energygram of ice requires 333.7 J of energy.

The entropy change is

J/K. 22.1K273

J 7.333===Δ

TQS

The entropy change is

K273T

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J/K.1J 300−=

−==Δ

QSh t J/K.1K300

ΔT

Shot

J/K.60K5

J 300=

+==Δ

TQScold

300KK5Tcold

Q

5 K

htt // t b / t h? X 6P tf23tQ39

http://www.youtube.com/watch?v=Xa6Pctf23tQ

Statistical Interpretation of Entropy

A microstate specifies the state of each constituent particle in a thermodynamic system. A macrostate is determined y yby the values of the thermodynamic state variables.

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smacrostate possible allfor smicrostate ofnumber totalmacrostate the toingcorrespond smicrostate ofnumber macrostate a ofy probabilit =

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The number of microstates for a given macrostate is related gto the entropy.

Ω= lnkS where Ω is the number of microstates.

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The Third Law of ThermodynamicsThe Third Law of Thermodynamics

It is impossible to cool a system to absolute zero.

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