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

Chemical Engineering Thermodynamics

Chemical Engineering Thermodynamics

Prepared by

Prepared by:: DrDr NNINIEKINIEK FajarFajar PuspitaPuspita M Eng AugustM Eng August 20201111 Prepared by

Prepared by:: Dr. Dr. NNINIEKINIEK FajarFajar PuspitaPuspita, M.Eng August, , M.Eng August, 20201111

2011Gs_III_Heat Effects 2011Gs_III_Heat Effects

1

Lesson Topics Descriptions

Lesson 3A

Lesson 3A Internal Energy & E h l

Mendiskusikan definisi energi dalam dan entalpi untuk

l b id l i d d Enthalpy real substances, gas-gas ideal, cairan dan padatan

incompresible.

Lesson 3B

Lesson 3B Thermo Properties: NIST WebBook

Mempelajari bagaimana menggunakan NIST Webbook untuk mencapai data termodinamika.

Lesson 3C

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

L 3D

L 3D H th ti l M l j i l hi t tik t k t k

Lesson 3D

Lesson 3D Hypothetical

Process Paths Mempelajari alur proses hipotetik untuk menentukan perubahan sifat termal dari keadaan awal ke keadaan akhir.

Lesson 3E

Lesson 3E Phase Changes Mempelajari bahwa alur proses hipotetik untuk alat yang bermanfaat dalam mempertimbangakan perubahan sifat terkait dengan perubahan fase.

Three Principle Phases, Revisited

Three Principle Phases, Revisited

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

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

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

with three different types of motion: vibration, rotation and translation.

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

y Molecules move randomly with all three types of motion, but they are much closer together and cannot travel very far between collisions

phase.

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

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

the vibrational, rotational and translational motion of molecules.

closer together and cannot travel very far between collisions.

3

U(T,P) for Real Substances

U(T,P) for Real Substances

` Real Substances:

` Kenaikan T meningkatkan gerakan molekuler a.l. vibrasi, rotasi dan

l i d j k ikk U

translasi, dan juga akan menaikkan U.

` Kenaikan P sedikit menurunkan U pada sebagian besar nilai T dan P. ` Fenomena ini terjadi karena interaksi molekuler yang kompleks.

U

U Energi Internal _{sangat dipengaruhi} UU

oleh T Energi Internal kurang dipengaruhi oleh P U = U(T, P) TT PP

U(T) for Ideal Gases

U(T) for Ideal Gases

Ideal Gases: Kenaikan T meningkatkan gerakan molekuler vibrasi, rotasi dan translasi. Kenaikan T juga menaikkan U.

Perubahan P tidak mempengaruhi U karena tidak adanya interaksip g y molekuler.

U = U(T)

5

U(T) for Incompressible Liquids

U(T) for Incompressible Liquids

Incompressible Liquids & Solids :

9Kenaikan T meningkatkan gerakan molekuler secara vibrasi, rotasi and translasi. Kenaikan T juga meningkatkan U.

9Perubahan P tidak mempengaruhi U karena volume molar dari senyawa incompressible tidak dipengaruhi oleh perubahan tekanan.

9Konsekuensinya, tidak ada perubahan pada ekstensinya dan sifat interkasi molekuler dan karena itu, U tidak berubah juga.

U = U(T)

Internal Energy_

Energi dalam

` (E _dalam) adalah total energi kinetik(Ek) danenergi potensial(Ep) yang ada di dalamsistem.

` Formula E = Ek + Ep.

` Namun karena besar energi kinetik dan energi potensial pada sebuah sistem ` Namun karena besar energi kinetik dan energi potensial pada sebuah sistem tidak dapat diukur, maka besar energi dalam sebuah sistem juga tidak dapat ditentukan, yang dapat ditentukan adalah besarperubahan energi dalam (E)suatu sistem.

` Perubahan energi dalam dapat diketahui dengan mengukurkalor(q) dan kerja(w), yang akan timbul bila suatu sistem bereaksi. Oleh karena itu, perubahan energi dalam dirumuskan dengan persamaan :

…

E = q - w

.

` q qÆ

` Jika sistem menyerap kalor, Æq +. ` Jika sistem mengeluarkan kalorÆq –

` w Æ

` Jika sistem melakukan kerja, Æw+

` Jika sistem dikenai kerja oleh lingkungan, Æ

w-7

Energi dalam

¾ Jadi bila suatu sistem

menyerap kalor dari

lingkungan sebesar 10 kJ, dan sistem tersebut juga

E = q - w

j g

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

q =+10 kJ w = - 6 kJ

w + (dilakukan) ¾ jika jumlah kalor yang masuk=

jumlah kerja yang dilakukan,

PERUBAHAN

PERUBAHAN ENERGIENERGI DALAMDALAM BERNILAIBERNILAI 0 0

SISTEM

SISTEM q + (masuk)

q –(dikeluarkan) ¾ dan

¾ jika jumlah kalor yang

dikeluarkan=jumlah kerja yang dikenakan pada sistem.

¾ Artinya, tidak adaPERUBAHAN ENERGI DALAMyang terjadi pada sistem.

Enthalpy

` Entalpi adalah istilah dalamtermodinamikayang menyatakan jumlah energi internal dari suatu sistem termodinamika ditambah energi yang digunakan untuk 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 k d (P )

` 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 kimia pada tekanan tetap.

` a. Pemutusan ikatan membutuhkan energi (= endoterm)

Contoh:H₂Æ 2H - A kJ ; ΔH= + A kJ

` b. Pembentukan ikatan memberikan energi (= eksoterm)

C h 2H Æ H + A kJ ΔH A kJ Contoh:2H Æ H₂+ A kJ ; ΔH = - A kJ

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

` Entalpi Pembentukan Standar ( ΔH_f):

ΔH untuk membentuk 1 mol persenyawaan langsung dari unsur-unsurnya yang diukur pada 298 o_{K dan tekanan 1 atm.}

` Contoh:

H

₂

(g) + 1/2 O

₂

(g)

Æ

H

₂

0 (l) ;

Δ

H

_f

= -241.8 kJ

11

` Entalpi Penguraian Standar:

ΔHdari penguraian 1 mol persenyawaan langsung menjadi unsur-unsurnya (= Kebalikan dariΔH pembentukan).

` Entalpi Pembakaran Standar (ΔHc ):

ΔH untuk membakar 1 mol suatu senyawa dengan O₂dari udara yang diukur pada keadaan standar (298 o_{K dan tekanan 1 atm).}

` Contoh:

CH

₄

(g) + 2O

₂

(g) Æ

CO

₂

(g) + 2H

₂

O(l) ;

Δ

Hc = -802 kJ

13

` Entalpi Reaksi:

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

k i h b l t d h persamaan reaksi harus bulat sederhana.

` Contoh:

2Al + 3H

₂

SO

₄

Æ

Al

₂

(SO

₄

)

₃

+ 3H

₂

;

Δ

H = -1468 kJ

http://wps.prenhall.com/wps/media/objects/3080/3154819/blb0507.html 14

` Entalpi Netralisasi:

ΔHyang dihasilkan (selalu eksoterm) pada reaksi penetralan asam atau basa.

` Contoh:NaOH(aq) + HCl(aq) Æ NaCl(aq) + H<sub>2</sub>O(l) ; ΔH= - 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 tersebut menjadi unsur-unsur pembentuknya.“

` Contoh:

N₂(g) + 3H₂(g) Æ 2NH₃(g) ; ΔH= - 92.220 J 2NH₃(g) Æ N₂(g) + 3H₂(g) ; ΔH= + 92.220 J

15

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

Entropy (S) of Various Substances at 298 o_{K dan tekanan 1 atm}

H

_f

G

S

Substance

Formula

Hfdan G (kJ/kmol) ; S (kJ/kmol.o_K) http://schoolworkhelper.net/2010/07/standard-enthalpies-of-formation/ 17

Table: Enthalpy of Formation (H_f), Gibbs Function (G), and Absolute Entropy (S) of Various Substances at 298 o_{K dan tekanan 1 atm}

H

_f

G

S

Substance

Formula

http://schoolworkhelper.net/2010/07/standard-enthalpies-of-formation/ Hfdan G (kJ/kmol) ; S (kJ/kmol.o_K) 18

Determine the standard heat of each of the following

reactions at 298.15 K (25°C)

Ref. Smith V Ness p. 144

21

Definition of Enthalpy

Definition of Enthalpy

` Enthalpy, H, is a thermodynamic property, or state variable, and it is defined by:

H U PV

Enthalpy Enthalpy (kJ) (kJ) Molar Enthalpy Molar Enthalpy (kJ/mole) (kJ/mole) Differential Form

Differential Form Integral FormIntegral Form

H = U + PV.

(kJ/mole) (kJ/mole) Specific Enthalpy Specific Enthalpy (kJ/kg) (kJ/kg) 22

H(T,P) for Real Substances

Real Substances

U is a strong function of T U is a weak function of P U is a weak function of P H is a function of both T & P

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

23

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 ideal gas is also a function of T only.

H(T,P) for Incompressible Liquids and

Solids

U is a function of T only, but H is a function of T H is a function of T

and also a weak function of P

25

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 enthalpyon both T and P. ` The conclusions that we drew from these two studies are summarized in the table, below. ` REAL GASES -Æ U is a strong function of T and a weak function of P.

H is a function of both T and P.

` IDEAL GASES -ÆU is a function of T only.

H is a function of T only.y

` Incompressible

Liquids and Solids -Æ 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 internal

energyand enthalpyfrom tables of data and databases.

Sample Data from NIST WebBook

27

Thermophysical Properties of Fluid Systems

` Density

` Enthalpy

•Choose the chemical species for which you want data. ` Enthalpy ` 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)

Ch d d h h

•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.

Liquid phase data

`

Data on saturation curve

29

What is a Reference State ?

` Every table or graph of

thermodynamic data has a standard state conventionor a reference state associated with it. But why ? ` It is not possible to determine the absolute value of U or H. We can only measure changes in U or H. In this regard, enthalpy and internal energy are much like altitude. ` The altitude of a mountain and the

altimeters on airplanes are referenced to mean sea level. 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 at which 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 interest

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 we expect for an incompressible liquid and for an ideal gas.

` Recall the conclusions we reached in the previous lesson:

31

Enthalpy & Internal Energy for Real

Substances

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

` 1 Units: oC atm kg/m3 kJ/kg ` 1. Units: oC, atm, kg/m3, kJ/kg

Type of data:Isothermal properties Standard State: ASHRAE Convention

T = 30o_{C, P = 1 to 10 atm by 0.5 atm}

` 2. Units: oC, atm, kg/m3, kJ/kg Type of data: Isobaric properties Standard State: ASHRAE Convention

P = 1 atm, T = 5 to 95o_{C by 5}o_C

` If you like, go ahead and obtain the data from NIST WebBookand plot the data in Excel to see the expected trends.

http://www.spiraxsarco.com/resources/steam- engineering-tutorials/steam-engineering-principles-

and-heat-transfer/entropy-a-basic-understanding.asp

33

The temperature/enthalpy diagram The enthalpy/pressure diagram

NIST Thermodynamic Data for Subcooled

Water

` Subcooled liquid tables for water obtained from the NIST WebBook ` The tables were imported into Excel and edited.

U(P) & H(P) for Subcooled Water at

Constant T

35

Comparison of Real and Incompressible

Substances

Real Liquid (water) Incompressible Liquid ( li id i l t l i ibl ) (no liquid is completely incompressible)

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.

U(T) and H(T) for Liquid Water at

Constant P

` Why does it seem like internal energy and enthalpy are the same here?

37

Comparison of U & H at Constant P

y Why does it seem like internal energy and enthalpy are the same here? 38

Thermodynamic Properties of

an Ideal Gas

` Ideal GasesÆU is a function of T only.

H is a function of T only.

` Now let's obtain data from the NIST WebBookfor hydrogen and compare the trends we observe to the behavior we would expect for an ideal gas.

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

Standard State: ASHRAE Convention

T = 30o_{C, P = 1 to 10 atm by 0.5 atm}

` 2. Units: oC, atm, kg/m3, kJ/kg Type of data: Isobaric properties Standard State: ASHRAE Convention

P = 1 atm, T = 5 to 95o_{C by 5}o_C

` If you like, go ahead and obtain the data from NIST WebBookand plot the data in Excel to see the expected trends.

39

Thermodynamic Properties of Hydrogen

` Thermodynamic data for hydrogen obtained from the NIST WebBook, then

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 gasover 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?

Comparison of U & H at Constant P

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

Internal energy isa function of

temperature. plus RT/MWgives

the specific enthalpy 43

` Does this plot show the expected trends f id l t for an ideal gas at constant pressure? ` The difference

between the H and the U curves grows linearly as temperature increases. ` Conclusion: Hydrogen does y g behave like an ideal gas at 1 atm between 5oC and 95oC

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 determine thermodynamic properties. First we showed, step-by-step, how to locate the NIST website and the NIST WebBook within the website

website and the NIST WebBook within the website.

` Be sure to bookmark the link to the Thermophysical Properties of Fluid Systems at

http://webbook.nist.gov/chemistry/fluid/.

` We considered how to specify what information we want to obtain such as the chemical 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. In

the next lesson, we will learn how to use NIST WebBookto obtain data on a new property called heat capacity.

45

The Relationship Between Heat and

Temperature

` 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 unique to 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 it will raise the temperature to about 20.2oC.

If we add 1 Joule of energy to one gram of liquid mercury at 20oC it will raise the temperature to about 27oC.

If we add 1 Joule of energy to one gram of air at 20oC it will raise If we add 1 Joule of energy to one gram of air at 20oC it will raise the temperature to about 21oC.

This capacity for absorbing heat is called the heat capacity or the specific heat.

Flip the page and we’ll show you the formal definitions for heat capacity and specific heat.

Then, we’ll show you how to use them to solve problems

Definition of Constant V and P Heat

Capacities

` 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 raise the temperature of a unit mass of a substance by one degree.

47

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 as polynomials in terms of T:

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

Gibbs Phase Rule

` where:

o_{F = C - P + 2}

` oF is the number of degrees of freedomor the number of intensive propertiesthat can be independently specified. All other intensive properties are then fixed and can be

determined.

` 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:

` Extensive Properties: Do depend on the size of the system ` Extensive Properties: Do depend on the size of the system.

Examples include:

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

` Therefore: o_{F = 1 - 1 + 2 = 2}

` We only need to fix or specify 2 state variables in order to completely define the state of our system.

49

Change in U as a Function of Ideal-Gas

Cv

Change in H as a Function of Ideal-Gas

Cp

51

Cp and Cv for Liquids & Solids

53

Ideal-Gas Cp Data from NIST WebBook

Here is our heat capacity polynomial

When you scroll down a bit farther on the Results Page of the NIST Webbook, you will find the constants for the heat capacity polynomial for Ammonia shown in a table like this:₅₄

3C-1 : Enthalpy Change of Ammonia Using

the IG Heat Capacity

` Determine the enthalpy change and internal energy change of ammonia in J/mole, as it is heated from 350 o_{K to 600} o_{K, using the ideal gas heat} capacity given by the Shomate Equation.

capacity given by the Shomate Equation.

57

Isothermal Vaporization of Water

Example #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. Determine the 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 200o_{C and 80% quality}

` b.) Freon 12 at -60o_{C 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 to

f

remember that if a quality is given, then the system contains a saturated mixture and you probably need to look up properties of both the saturated liquid and the saturated vapor.

`

Given :

`

a.) Water :

` T 200 oC ` T 200 oC ` x 0.80 kg vap/kg tot `

b.) R-12 :

` T -60 oC ` x 0.90 kg vap/kg tot `

c.) NH

₃

:

` P 500 kPa ` P 500 kPa ` x 0.85 kg vap/kg tot

`

Find : V ??? m

3

/kg for each of the three parts of this

problem.

59

` Solution Part a.) Data from the Saturation Temperature Table of the Steam Tables at 200o_C:

` Vsat liqq0.001156 m3/kg

` V<sub>sat vap</sub>0.1274 m3/kg ` P* 1554 kPa

` The key equation for this problem is the relationship between the properties of a saturated 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 3_/k

` V 0.1022 m3/kg

` Notice that I kept 4 significan figures in this answer istead of the usual 2 or 3 because there 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 2 significant digits in the quality. I have assumed that there are more than 2, really 4 or more, digits in the quality. This may not be a good assumption.

` Part b.) Data from the Saturation Temperature Table of the R12 Tables at -60o_C:

` V<sub>sat liq</sub>0.000637 m3<sub>/kg</sub> ` V_{sat vapsat vap}0.63791 m3/kg g ` 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 at 500 kPa:

` V<sub>sat liq</sub>0.00158 m3/kg ` Vsat vapsat vap0.25032 m3/kg g ` Tsat4.1396 oC 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.

61

Lesson Summary_Lesson 3C

` CHAPTER 3, LESSON C - HEAT CAPACITIES

` In this lesson, we learned about new properties called heat capacityand specific heat. ` Heat capacity is the energy required to raise the temperature of a mole of a substance by one

degree. Specific heat is the energy required to raise the temperature of a unit mass of a substance by one degree

by one degree.

` Next, we studied the heat capacity of an ideal gas. We used the superscript o to identify heat

capacities that apply only to ideal gases. We derived the following relationships for ideal gases:

` Ideal-gas heat capacitiesare 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 from

theNIST WebBook.

What is a State ?

` Gibbs Phase Rule ` oF = C - P + 2

` F = derajat kebebasan

` C Jumlah senyawa kimia didalam At a given state all the properties

Initial State

1

` C = Jumlah senyawa kimia didalam

sistem

` P = Jumlah fase dalam sistem

At a given state all the properties of a system are fixed and can be measured or calculated

Ch3, Lesson D, Page 1 - What is a State ? Chapter 1

menjelaskan keadaan yang diberikan

seluruh sifat materi didalam batasan sitsem yang ditetapkan Chapter 2

Gibbs Phase Rule Æ

untuk menetapkan berapa banyak sifat-sifat intensif untuk menspesifikasi keadaan sistem

63

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

kebebasanÆmaka 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

F 2 Æd t di ifik i 2 i b l i t if ` F=2 Ædapat dispesifikasi 2 variabel intensif.

` Yaitu T dan P

` Karena bermanfaat dan mudah untuk mengukur.

` Kemungkinan juga densitas dan entalpi spesifik karena juga sifat

What Happens When a Property of a State

Changes ?

` When the value of a ` When the value of a property changes, the system is at a new state

1

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 ? If any property of the system changes then If any property of the system changes, then the system is in a different state.

65

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 processg p

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 the

properties of a system are fixed and can be measured or calculated.

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 initial state and instantly take the values associated with the final state. The properties of the system change smoothly as the process progresses from the initial state to the final state.

The series of states in which the system exists during the process is called the process path.

State Variables & Hypothetical Process

Paths

`

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

Final State state during a process.

The system occupies an infinite number of states between the initial state and the final state. We use a smooth curve to represent the smooth path.

Initial State Changes in state variables are not path dependent.

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

The trick is to choose a path that makes the problem easy to solve. 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 which 3. Real substances for which tables

f h d

of thermodynamic properties are available.

Ideal Gas: Change in H Using an HPP

What hypothetical process path

will allow us to easily calculate the change in enthalpy of the air?

Let's consider a process in which air changes from the initial state to the Let s consider a process in which air changes from the initial state to the 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

Change in H for an Ideal Gas at Constant

T

Therefore, state 1 is an ideal gas state. From state 1 to state 2, the pressure decreases but we are

t t t at constant temperature. Recall from Lesson A that for an ideal gas:

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Change in H for an Ideal Gas at Constant

T

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

Change in H for an Ideal Gas at Constant

P

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 at a 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 Cpfrom T2 to T3.

Now, all we need to do is obtain ideal gas Cpdata for air. Does that sound familiar ?

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Heat Capacity Polynomial Data from

NIST WebBook

Summary of HPP for an Ideal Gas

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Ideal Gas HPP Simplification

Ch3, Lesson D, Page 13 - Ideal Gas HPP Simplification

So, we figured out that for a process in which the initial and final states are both ideal gas 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 integrate the 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.

Properties of Solids & Liquids

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HPP's for Liquids and Solids

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 as

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Thermodynamic Properties of Real

Substances

‰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 calculate the difference between them

calculate the difference between them.

‰In this course, this is the only method presented that is applicable for non-ideal gases !

Final State

` Use themodynamic tables for liquids whenever possible because they are more 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 ?

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.

By now, you probably have some intuition about the temperatures and pressure at which gases can be assumed to y ow, you p obab y ave so e tu t o about t e te pe atu es a p essu e at w c gases ca be assu e to be ideal.

If you check the molar volumes at the initial and final states, you will see that the initial state can be safely assumed to 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 find something interesting.

Flip the page and let’s see. 81

HPP for a Real Substance

Summary of the Solution for a Real

Substance

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What If We Treated Steam as an Ideal

Gas?

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 defined a process path as the actual series of states that the system passes through as it moves from the initial state to the final state during a g g process. We emphasized that state variables are not dependent on the actual process path and therefore a hypothetical process path (HPP) can be used to determine changes in thermodynamic properties during a process. We focused on determining the change in enthalpy and internal energy for processes.

` You must be able to determine and ` for:

1. Ideal Gases

2 Li id d lid f hi h 2. Liquids and solids for which

3. 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 to determine ΔU or ΔH. We found that Cp data for solids and liquids is limited and Cv data is not available at all. For liquids, the Cp equation is often linear and for solids, a single Cv value is the norm.

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, liquids and solids for which is constant and real substances for which we liquids and solids for which is constant, and real substances for which we have thermodynamic tables.

Final State

PHASE CHANGE BUT only in one phase !

Initial State

PHASE CHANGE

In this lesson, we will learn how to use

hypothetical process paths to determine the change in properties associated with a phase change.

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What is Enthalpy of Vaporization ?

` Vaporization (or boiling): Saturated Liquid Saturated Vapor

… Condensation: Saturated Vapor Saturated Liquid … Condensation: Saturated Vapor Saturated Liquid

Enthalpy of Vaporization ( latent heat of

vaporization):

the amount of energy that must be added to the system to be added to the system to convert 1 mole of saturated liquid into 1 mole of saturated vapor

Is the Heat of Vaporization Constant

` Would it take the same amount of energy to boil 1 kg of water at

sea level it would at an elevation of 14,000 ft?

` What is the Heat of Vaporizationis at the critical point?

Berapa banyak energi yang harus diberikan untuk menguapkana kg cair jenuh jika proses terjadi padaT& Pkritis ?

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Saturation, Vaporization and Vapor

Pressure

` 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 temperature

P* = Vapor pressure or saturation pressure

Heat of Vaporization from the NIST

WebBook

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

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

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Vaporization of Subcooled Water

Subcooled Liquid : T < Tsat

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 of

Subcooled Water

We determined the latent heat of vaporization of water at 100o_{C and 1 atm on page 5of this lesson:}

On the previous page, we determined the enthalpy change associated with heating of water from 75o_{C to}

100o_{C 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:

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 then determine 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!

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The Clapeyron Equation

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

` The Clapeyron Equationprovides such a thermodynamic relationship ` The Clapeyron Equationprovides such a thermodynamic relationship.

Where:

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

The Clausius-Clapeyron Equation

We will start with the Clapeyron Equation

At moderatepressure, 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

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 properties that occur when melting and sublimationphase changes occur in a process.

` Latent Heat of Vaporization

` amount of energy absorbed during vaporization ` amount of energy released during condensation

` Latent Heat of Fusion

` amount of energy absorbed during melting ` amount of energy absorbed during melting ` 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 membantu menetapkan perubahan-perubahan dalam sifat termodinamis selama proses perubahan fase. p p p p p Kami menentukan perubahan dalam entalpi karena variabel tersebut merupakan variabel keadaan didalam proses perubahan fase.

` Kami fokus belajar pada proses penguapan. Panas laten penguapan yaitu jumlah energi yang

harus menjadi input agar supaya menguapkan a mol atau a kg cairan. Sumber data panas penguapan yatiu

` NIST dll dan data dari NIST WebBook menunjukkan bahwaΔHvap turun sebagaimana

kenaikan temperatur terhadap Tc.

` Kami menggunakan HPP dan NIST WebBook untuk menghitung perubahan entalpi terkait

d i j h d i di i d k 2 t k ik b b d

dengan penguapan cair jenuh dan cair dingin dengan menggunakan 2 teknik yang berbeda.

` Kemudian, kami mendiskusikan hubungan antara Tsat, P* danΔHvap. Kami mulai dari Clapeyron Equation dan turunan Clausius-Clapeyron Equation.

` Kami mendiskusikan penggunaan-penggunaan Clausius-Clapeyron Equation dan

memperkenalkan perhitungan untuk memprediksi tekanan uap jenuh sebagai fungsi T.