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Chapter 2 The First Law: the concepts Thermodynamics : the study of the transformations of energy , enables us to discuss all these matters quantitatively and to make useful prediction . Conservation of energy in the universe . Universe = System + Surrounding

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slide1

Chapter 2 The First Law: the concepts

Thermodynamics: the study of thetransformations of energy, enables

us to discuss all these mattersquantitatively and to make useful prediction.

Conservation of energy in the universe.

Universe = System + Surrounding

System: isthe part of the worldin which we have a special interest.

Surrounding: is where we make our measurement.

slide2

The basic concepts:

Open system: Matter and energycan be

transferred between system

and surrounding.

Close system: only energycan be transferred.

Isolated system: can exchangeneither energy

no matterwith its surrounding.

slide3

2.1 Work , Heat, and Energy:

Internal energy change = Work + Heat (measurable)

Work: is done when an object is movedagainst an opposing force.

W > 0: when work is done on the (isolated) system by the surrounding.

W < 0: when the system does work to the surrounding.

Energy of a system is its capacity to do work.

Heat: the energy of a systemchanges as a result of temperature difference.

Heat transfer direction isdependent on temperature.

q > 0: when the system absorbs the heat from surroundings. (Endothermic)

q < 0: when the system releases the heat to surrounding. (Exothermic)

slide4

Diathermic: A boundary that does permit

energy transfer as heat.

Adiabatic: A boundary that does not permit

energy transfer as heat.

slide5

Consider a reaction (e.g. vaporization of water) occurs in the system:

In anadiabatic container:

(a). An endothermic reaction;

T of the system decreases.

(b). An exothermic reaction ; T raises.

In andiathermic system:

(c). An endothermic reaction, T = constant,

q is adsorbed from surrounding.

(d). An exothermic reaction, T = constant,

q is released to surrounding.

slide6

Molecular interpretation 21.

In molecular terms, heat is the transfer of energy thatmakes use of

chaotic motion. Chaotic motion of molecules is called thermal motion.

In a hotter system (or surrounding), the thermal motion is more vigorous.

Work is the transfer of energy thatmakes use of organized motion.

Work: energy transfer making use of theorganized motionof particles.

Heat: energy transfer making use of thethermal motion.

Work and heat can coexist in a process or chemical reaction.

slide7

When a system heats its surrounding, molecules of the system stimulate the thermal

motion of the surroundings.

When a system does work, it stimulates orderly motion in the surroundings. .

slide8

2.2 The Internal Energy:

In thermodynamics, the total energy of a system called its

internal energy, U.

Internal energy =kinetic energy + potential energy

ΔU = Uf – Ui : the change in internal energy

The internal energy is astate function, that its value depends only on the

current state of the system and isindependent of pathway.

The internal energy is anextensive property.

(Extensive property changes with the amount of substance)

The unit of the internal energy: joule (J)

Molar internal energy: kJ/mol

slide9

Molecular interpretation 2.2

A molecule has a certain number of degree of freedom.

Equipartitiion theorem of classical mechanics:

For a collection of particles at thermal equilibrium at a temperature, the

average energy of each quadratic contribution (degree of freedom) to the

energy is same and equal to ½ kT.

For the case of a monatomic perfect gas at a temperature T,

Ek = ½ mvx2 + ½ mvy2 + ½ mvz2 (translation energy)

(no potential energy exists in the perfect gases)

Um = Um(0) + 3/2 RT

Um(0): is the molar internal energy at T = 0.

slide10

When the gas of polyatomic molecules (nonlinear moleucle), the rotation

(Rx, Ry, Rz) energy is an additional contribution of 3/2RT to the internal energy:

Um = Um(0) + 3RT

A linear molecule can rotate only around

two axes, so it has two rotational modes of

motion, each contributing ½ kT to the internal

energy.

Um = Um(0) + 5/2RT

slide11

The internal energy of a system may be changedether by doing work on

the system or by heating it.

Heat and work areequivalent ways of changing a system’s internal energy.

ΔU = q + w

In acquisitive convention,

w > 0; q > 0 if energy is transfer to the system as work or heat

w < 0; q < 0 if energy is lost from the system as work or heat.

In scientific representations,sign + quantity.

 If a system isolated from its surroundings, then no change in internal energy takes place.

First Law of Thermodynamics:

The internal energy of an isolated system is constant.

slide12

The mechanical definition of heat

In adiathermic system,ΔU is the same as in adiabatic,

but we might find that thework must do is

not the sameas that in adiabatic.

The difference is definedas the heat

adsorbed by the system in the process:

Path I: adiabatic process

ΔU = q + w = Wad

Path II : diathermic process

q = U – w = Wad – w

Exp. Wad = 42 kJ; w = 50 kJ q = wad – w = – 8 kJ

Path I

Path II

(W dia)

(W ad)

slide13

2.3 Expansion work:

In an infinitesimal change:

dU = dq + dw

In physics:

dw = – Fdz (against the opposing force)

For expansion work,

dw = – pex dV

When the volume changes from Vi to Vf

W =–  pex dV

vf

vi

slide14

Other types of work: non-expansion work or additional work

Work = intensive factor (pressure) x extensive factor (volume)

slide15

(b). Free expansion: System expands to a vacuum.

pex = 0 ; w = 0.

(C). Expansion against constant pressure:

W = –  pex dV = – pex (Vf – Vi) = – pexΔV

A p, V-graph used to compute expansion

work is called anindicator diagram.

slide16

(d). Reversible expansion:

A reversible change: is a change that can bereversed by an infinitesimal

modification of a variable.

Equilibrium: if an infinitesimal changes in the conditions inopposite

directions results in opposite change in its state.

In the view of the work (reverse):

dw = – pex dV = – pdV

W = – pex dV

slide17

(e). Isothermal reversible expansion:

An isothermal reversible expansion of the perfect gas: PV = nRT

Wrev = – pex dV = –nRT ln (Vf/Vi)

| Wrev| > | Wone-step|

Themaximum workavailable from a system operating between specified initial and final states and passing along a specified path is obtained when the change takes placereversibly.

Vf

Vi

slide18

One-step expansion (膨脹) Pex = 1/4 P1

Work = – Pex –ΔV = – ¼ P1 (4V1 – V1) = – 3/4 P1V1

P1; V1

¼ P1; 4V1

slide19

Infinite-step (無限多步)expansion: The external pressure is always almost exactly equal to the pressure of the gas. Reversible process:P~Pex P = Pex+ΔP (P  0) ∣Work∣=∫Pex dV P ~ PexPex ~ P= RT/V∣W∞∣=∣Wrev∣=∫nRT/V dV

slide21

2.4 Heat transaction:

dU = dq + dwexp + dwe

At constant V (dwexp = 0) and no

other kind worked produced (dwe = 0).

dU = dqv  U = qv

(a). Calorimeter:

The most common device for

measuring ΔU is the adiabatic bomb

calorimeter.

Constant-volume container.

slide22

No net loss of heat from the calorimeter,

the calorimeter is adiabatic.

For a calorimeter:

q = CΔT; C: calorimeter constant

(b) Heat capacity:

The internal energy of a substance

increases when its temperature is raised.

The slope of tangent to the curve at any

temperature is called the heat capacity of

the system.

Cv = (U/T)v

At constant volume

slide23

Heat capacity areextensive properties.

However,the molar heat capacity at

constant volume is intensive.

Unit: kJmol-1K-1

Specific heat capacity: Unit: kJK-1g-1

In general, internal energies depend on the

temperature and decrease at low temperature.

For perfect gas: Cv is T-independent

Cv = 3/2 R (monatomic molecules)

Cv = 3 R (polyatomic molecules)

slide24

At constant volume:

dU = Cv dT

If Cv is T-independent, ΔU = CvΔT

For measuring ΔU, heat supplied at constant volume.

qv = CvΔT = ΔU

2.5 Enthalpy:

When the system is free to change its volume,dU  dq.

The heat supplied at constant p is equal the change in another

thermodynamic property, enthalpy H.

slide25

At constant p, some of energy supplied

as heat may escape into the surrounding

as work.

H = U + PV

H is a state function.

dH = dq (at pex= constant, no additional work)

dH = dU + pdV + Vdp

= dq – pdV + pdV + Vdp = dq + vdP

At constant p: dH = dqp

ΔH = qp

slide26

(b). The measurement of an enthalpy change

An enthalpy change can be measured

calorimetrically by monitoringthe temperature changeat constant p. (isobar calorimeter)

Because solids and liquids have small molar

volume,

Hm = Um + pVm≈ Um

ΔHm = ΔUm (solid and liquid)

For the perfect gas,H = U + pV = U + nRT

The change of enthalpy in a reaction

that produce or consumes gas is:

ΔH = ΔU + Δ(nRT)

slide27

(c). The variation of enthalpy with temperature:

H = f (T, p ,V)

The enthalpy increase with the temperature.

Heat capacity at constant pressure (Cp):

theslopeof the tangent to a plot of

enthalpy against T.

Cp = (H/T)p

Cp,m: heat capacity per mole of material.

dH = CpdT (at constant p)

If Cp is a constant:ΔH = CpΔT

at constant pressure

slide28

A common approximate empirical expression:

Cp,m = a + bT + c/T2

H(T2) – H(T1) = a (T2 – T1) + ½ b(T22–T12) – c(1/T2 – 1/T1)

slide29

The relationship between heat capacities:

q = ΔU – w = ΔU + pexdV

At constant V, w = 0;

At constant P, w = –pexdV

qp > qv ;Cp > Cv

For the perfect gas:

Cp – Cv = nR

For 1 mole perfect gas:

Cp – Cv = R

slide30

I 2.1 Differential Scanning Calorimeter (DSC)

A DSC measures the energy transferred as heat to or from a sample at a

constant pressure during a physical or chemical change.

A DSC consists of two small compartments that are heated electrically at a

constant rate. T = T0 + T

slide31

To maintain the same temperature in both compartments, excess energy is

transferred as heat to or from the sample during the process.

If no physical or chemical change occurs,

qp = CpT (Cp independent of temperature)

The chemical or physical process requires

the transfer of qp + qp,ex

qp,ex: excess energy transferred as heat to

attain the same change in temperature

qp + qp,ex = (Cp + Cp,ex)T

 Cp,ex = qp,ex/ T = qp,ex/t = Pex/

Pex = the excess electrical power

necessary to equalize the temperature

slide32

A DSC trace, also call a thermogram,

consists of a plot of Pex(Cp,ex) against T.

The enthalpy change associated with the process is

H = T1Cp,ex dT

2.6 Adiabatic Changes:

Adiabatic process:

A process in which no energy as heat flows

into or out of the system.

q=0 ; ΔE=q + W= W;

For an ideal gas:

dE = nCvdT ; W= – Pex dV

T2

slide33

For a reversible, adiabatic expansion-compression of an ideal gas.

dE=nCvdT= –Pex dV= –nRT / V dV

Cv / T dT= –R / V dV

Form T1 to T2 (infinitesimal changes)

Cv∫1/ T dT= –R∫1/ V dV; Cv㏑T2 / T1= – R㏑V2 / V1

T2 / T1= (V1 / V2)r-1; r=Cp / Cv

T1V1r-1 = T2V2r-1

PV=nRT ; T2 / T1= P2V2 / P1V1

P1V1r= P2V2r

slide34

For isothermal expansion

(At const. T,ΔE=0)

P1V1= P2V2= constant

For adiabatic expansion

(q=0,T will change)

P1V1r= P2V2r=constant

Cp/Cv > 1

slide35

Thermochemistry

The study of theheat produced or requiredby chemical reactions.

Internal energy = Kinetic energy (T) + Potential energy

Potential energy = Physical potential + Chemical potential

Physical potential is from intermolecular interaction.

Chemical potential is ascribed to intramolecular chemical bond.

For chemical reactions (Thermochemistry):

Vessel + content (chemicals): system

In general, chemical reaction takes place at constant p

qp = ΔH

slide36

At constant pressure:

An endothermicprocess (q > 0) hasΔH > 0.

Anexothermic process (q < 0)is one for whichΔH < 0.

2.7 Standard enthalpy changes: H(T, P, n…….)

Thestandard stateof a substance at a specified temperature

is its pure format 1 bar.

Thestandard enthalpy change for a reactionor a physical process is the

difference between the enthalpy of the products instandard statesand the

enthalpy of the reactants in their standard states, all at the same specified

temperature.

slide37

Standard enthalpy of vaporization:

H2O(l) H2O(g) ΔvapHØ (373K) = + 40.66 kJ/mol

Standard enthalpy may be reportedfor any temperature.

The conventional temperature =298.15 K

(a). Enthalpy of physical change:

Standard enthalpy of transition (a change of physical state): ΔtrsHØ

slide38

Enthalpy is a state function: (path-independent)

ΔsubHØ = ΔfusHØ + ΔvapHØ

(at same T)

ΔHØ(AB) = – ΔHØ(BA)

slide40

(b). Enthalpies of Chemical Change:

Standard reaction enthalpy, ΔrHØ, is the change in enthalpy when reactants in their standard states change to products in their standard states.

CH4(g) + 2 O2 (g) CO2(g) + 2H2O(l) ΔrHØ = – 890 kJ/mol

Thermochemical equation: a chemical equation

+ standard reaction enthalpy

2 A + B 3C + D ; ΔrHØ = ?

ΔrHØ = ∑ v HmØ – ∑ v HmØ

The general form:

ΔrHØ = ∑ vJHmØ(J) ; vJ: stoichiometric number;

HmØ(J): the standard molar enthalpy of a species J at standard temperature.

reactants

products

slide41

Somestandard reaction enthalpies have special names and a particular significant.

The standard enthalpy of combustion cHØ.

C6H12O6(s) + 6O2 6CO2(g) + 6 H2O(l) cHØ = – 2808 kJmol-1

slide42

I2.2 Food and Energy Reserve:

The thermodynamically properties of fuels and foods are commonly

discussed in term of their specific enthalpy (cH/M).

A typical 18 – 20 year old man requires a daily input of about 12 MJ;

a woman needs about 9.0 MJ.

slide43

Glucose has a specific enthalpy of 16 kJg-1.

Digestible carbohydrates has a specific enthalpy of 17 kJg-1.

Indigestible cellulose helps to move digestion products through the intestine.

Fats (long-chain esters) have a specific enthalpy of 38 kJg-1.

The specific enthalpyHydrocarbon oils used as fuel is 48 kJg-1.

In Arctic(北極) species, the stored fat acts as layer of insulation.

In desert species, the fat is also a source of water, one of its oxidation products.

Proteins are also used as a source of energy, but their components (amino acids) are often used to construct other protein instead.

Radiation is one means of discarding heat; another is evaporation and energy demands of vaporization of water (2.4 kJg-1).

Vigorous excises  1 – 2 dm3 of perspired water per hour  2.4 – 5.0 MJh-1.

slide44

(c). Hess’s law:

The standard enthalpy of an overall reaction is thesumof the standard enthalpies of the individual reactionsinto which a reaction may be divided.

C(s) + 2 H2(g)CH4(g)ΔHf0 = ?

C(s) + O2(g)CO2(g) ΔH10 = -394 kJ

2 H2(g) + O2(g) H2O(l) ΔH20= -572 kJ

CH4(g)+ 2O2(g) CO2(g) + H2O(l) ΔH30 = -891 kJ

ΔH0 = -ΔH30 + ΔH10+ ΔH20 = -75 kJ

The importance of Hess’s law is that information about a reaction of interest, which may be difficult to determined directly, can assembled from information on other reactions.

slide45

Illustration:

N2 + 3 H2 2 NH3

v(N2) = –1 ; v(H2) =–3; v(NH3) = +2

ΔrHØ = 2HmØ(NH3) – {(HmØ(N2) + 3 HmØ(H2)}

However, how to have the value of HmØ(NH3) ???

slide46

2.8 Standard Enthalpies of formation:

Standard enthalpy of formation ΔfHØ: the standard reaction enthalpy for the formation of the compound from its elementsin their reference states.

Reference state of an element: its most stable state

at the specified temperature and 1 bar.

6C(s, graphite) + 3 H2(g) C6H6(l) ΔH = + 49.0 kJ/mol

The standard enthalpy offormation of liquid benzeneat 298 K is + 49.0 kJ/mol.

The standard enthalpies of formation of elements in their reference states are zero at all temperatures because they are the enthalpies of such “null” reactions as N2(g)  N2(g).

slide47

Definitions of Standard State:

1. Gas : 1 atm (or 1 bar = 105 Pa).

2. In solution, concentration = 1.0 M at 1 atm.

3. A pure substance in condensed state, pure liquid or solid.

4. Element: it exists (most stable) under conditions of 1 atm and the temperature of interest (25 oC).

Notice the physical state or the structure of reactant and product:

ΔHof of H2O(g)  ΔHof of H2O (l)

ΔHof of C(graphite) = 0 ΔHof of C(diamond) = 2 kJ/mol

ΔHof of many compounds are listed in Table 2.5 and 2.6

slide48

Enthalpies of formation (Hof) are always given per mole of product with

the product in its standard state.

slide49

(a). The reaction enthalpy in terms of enthalpies of formation:

Conceptually, we can regard a reaction as

proceeding by decomposing the reactants into

their elements and then forming those

elements into the products.

ΔrHØ = ∑ vJΔHfØ(J)

Δ HfØ(J): the standard enthalpy

of formation of a species J

at interested temperature.

slide50

Reference (standard) state  Set up a scale for measuring thermodynamic properties.

Hof of element at standard state  0

Reactants Elements Products

Hof (reactants ) Hof = 0 Hof (products)

slide51

Using the Hof to calculate the standard enthalpy change of a reaction.

Exp. CH4(g) + 2 O2 (g)  CO2(g) + 2 H2O(l) Ho = ?

CH4(g)  C(s) + 2 H2(g) aHo = 75 kJ (-Hof of methane )

C(s) + O2(g)  CO2(g) bHo = -394 kJ (Hof of CO2)

2 x ( 2 H2 + ½ O2(g)  H2O(l) ) dHo = 2 x (-286 kJ) (Hof of H2O(l))

slide52

(b). Group contribution:

Chemical reaction could be considered as theformation and breaking of bonds.

Mean bond enthalpies, ΔH(A-B). This procedure is notoriously unreliable because thebond energy changes with many factors.

Thermochemical groups approach: a molecular is regarded as

being built up of the thermochemical groups.

slide53

ΔfHØ (C6H14(l)) = ? , at 298 K

ΔHfØ (C6H14(g)) = 2 (-42.17) + 4(-20.7) /Jmol-1

ΔcondenseH of C6H14(g) = -28.9/Jmol-1

slide54

2.9 The temperature dependence of reaction enthalpies

For calculating ΔrH at different temperature,

H(T2) = H(T1) +  Cp dT

Kirchhoff’s law:

ΔrHØ(T2) = ΔrHØ(T1) + ΔrCpØ dT

ΔrCpØ = ∑ vJCp,mØ(J)

Cp,m could be considered as temperature-independent because its deviation is much smaller compared to the chemical reaction enthalpy,ΔrH.

slide55

f(x, y); df = (f/x)ydx + (f/y)xdy

If f = a x3y + by2 ; df = (3ax2y)dx + (ax3 + 2by)dy

2f/xy = 2f/yx = 3ax2

Relation no. 1. when x is changed at constant z

(f/x)z = (f/x)y + (f/y)x(y/x)z

Relation no. 2. (the inverter):

(x/y)z = 1/(y/x)z

Relation no. 3. (the permuter):

(x/y)z(z/y)x(x/z)y = – 1 (Euler’s chain relation)

Relation no. 4. This relation establishes whether or not df is an exact diffraction.

df = g(x, y)dx + h(x, y)dy is exact if (g/y)x = (h/x)y

slide56

A number of relations between the experimental observables on

thermodynamics by exploring the mathematical consequences of these

facts.

State function: independent of how a sample is prepared.

Path function: related to the preparation of the state

State functions and exact differentials (正合微分):

Using a proper mathematical derivations, we can combine

measurements of different properties to obtain the value of a property

we require.

slide57

3.1 State function:

Path 1: adiabatic process

Path 2: nonadiabatic process

Uf – Ui = w1 + q1 (q1 = 0)

= w2 + q2 (q20 )

U is a property of the state.

w is a property of the path.

slide58

State function:

Exact differential is an infinitesimal quantity, which when integrated, gives a result that is independentof the path between the initial and final state.

ΔU = dU; dU is an exact differential

Heat (q):

q =  dq

Because heat is not a state function, the heat can not be expressed as qf – qi or Δq.

This path-dependence is expressed by saying that dq is an inexact differential.

For inexact differential, dq is written as đq.

The work (w) is an inexact differential as well as heat.

f

i

f

i, path

slide59

(b). Change in internal energy:

U = f (V, T) for one mole molecules in a close system.

With an infinitesimal change of V, T

( The second-order infinitesimal could be neglected.)

U’ = U + (U/V)TdV + (U/T)VdT

dU = (U/V)TdV + (U/T)VdT

Any infinitesimal change in the internal energy is proportional to the infinitesimal changes of volume and temperature, the coefficients of proportionality being the partial derivatives.

dU = (U/V)TdV + CVdT

slide62

Internal pressure:

  • πT = (U/V)T
  • For attractive force between the particles,
  • For perfect gas, no interaction between particles.
  • For repulsion force,
  • T = (U/V)T < 0

πT = (U/V)T > 0

πT = 0

πT = 0

slide63

(c). The Joule experiment:

The gas expands from a high pressure system (22 atm) to a vacuum.

Due to expansion to vacuum (Pex = 0),

w = 0

ΔU = q = πTΔV + CvΔT

His result: ΔT of bath = 0

However, this experiment was crude.

Cv and πT of a gas is extremely smallerthan that of a liquid in bath.

slide64

(d). Changes in internal energy at constant pressure:

dU = πTdV + CvdT

Divided by dT and at constant p

(U/T)p = πT(V/T)p + Cv

Expansion coefficient, α

α = 1/V(V/T)p

A large value of α means that the volume

responses strongly to temperature change.

For the ideal gas:

α = 1/T

slide65

The general equation for the closed and constant-composition system

(U/T)p = απTV + Cv

Cv, α, πT could be measured in the individual experiment.

For the perfect gas:

πT = 0, (U/T)p = Cv = (U/T)v

Only at constant V, ΔU = qv.

3.2 The temperature dependence of the enthalpy (H):

At constant pressure, ΔH = qp

H = H (p, T)

slide66

(a). Changes in the enthalpy at constant volume:

dH = (H/p)T dp + (H/T)p dT

dH = (H/p)T dp + CpdT

(H/T)v = (1- α/T) Cp

Justification:

(p/T)V (T/V)p (V/p)T = –1 ;

1. (p/T)V = - 1/(T/V)p (V/p)T = - (V/T)p/(V/p)T = α/T

2. (H/p)T = - 1/(p/T)H (T/H)p = - (T/p)H (H/T)p = –Cp

Isothermal compressibility, T T = – 1/V(V/p)T

Joule-Thomson coefficient, = (T/P)H

In the above equation, all the quantities can be measured in suitable experiments.

slide67

(b). The isothermal compressibility, T T = – 1/V(V/p)T

The negative sign in the definition of T ensures that T is positive.

The T is obtained from the slope of the plot of volume against pressure at constant temperature.

For a perfect gas:

T = – 1/V [(nRT/p)/p]T

= –nRT/V(–1/p2) = 1/p

The higher pressure of the gas, the lower it compressibility.

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3.3 The relation between Cv and Cp:

Cp – Cv = work needed to change the volume of the system

to maintain the constant pressure

= (1). work of driving back the atmosphere, + (2). work of stretching the bondsor changing intermolecular interactions in the materials.

For perfect gas (no potential energy), the second term makes no contribution.

(a). The relation for a perfect gas:

Cp – Cv = (H/T)p – (U/T)v = (U/T)p + nR – (U/T)v = nR

(H = U + PV = U + nRT; (U/T)p = (U/T)v)

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(b). The general case:

Cp – Cv = α2TV/T

Justification (3.2):

Cp – Cv = (U/T)p + ((pV)/T)p – (U/T)v

( (U/T)p = απTV + Cv; ((pV)/T)p= p (V/T)p = αpV )

Cp – Cv = α(p + πT)V ; (πT = T (p/T)v – p)

Cp – Cv = αTV(p/T)v

For the liquids and solids, the thermal expansivities, α, are small, Cp ≈ Cv.

However, if T is small, the difference between the two heat capacity

would be apparent.

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(c). The Joule-Thomson effect:

For achieving the constrain ofconstant H,

(Pi, Vi, Ti)  (Pf, Vf, Tf)

The process was adiabatic (q = 0).

They observeda lower temperature on the low-pressure side, the difference in temperature being proportional to the pressure difference.

ΔT ΔP

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The process was adiabatic (q = 0).

Uf – Ui = w = –PfVf – (– PiVi)

Uf + PfVf = Ui + PiVi

or Hf = Hi

This process is an isoenthalpic process.

The ΔT/ΔP ratios had been measured.

The  value could be obtained from the intercept (ΔP 0) of the plot ΔT/ΔPvs. ΔP  (P/T)H.

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In the modern method, the  has been indirectly measured.

T = (H/p)T : isothermal Joule-Thomson coefficient

T = –Cp ; (H/p)T = –Cp (eq. 15)

The value of T can be obtained from the limit value of ΔH/ΔP as ΔP  0.

Real gases have nonzero

Joule-Thomson coefficients and

depending on the identity of gas,

p, T, V.

The  can be ether + or –.

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Measurement of T:

The cooling effect is exactly offset by an electronic heater.

The energy provided by the heater = qp = ΔH

(Pi, Ti)

(Pf, Tf); Tf = Ti = constant

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The  can be ether + or –.

If  > 0, cooling effect,

(T/P)H > 0, gas cools on expansion.

(Attraction force)

When  < 0, heating effect,

(T/P)H < 0, gas heats on expansion.

(Repulsion force)

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Inversion temperature (TI):

The  value change the sign just below the upper TI or above the lower TI.

A gas generally has two inversion temperatures, one at high temperature and the other at low.

For a perfect gas,  = 0, the temperature of a perfect gas is unchanged by Joule-Thomson expansion.

 value is dependent on the intermolecular force.

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Linde refrigerator:

Makes use of Joule-Thomson expansion to liquefy gases.

At  > 0,

1. The gas at high pressure is allowed to expand through the throttle. (ΔP ; ΔT)

2. The cool gas cools the high-pressure gas, which cools still further as it expands.

3. Eventually, liquefied gas drips from the throttle.

Throttle

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Molecular interpretation 2.3:

The equipartition theorem: the mean kinetic energy of molecules in gas is proportional to the temperature.

When a ball is thrown into the air, as it rises it slows in response to the gravitational attraction. Kinetic energy  potential energy.

As the gas expands, the molecules move apart to fill the available volume, struggling as they do so against the attraction of their neighbors.

Kinetic energy  potential energy (greater separation).

The cooling effect, which corresponds to  > 0, is observed when attractive force are dominant (Z < 1)

 When repulsions are dominant (Z > 1), the Joule-Thomson effect results in gas becoming warmer ( < 0 ).

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