Thermotk 2011gs III Heat Effects

8/20/2019 Thermotk 2011gs III Heat Effects

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Chemical Engineering ThermodynamicsChemical Engineering Thermodynamics .. , . ,, . ,

2011Gs_III_Heat Effects2011Gs_III_Heat Effects

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Lesson Topics Descriptions

Lesson 3ALesson 3A Internal Energy & Mendiskusikan definisi energi dalam dan entalpi untuknt a py rea su stances , gas-gas ea , ca ran an pa atan

incompresible .

Lesson 3BLesson 3B ThermoProperties: NIST

WebBook

Mempelajari bagaimana menggunakan NIST Webbook

untuk mencapai data termodinamika.

Lesson 3CLesson 3C Heat Capacities Mempelajari 2 definisi kapasitas panas: kapasitas panasvolume konstan dan kapasitas panas tekanan konstan.

essonesson ypo e caProcess Paths

empe a ar a ur proses po e un u menen u an

perubahan sifat termal dari keadaan awal ke keadaanakhir.

Lesson 3ELesson 3E Phase Changes Mempelajari bahwa alur proses hipotetik untuk alatyang bermanfaat dalam mempertimbangakan

perubahan sifat terkait dengan perubahan fase.

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Three Principle Phases, Revisited Three Principle Phases, Revisited

In Chapter 2, Lesson B, we discussed the difference between the three

principle phases: gas, liquid, and solid. Just to refresh your memory, move themouse pointer over the sketch of each phase to see a description of that

hase.

We also learned that, in all the phases, molecules move randomly

with three different types of motion: vibration, rotation andtranslation.

Molecules move randomly with three different types of motion:vibration, rotation and translation. Molecules are separated bylarge distances and travel a long way between collisions.

Molecules move randomly with all three types of motion, but they are much

Atoms or molecules have all three types of motion, but they are very closetogether. As a result, they cannot travel far at all before they collide. Eachmolecule moves about within a small space and does not tend to wander.

The internal energy of the system is defined as the sum of the kinetic energies in

the vibrational, rotational and translational motion of molecules.

.

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U(T,P) for Real SubstancesU(T,P) for Real Substances

Real Substances: Kenaikan T meningkatkan gerakan molekuler a.l. vibrasi, rotasi dan

trans as , an uga a an mena an . Kenaikan P sedikit menurunkan U pada sebagian besar nilaiT dan P.

Fenomena ini terjadi karena interaksi molekuler yang kompleks.

UUUUEnergi Internal

sangat dipengaruhi

oleh T

Energi Internal

kurang dipengaruhioleh P

U = U(T, P)

U as a Function of T at Constant P U as a Function of P at Constant T

TT PP4


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U(T) for Ideal GasesU(T) for Ideal Gases

Ideal Gases: Kenaikan T meningkatkan gerakan molekuler vibrasi, rotasi dan translasi.KenaikanT juga menaikkan U.Perubahan P tidak mempengaruhi U karena tidak adanya interaksi

molekuler.U = U(T)

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U(T) for Incompressible LiquidsU(T) for Incompressible Liquids

Incompressible Liquids & Solids :Kenaikan T meningkatkan gerakan molekuler secara vibrasi, rotasi and translasi.

KenaikanT juga meningkatkan U.

Perubahan P tidak mempengaruhi U karena volume molar dari senyawa incompressibletidak dipengaruhi oleh perubahan tekanan.Konsekuensinya, tidak ada perubahan pada ekstensinya dan sifat interkasi molekuler dan

karena itu, U tidak berubah juga.

U = U(T)

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Internal Energy_ Energi dalam (E dalam) adalah total energi kinetik (Ek) dan energi potensial (Ep) yang ada di

dalam sistem. Formula E = Ek + Ep.

tidak dapat diukur, maka besar energi dalam sebuah sistem juga tidak dapatditentukan, yang dapat ditentukan adalah besar perubahan energi dalam(E) suatu sistem.

Perubahan energi dalam dapat diketahui dengan mengukur kalor (q) dankerja (w), yang akan timbul bila suatu sistem bereaksi. Oleh karena itu,perubahan energi dalam dirumuskan dengan persamaan :

E = q - w. q

Jika sistem menyerap kalor, q +.

Jika sistem mengeluarkan kalor q –

w Jika sistem melakukan kerja, w+

Jika sistem dikenai kerja oleh lingkungan, w-

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Energi dalam

Jadi bila suatu sistemmenyerap kalor darilingkungan sebesar 10 kJ, dansistem tersebut juga

E = q - w

melakukan kerja sebesar 6 kJ,maka perubahan energidalam-nya akan sebesar 16 kJ.

q =+10 kJ w = - 6 kJ

w + (dilakukan) jika jumlah kalor yang masuk = jumlah kerja yang dilakukan,

PERUBAHANPERUBAHAN ENERGIENERGI DALAMDALAM BERNILAIBERNILAI 00

SISTEM

SISTEMq + (masuk)

q – (dikeluarkan)

dan

jika jumlah kalor yang

dikeluarkan = jumlah kerja yangdikenakan pada sistem.

Artinya, tidak ada PERUBAHAN ENERGI DALAM yang terjadi pada sistem.

SISTEM w - (dikenakan)

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Enthalpy Entalpi adalah istilah dalam termodinamika yang menyatakan jumlah energi

internal dari suatu sistem termodinamika ditambah energi yang digunakanuntuk melakukan kerja.

Entalpi tidak bisa diukur, tetapi nilai perubahannya bisa dihitung.

Secara matematis, perubahan entalpi dapat dirumuskan sebagai berikut:

ΔH = ΔU + PΔV

di mana:

H = entalpi sistem (joule)

U = energi dalam (joule)

P = tekanan dari sistem (Pa)

V = volume sistem (m3)

http://id.wikipedia.org/wiki/Entalpi"9

Enthalpy reaksi

Entalpi = H

= Kalor reaksi pada tekanan tetap

Perubahan entalpi adalah

perubahan energi yang menyertai peristiwa perubahan kimiapada tekanan tetap.

a. Pemutusan ikatan membutuhkan energi (= endoterm)

Contoh: H2 2H - A kJ ;ΔH= + A kJ

b. Pembentukan ikatan memberikan energi (= eksoterm) onto : 2 ; = -

http://id.wikipedia.org/wiki/Entalpi"10


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Entalpi Pembentukan Standar ( Hf ):

H untuk membentuk 1 mol persenyawaan langsung dari unsur-unsurnyayang diukur pada 298 oK dan tekanan 1 atm.

Contoh: H2(g) + 1/2 O2(g) H20 (l) ; ΔHf = -241.8 kJ

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Entalpi Penguraian Standar:

H dari penguraian 1 mol persenyawaan langsung menjadi unsur-unsurnya

(= Kebalikan dari ΔH pembentukan).

Contoh: H2O (l) H2(g) + 1/2 O2(g) ; ΔH = +241.8 kJ

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Entalpi Pembakaran Standar (

Hc ):

H untuk membakar 1 mol suatu senyawa dengan O2 dari udara yangdiukur pada keadaan standar (298 oK dan tekanan 1 atm).

Contoh: CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ; Hc = -802 kJ

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Entalpi Reaksi:

H dari suatu persamaan reaksi, di mana zat-zat yang terdapat dalampersamaan reaksi dinyatakan dalam satuan mol dan koefisien-koefisien

persamaan rea s arus u a se er ana.

Contoh: 2Al + 3H2SO4 Al2(SO4)3 + 3H2 ; H = -1468 kJ

http://wps.prenhall.com/wps/media/objects/3080/3154819/blb0507.html14


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Entalpi Netralisasi:

H yang dihasilkan (selalu eksoterm) pada reaksi penetralan asam ataubasa.

Contoh: NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) ; H = - 890.4 kJ/mol

Hukum Lavoisier-Laplace:

"Jumlah kalor yang dilepaskan pada pembentukan 1 mol zat dari unsur-

unsurya = jumlah kalor yang diperlukan untuk menguraikan zat tersebutmenjadi unsur-unsur pembentuknya.“

Contoh:

N2(g) + 3H2(g) 2NH3(g) ; H = - 92.220 J2NH3(g) N2(g) + 3H2(g) ; H = + 92.220 J

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Table: Enthalpy of Formation (

Hf )

http://wps.prenhall.com/wps/media/objects/3080/3154819/blb0507.html16


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Table: Enthalpy of Formation (Hf ), Gibbs Function (G), and Absolute

Entropy (S) of Various Substances at 298 oK dan tekanan 1 atm

Hf G SSubstance Formula

Hf dan G(kJ/kmol) ;

S (kJ/kmol.oK)

http://schoolworkhelper.net/2010/07/standard-enthalpies-of-formation/17

Table: Enthalpy of Formation (Hf ), Gibbs Function (G), and AbsoluteEntropy (S) of Various Substances at 298 oK dan tekanan 1 atm

Hf G SSubstance Formula

http://schoolworkhelper.net/2010/07/standard-enthalpies-of-formation/

Hf dan G

(kJ/kmol) ;

S (kJ/kmol.oK)

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Determine the standard heat of each of the following

reactions at 298.15 K (25°C)

Ref. Smith V Ness p. 14419

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Definition of Enthalpy Definition of Enthalpy

Enthalpy, H, is a thermodynamic property, or state variable, and it is definedby:

EnthalpyEnthalpy(kJ)(kJ)

Molar EnthalpyMolar Enthalpy

Differential FormDifferential Form Integral FormIntegral Form

= + .

Specific EnthalpySpecific Enthalpy

(kJ/kg)(kJ/kg)

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H(T,P) for Real Substances

Real SubstancesU is a strong function of T

H is a function of both T & P

Note : At most, but not all values of T & P, H increases as P increases

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H(T) for Ideal Gases

Ideal Gases

Molar enthalpy:

Specific enthalpy:

Since U of an ideal gas is a function of T only and R and MW are constants, H of an idealgas is also a function of T only.

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H(T,P) for Incompressible Liquids andSolids

U is a function of T only, but

H is a function of T

and also a weak function of P

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Lesson Summary_Lesson 2A

CHAPTER 3, LESSON A - INTERNAL ENERGY & ENTHALPY

In this lesson we studied the functional dependence of internal energy, U, on both T and P.

We considered three cases: real gases, ideal gases, and liquids and solids.

Next, we defined a new state variable called enthalpy, H.

We then considered and discussed the functional dependence of enthalpy on both T and P.

The conclusions that we drew from these two studies are summarized in the table, below.

REAL GASES - U is a strong function of T and a weak function of P.H is a function of both T and P.

IDEAL GASES - U is a function of T only.

H is a function of T onl . Incompressible

Liquids and Solids - U is a function of T only.H is a function of both T and P.

In the next lesson, we will learn how to obtain values for both specific and molar internalenergy and enthalpy from tables of data and databases.

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Sample Data from NIST WebBook

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Thermophysical Properties of Fluid Systems

Density •Choose the chemical species for which youwant data.

Specific volume

Entropy

Cp

Internal energy

Cv

Speed of Sound

(Example: Water)•Choose the system of units that you would

like to use.(Example: K, MPa, kJ...)

•Choose the type of thermodynamic data you

want to obtain.(Example: Saturation Properties - Temperature

Increments) •Choose a standard state convention, which we

call a reference state. (We will learn more

about reference states later in this lesson.)•When you have finished steps 1-4 click the

"Press to Continue" button.

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Liquid phase data

Data on saturation curve

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What is a Reference State ?

Every table or graph ofthermodynamic data has a standardstate convention or a referencestate associated with it. But why ?

It is not possible to determine theabsolute value of U or H. We canonly measure changes in U or H. Inthis regard, enthalpy and internalenergy are much like altitude.

The altitude of a mountain and thealtimeters on airplanes are

referenced to mean sea level.A reference state is also needed when tabulating the measured changes in U or H. A

reference state is usually specified by the temperature, pressure, and the phase atwhich either U or H is zero.

Then, U and H of every other state is equal to the change in U or H as the substance

changes from the reference state to the state of interest30


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Review : U & H as Functions of T & P

In this part of the lesson, we will plot enthalpy and internal energy data

from the NIST WebBook . We will compare the results to what weexpect for an incompressible liquid and for an ideal gas.

Recall the conclusions we reached in the previous lesson:

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Enthalpy & Internal Energy for RealSubstances

Now, we can obtain data from the NIST WebBook for water to show thetrends observed for a real liquid. We will obtain two tables of data:

. , , ,Type of data: Isothermal propertiesStandard State: ASHRAE Convention

T = 30oC, P = 1 to 10 atm by 0.5 atm

2. Units: oC, atm, kg/m3, kJ/kgType of data: Isobaric propertiesStandard State: ASHRAE Convention

P = 1 atm, T = 5 to 95o

C by 5o

C

If you like, go ahead and obtain the data from NIST WebBook and plotthe data in Excel to see the expected trends.

ASHRAE_American Society for Heating Refrigeration Air Conditioning Engineer

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http://www.spiraxsarco.com/resources/steam-

engineering-tutorials/steam-engineering-principles-and-heat-transfer/entropy-a-basic-understanding.asp

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The temperature/enthalpy diagramThe enthalpy/pressure diagram

NIST Thermodynamic Data for SubcooledWater

Subcooled liquid tables for water obtained from the NIST WebBook

The tables were imported into Excel and edited.

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U(P) & H(P) for Subcooled Water atConstant T

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Comparison of Real and IncompressibleSubstances

Real Liquid (water) Incompressible Liquid

no qu s comp e e y ncompress e

Incompressible Liquid: U at constant T is not a function of P. H increases as P increases.Real Liquid : U can increase slightly, or like water, U can decrease slowly as P increases.

H increases as P increases.36


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U(T) and H(T) for Liquid Water atConstant P

Why does it seem like internal energyand enthalpy are the same here?

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Comparison of U & H at Constant P

Why does it seem like internal energyand enthalpy are the same here?

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Thermodynamic Properties of an Ideal Gas

Ideal Gases U is a function of T only.

H is a function of T only.

Now let's obtain data from the NIST WebBook for hydrogen andcompare the trends we observe to the behavior we would expect for anideal gas.

1. Units: oC, atm, kg/m3, kJ/kgType of data: Isothermal properties

Standard State: ASHRAE ConventionT = 30oC, P = 1 to 10 atm by 0.5 atm

2. Units: oC, atm, kg/m3, kJ/kgType of data: Isobaric properties

Standard State: ASHRAE ConventionP = 1 atm, T = 5 to 95oC by 5oC

If you like, go ahead and obtain the data from NIST WebBook and plotthe data in Excel to see the expected trends.

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Thermodynamic Properties of Hydrogen

Thermodynamic data for hydrogen obtained from the NIST WebBook , thenimported into Excel and edited.

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U and H of Hydrogen at Constant T

Hydrogen data obtained from NIST WebBook , then imported into Excel, edited,

and plotted.

To an accuracy of three significant figures in and ,

hydrogen can be considered an ideal gas over this range of P.41

U and H of Hydrogen at Constant P

Hydrogen data obtained from NIST WebBook , then imported into Excel ,edited, and plotted.

Does this plot show the expected trends for

an ideal gas at constant pressure?42


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Comparison of U & H at Constant P

Does this plot show the expected trends for an ideal gas at constant pressure?

Internal energy is a function oftemperature. plus RT/MW gives

the specific enthalpy

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Does this plotshow theexpected trendsor an ea gas aconstant pressure?

The differencebetween the Hand the U curvesgrows linearly astemperatureincreases.

Conclusion:

Hydrogen doesbehave like anideal gas at1 atm between5oC and 95oC

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Lesson Summary_Lesson 3B

CHAPTER 3, LESSON B - THERMO PROPERTIES: NIST WEBBOOK

In this lesson we learned that NIST WebBook is a valuable tool to determinethermodynamic properties. First we showed, step-by-step, how to locate the NIST

.

Be sure to bookmark the link to the Thermophysical Properties of Fluid Systems athttp://webbook.nist.gov/chemistry/fluid/.

We considered how to specify what information we want to obtain such as thechemical species, phase, units for properties, and the reference state. We learned

that a reference state is needed when tabulating the measured changes in internal

energy or enthalpy. Often a reference state is fixed by the temperature, pressure,and phase.

We then plotted enthalpy and internal energy data obtained from the NIST

WebBook for an incompressible liquid, an ideal gas, and a real substance. From

this we reinforced internal energy and enthalpy trends we learned in lesson B.

We will be using NIST WebBook to obtain most of the data for this course. Inthe next lesson, we will learn how to use NIST WebBook to obtain data on a

new property called heat capacity.

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The Relationship Between Heat andTemperature

When heating a substance, we must add a certain amount of heat to raise

the temperature a specified amount. This capacity to take in heat is uniqueto each substance and is defined with the symbol C.

Ch3 Lesson C Page 1 - The Relationship Between Heat and

Temperature

If we add a Joule of energy to one gram of liquid water at 20oC itwill raise the temperature to about 20.2oC.If we add 1 Joule of energy to one gram of liquid mercury at 20oC itwill raise the temperature to about 27oC.

the temperature to about 21oC.This capacity for absorbing heat is called the heat capacity or thespecific heat.Flip the page and we’ll show you the formal definitions for heatcapacity and specific heat.Then, we’ll show you how to use them to solve problems

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Definition of Constant V and P HeatCapacities

The molar heat capacity of a substance is defined as the energy required to

raise the temperature of a mole of a substance by one degree.

The specific heat of a substance is defined as the energy required to raisethe temperature of a unit mass of a substance by one degree.

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Ideal Gas Specific Heats

Recall from the previous lesson that the internal energy of an ideal gas is a

function of T only: and

Therefore, the heat capacities (or specific heats) for ideal gases reduce to:

Often, ideal gas specific heats and heat capacities are approximated aspolynomials in terms of T:

where a, b, c, and d are constants for agiven substance

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Gibbs Phase Rule

where:

oF = C - P + 2

oF is the number of degrees of freedom or the number of intensive propertiesthat canbe independently specified. All other intensive properties are then fixed and can bedetermined.

C is the number of different chemical substances in the system.

P is the number of distinct phases within the system.

Intensive Properties: Do not depend on the size of the system.Examples include:

.Examples include:

Since we are dealing with pure substances that exist in a single-phase,(C = 1 and P =1),

Therefore: oF = 1 - 1 + 2 = 2 We only need to fix or specify 2 state variables in order to completely define the state of our

system.

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Change in U as a Function of Ideal-GasCv

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Change in H as a Function of Ideal-GasCp

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Cp and Cv Relationship for an Ideal Gas

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Cp and Cv for Liquids & Solids

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Ideal-Gas Cp Data from NIST WebBook

Here is our heatcapacity polynomial

When you scroll down a bit farther on the

Results Page of the NIST Webbook, you will

find the constants for the heat capacitypolynomial for Ammonia shown in a table like

this:54


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3C-1 : Enthalpy Change of Ammonia Using

the IG Heat Capacity

Determine the enthalpy change and internal energy change of ammoniain J/mole, as it is heated from 350 oK to 600 oK, using the ideal gas heatca acit iven b the Shomate E uation.

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Isothermal Vaporization of WaterExample #3C-1

The temperature of 10 lbm of water is held constant at 205°F.The pressure

is reduced from a very high value until vaporization is complete.Determinethe final volume of the steam in ft3.

2C-1 : Specific Volume of Saturated Mixtures 4 pts

Calculate the specific volume for the following situations:

a.) Water at 200oC and 80% quality

b.) Freon 12 at -60oC and 90% quality

c.) Ammonia at 500 kPa and 85% quality

Read : This is an exercise designed to drive home the meaning and use of

the new concept of the quality of a saturated mixture. It is crucial toremember that if a quality is given, then the system contains a saturatedmixture and you probably need to look up properties of both the saturated

liquid and the saturated vapor.

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Given : a.) Water :

o

x 0.80 kg vap/kg tot

b.) R-12 :

T -60 oC


c.) NH3 :

a


Find : V ??? m3

/kg for each of the three parts of thisproblem.

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Solution Part a.) Data from the Saturation Temperature Table of the Steam Tables at200oC: Vsatliq 0.001156 m

3/kg

Vsatvap 0.1274 m3/kg

P* 1554 kPa

The key equation for this problem is the relationship between the properties of asaturated mixture and the properties of saturated liquid and vapor and the quality.

Eqn 1

Now, we can plug numbers into Eqn 1 to answer this part of the question.

V 0.1022 m /kg

Notice that I kept 4 significan figures in this answer istead of the usual 2 or 3 becausethere are 4 significant digits in the sat'd liquid and sat'd vapor values.

Perhaps I should have only retained 2 significant digits because there only appear to be 2significant digits in the quality. I have assumed that there are more than 2, really 4 ormore, digits in the quality. This may not be a good assumption.

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Part b.) Data from the Saturation Temperature Table of the R-12 Tables at -

60oC: Vsatliq 0.000637 m

3/kg

V 0.63791 m3/k

P* 22.6 kPa

Now, we can plug numbers into Eqn 1 to answer this part of the question. V 0.57418 m3/kg

Significant figures are a bit tricky here.

Part c.) Data from the Saturation Pressure Table of the Ammonia Tables at500 kPa: Vsatliq 0.00158 m

3/kg

V 0.25032 m3/k

Tsat 4.1396oC Now,

we can plug numbers into Eqn 1 to answer this part of the question. V 0.21301 m3/kg

Significant figures are a bit tricky here as well.

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Lesson Summary_Lesson 3C CHAPTER 3, LESSON C - HEAT CAPACITIES

In this lesson, we learned about new properties called heat capacity and specific heat.

Heat capacity is the energy required to raise the temperature of a mole of a substance by onedegree. Specific heat is the energy required to raise the temperature of a unit mass of a substance

.

Next, we studied the heat capacity of an ideal gas. We used the superscript o to identify heatcapacities that apply only to ideal gases. We derived the following relationships for ideal gases:

Ideal-gas heat capacities are often expressed as polynomials, such as:

where a, b, c, and d are constants for a given substance.

Next, we considered the heat capacities of liquids and solids.We found that over a moderate range of temperatures and pressures.

Finally, we learned how to obtain heat capacity polynomials for a variety of substances fromtheNISTWebBook .

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What is a State ?

Gibbs Phase Rule

oF = C - P + 2

F = derajat kebebasan

Initial State1

= um a senyawa m a a amsistem

P = Jumlah fase dalam sistem

of a system are fixed and can bemeasured or calculated

Ch3 Lesson D Page 1 - What is a State ?

Chapter 1

menjelaskan keadaan yang diberikanseluruh sifat materi didalam batasan sitsem yang ditetapkan

Chapter 2Gibbs Phase Ruleuntuk menetapkan berapa banyak sifat-sifat intensif untuk menspesifikasi keadaan sistem

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What is a State ?

Untuk menggunakan Gibbs Phase Rule, harus tahu spesies kimia apa

yang ada dalam sistem dan juga fase apa yang ada dalam sistem.

Jika jumlah sifat-sifat intensif yang telah dispesifikasi= jumlah derajat

kebebasan maka sifat-sifat intensif sistem tersebut dapat ditentukan.

Jika air murni ada dalam sistem, maka

C=air murni = 1

P = fase = cair = 1

F=C-P+2

=1-1+2 = 2

= apat spes as var a e ntens .

Yaitu T dan P

Karena bermanfaat dan mudah untuk mengukur.

Kemungkinan juga densitas dan entalpi spesifik karena juga sifatintensif.

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What Happens When a Property of a StateChanges ?

property changes, thesystem is at a new state1

2

At a given state all the properties

of a system are fixed and can be

measured or calculated

Initial State

Ch3 Lesson D Page 2 - What Happens

When a Property of a State Changes ?

any

any

proper y o e sys em c anges, en

the system is in a different state.

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What is a Process Path ?

Process Path:A process path is the actual series of states that the system passes through as it moves

from the initial state to the final state during a process

Initial State

When the value of a property changes, the

system is at a new state

Ch3 Lesson D Page 3 What is a Process Path ?

At a given state all theproperties of a system

are fixed and can be

measured orcalculated.

Ch3 Lesson D Page 3 - What is a Process Path ?

A system can continue to change one or more of its properties.As it does so, it moves from one state to another.The series of states through which a system moves.During a process, a system moves from an initial state to a final state,but its properties do not abruptly jump from the values at the initialstate and instantly take the values associated with the final state.The properties of the system change smoothly as the process progressesfrom the initial state to the final state.The series of states in which the system exists during the process iscalled the process path.

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State Variables & Hypothetical ProcessPaths

Process Path A process path is the actual series of states that the systempasses through as it moves from the initial state to the finalstate durin a rocess

Final State

.

The system occupies an infinite number of states between the

initial state and the final state. We use a smooth curve torepresent the smooth path.

Initial State Changes in state variables are not path dependent.

Therefore, when solving problems we can choose any path thatconnects the initial and final states.

The trick is to choose a path that makes the problem easy tosolve. This path is called a hypothetical process path (HPP).

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You Must Be Able to …

Final State

Initial State 1. Ideal Gases

2. Liquids and solids for which3. Real substances for which tables

of thermodynamic properties

are available.

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Ideal Gas: Change in H Using an HPP

What hypothetical process pathwill allow us to easily calculate the change

in enthalpy of the air?

'

final state, as shown in the diagram above.Determine the change in the molar enthalpy of the air for this process.

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HPP for an Ideal Gas

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Change in H for an Ideal Gas at Constant

T

Therefore, state 1 isan ideal gas state.

From state 1 to state2, the pressure

decreases but we are

at constanttemperature.

Recall from Lesson A

that for an ideal gas:

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Change in H for an Ideal Gas at ConstantT

From state 3 to state 4 the pressure increasesbut we are at constant temperature.

Therefore, state 4 is also an ideal gas state72


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Change in H for an Ideal Gas at ConstantP

From state 2 to state 3, the temperature

increases at constant pressure.

Therefore:

Step 2-3 involves an ideal gas being heated from 25oC to 125oC ata constant pressure.That’s exactly like what we studied in the previous lesson !So, ΔH-wiggle 2-3 is just the integral of the ideal gas Cp from T2 toT3.Now, all we need to do is obtain ideal gas Cp data for air. Doesthat sound familiar ?

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Heat Capacity Polynomial Data fromNIST WebBook

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Summary of HPP for an Ideal Gas

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Ideal Gas HPP Simplification

Ch3 Lesson D Page 13 - Ideal Gas HPP SimplificationSo, we figured out that for a process in which the initial and final states are both idealgas states, changes in pressure have no affect on the change in enthalpy.

Therefore, to determine ΔH-wiggle for these processes, all we need to do is integratethe ideal gas heat capacity from the initial temperature to the final temperature.

Cool. That’s not so hard.

What do we do if one or more of our states is NOT an ideal gas state ?Flip the page and we will begin to deal with that problem.

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Properties of Solids & Liquids

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HPP's for Liquids and Solids

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Heat Capacities of Solids and Liquids

Solids and Liquids:

If a polynomial for the heat capacity is available

is is generally of the form:

Often, just one value of is known. In this case

use the known value as79

Thermodynamic Properties of RealSubstances

The best way to determine a value for a change in any property is to look up the

value of the property for the initial and final states in a thermodynamic table and.

In this course, this is the only method presented that is applicable for non-idealgases !

Final State

Use themodynamic tables for liquids whenever possible because they aremore accurate than the methods discussed on the previous three pages

How much error is involved if we assume a gas to be ideal when it is not ?

Initial State

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Change in H for a Real Gas: Steam

Ch3, Lesson D, Page 18 - Change in H for a Real Gas: Steam

Let’s consider a process in which steam is compressed from 100 kPa to 1 MPa.During this process, the temperature of the steam rises from 100oC to 200oC.B now, ou robabl have some intuition about the tem eratures and ressure at which ases can be assumed tobe ideal.If you check the molar volumes at the initial and final states, you will see that the initial state can be safely assumedto be an ideal gas state, but the final state CANNOT.The easiest way to determine ΔH-hat for this process is just to look up H-hat for the initial and final states in the

NIST WebBook, and that is exactly what you should do in practice.But just to make a point, let’s construct an HPP and do a little extra work on this problem. We might findsomething interesting.Flip the page and let’s see.

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HPP for a Real Substance

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Summary of the Solution for a RealSubstance

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What If We Treated Steam as an IdealGas?

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Error from the Ideal Gas Assumption

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Lesson Summary_Lesson 3D CHAPTER 3, LESSON D - HYPOTHETICAL PROCESS PATHS

We began this lesson by refreshing your memory about states and defineda process path as the actual series of states that the system passesthrough as it moves from the initial state to the final state during aprocess. We emphasized that state variables are not dependent on theactual process path and therefore a hypothetical process path (HPP) canbe used to determine changes in thermodynamic properties during aprocess. We focused on determining the change in enthalpy and internalenergy for processes.

You must be able to determine and

for:1. Ideal Gases

2. qu s an so s or w c3. Real substances for which tables of thermodynamic properties are available.

For ideal gases, all you need to do is integrate Cv or Cp from T1 to T2 todetermine ΔU or ΔH. We found that Cp data for solids and liquids islimited and Cv data is not available at all. For liquids, the Cp equation isoften linear and for solids, a single Cv value is the norm.

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HPP's for Phase Changes

In the previous lesson, we used hypothetical process paths

to make it easier to determine the change in the properties of an ideal gas,,

have thermodynamic tables.

Final State

BUT only in one phase !

Initial State

In this lesson, we will learn how to usehypothetical process paths to determine thechange in properties associated with a phasechange.

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What is Enthalpy of Vaporization ?

Vaporization (or boiling): Saturated Liquid Saturated Vapor

Enthalpy of Vaporization

( latent heat ofvaporization):the amount of energy that must

be added to the s stem toconvert 1 mole of saturatedliquid into 1 mole of saturated

vapor

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Is the Heat of Vaporization Constant

Would it take the same amount of energy to boil 1 kg of water atsea level it would at an elevation of 14,000 ft?

What is the Heat of Vaporization is at the critical point ?

Berapa banyak energi yangharus diberikan untuk

menguapkan a kg cair jenuh jika proses terjadi

padaT & P kritis ?

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Saturation, Vaporization and VaporPressure

The latent heat of vaporization

didefiniskan sebagai jumlah energi yang diinginkan untuk menguapkan cairan secara sempurna pada Tsat (temperatur jenuh)dan P* (uap atau tekanan jenuh).

Tsat = Saturation temperatureP* = Vapor pressure or saturation pressure

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Heat of Vaporization from the NISTWebBook

In Chapter 3, Lesson C we learned how to obtain data (including enthalpy) fromNIST WebBook

To determine the latent heat of vaporizationwe the difference between the two enthalpy values.

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Vaporization of Subcooled Water

Subcooled Liquid : T < Tsator: P > P*

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Heat Liquid Water to the Tsat

Evaluated at:

Integration using Excel, followed by a unit conversion,

yields

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Enthalpy Change for Vaporization ofSubcooled Water

We determined the latent heat of vaporization

of water at 100oC and 1 atm on page 5 of this lesson:

On the previous page, we determined the enthalpy

change associated with heating of water from 75oC to

100oC at a constant pressure of 1 atm:

The total change in enthalpy associated with the change

from State 1 to State 3 is the sum of these two values:

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Vaporization of Subcooled Water:Shortcut Method

Since we have NIST WebBook

as a tool, we can determine the enthalpy at State 1 and State 3 and thendetermine the difference. We get the following values:

Enthalpy of Vaporization

The change in enthalpy is:

This is the same answer obtained from using the hypothetical process path !95

The Clapeyron Equation

For many processes, such as distillation, it is very important to know howvapor pressure, P*, depends on temperature, T.

.

Where:

The equation also allows us to determine the the latent heat of

vaporization

when we know , P*, and T for the phase change96


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The Clausius-Clapeyron Equation

We will start with the Clapeyron Equation

At moderate pressure, not close to the critica point,

Therefore:

If we also assume we have an ideal gs because we

are at moderate pressure, we can substitute:

A little algebra and calculus helps us put the equation

into a form that we can easily integrate:

Integrating, we obtain the Clausius-Clapeyron Equation:

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Using The Clausius-Clapeyron Equation

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Other Latent Heats

This lesson has focused on vaporization because it is the most important

phase change in this course.

HPP's can be used in the same way to calculate the change in propertiesthat occur when melting and sublimation phase changes occur in aprocess.

Latent Heat of Vaporization

amount of energy absorbed during vaporization

amount of energy released during condensation

Latent Heat of Fusion

amount of energy released during freezing

Latent Heat of Sublimation

amount of energy absorbed during sublimation

amount of energy released during desublimation

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Lesson Summary_Lesson 3E CHAPTER 3, LESSON E - PHASE CHANGES

Pada pelajaran ini, kami belajar bagaimana variabel keadaan berubah selama perubahan fase.Kami menggunakan alur proses hipotetik (hypothetical process paths_ HPP's) untuk membantumenetapkan perubahan-perubahan dalam sifat termodinamis selama proses perubahan fase.Kami menentukan perubahan dalam entalpi karena variabel tersebut merupakan variabelkeadaan didalam proses perubahan fase.

Kami fokus belajar pada proses penguapan. Panas laten penguapan yaitu jumlah energi yangharus menjadi input agar supaya menguapkan a mol atau a kg cairan. Sumber data panaspenguapan yatiu

NIST dll dan data dari NIST WebBook menunjukkan bahwa ΔHvap turun sebagaimanakenaikan temperatur terhadapTc.

Kami menggunakan HPP dan NIST WebBook untuk menghitung perubahan entalpi terkait engan penguapan c a r enu an ca r ng n engan mengguna an e n yang er e a.

Kemudian, kami mendiskusikan hubungan antara Tsat, P* dan ΔHvap. Kami mulai dariClapeyron Equation dan turunan Clausius-Clapeyron Equation.

Kami mendiskusikan penggunaan-penggunaan Clausius-Clapeyron Equation danmemperkenalkan perhitungan untuk memprediksi tekanan uap jenuh sebagai fungsi T.

2 perubahan fase yang umum : sublimation dan fusion

Thermotk 2011gs III Heat Effects

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