Solutions to Problem Set 6

Similar documents
The Second Law of Thermodynamics (Chapter 4)

Problem: Calculate the entropy change that results from mixing 54.0 g of water at 280 K with 27.0 g of water at 360 K in a vessel whose walls are

Last Name or Student ID

Chem 75 Winter, 2017 Practice Exam 1

Solutions to Problem Set 9

Thermodynamic Variables and Relations

Module 5 : Electrochemistry Lecture 21 : Review Of Thermodynamics

MME 2010 METALLURGICAL THERMODYNAMICS II. Fundamentals of Thermodynamics for Systems of Constant Composition

Homework Problem Set 8 Solutions

Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry

1 mol ideal gas, PV=RT, show the entropy can be written as! S = C v. lnt + RlnV + cons tant

dg = V dp - S dt (1.1) 2) There are two T ds equations that are useful in the analysis of thermodynamic systems. The first of these

Enthalpy and Adiabatic Changes


Ch. 19 Entropy and Free Energy: Spontaneous Change

Lecture 4 Clausius Inequality

I affirm that I have never given nor received aid on this examination. I understand that cheating in the exam will result in a grade F for the class.

CHEMICAL ENGINEERING THERMODYNAMICS. Andrew S. Rosen

Physics is time symmetric Nature is not

Class XI Chapter 6 Thermodynamics Chemistry

Lecture 6 Free energy and its uses

Physical Chemistry I Exam points

Lecture 4 Clausius Inequality

CHAPTER 3 LECTURE NOTES 3.1. The Carnot Cycle Consider the following reversible cyclic process involving one mole of an ideal gas:

Chem Lecture Notes 6 Fall 2013 Second law

Thermodynamic condition for equilibrium between two phases a and b is G a = G b, so that during an equilibrium phase change, G ab = G a G b = 0.

Chapter 5. Simple Mixtures Fall Semester Physical Chemistry 1 (CHM2201)

ESCI 341 Atmospheric Thermodynamics Lesson 12 The Energy Minimum Principle


Thermodynamics. Chem 36 Spring The study of energy changes which accompany physical and chemical processes

Practice Examinations Chem 393 Fall 2005 Time 1 hr 15 min for each set.

Identify the intensive quantities from the following: (a) enthalpy (b) volume (c) refractive index (d) none of these

Physical Biochemistry. Kwan Hee Lee, Ph.D. Handong Global University

Chemistry. Lecture 10 Maxwell Relations. NC State University

Concentrating on the system

The Chemical Potential

Thermodynamics. For the process to occur under adiabatic conditions, the correct condition is: (iii) q = 0. (iv) = 0

Some properties of the Helmholtz free energy

OCN 623: Thermodynamic Laws & Gibbs Free Energy. or how to predict chemical reactions without doing experiments

Classical Thermodynamics. Dr. Massimo Mella School of Chemistry Cardiff University

NCERT THERMODYNAMICS SOLUTION

where R = universal gas constant R = PV/nT R = atm L mol R = atm dm 3 mol 1 K 1 R = J mol 1 K 1 (SI unit)

THERMODYNAMICS I. TERMS AND DEFINITIONS A. Review of Definitions 1. Thermodynamics = Study of the exchange of heat, energy and work between a system

1. State in your own terms what is the first law of thermodynamics, a closed system, an isolated system, surroundings, heat, work, and energy.

Lecture 6 Free Energy

For more info visit

Chapter 3. Property Relations The essence of macroscopic thermodynamics Dependence of U, H, S, G, and F on T, P, V, etc.

From what we know now (i.e, ΔH and ΔS) How do we determine whether a reaction is spontaneous?

Phase Diagrams. NC State University

ln( P vap(s) / torr) = T / K ln( P vap(l) / torr) = T / K

du = δq + δw = δq rev + δw rev = δq rev + 0

Liquids and Solids. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry S H 2 = S H 2 R ln P H2 P NH

The Standard Gibbs Energy Change, G

Potential Concept Questions

MS212 Thermodynamics of Materials ( 소재열역학의이해 ) Lecture Note: Chapter 7

Outline Review Example Problem 1 Example Problem 2. Thermodynamics. Review and Example Problems. X Bai. SDSMT, Physics. Fall 2013

Chapter 19 Chemical Thermodynamics

Challa Vijaya Kumar University of Connecticut Module 4. Physical Chemistry 1 (Thermodynamics) Module 4. Open Source Textbook. Challa Vijaya Kumar

Lecture 3 Evaluation of Entropy

Chem 75 February, 2017 Practice Exam 2

Chapter 19 Chemical Thermodynamics Entropy and free energy

Reversible Processes. Furthermore, there must be no friction (i.e. mechanical energy loss) or turbulence i.e. it must be infinitely slow.

THERMODYNAMICS. Dr. Sapna Gupta

Outline Review Example Problem 1. Thermodynamics. Review and Example Problems: Part-2. X Bai. SDSMT, Physics. Fall 2014

Lecture 3 Clausius Inequality

Pressure Volume Work 2

You MUST sign the honor pledge:

Chemical Thermodynamics

Chemistry 163B Absolute Entropies and Entropy of Mixing

CHEMISTRY 202 Practice Hour Exam II. Dr. D. DeCoste T.A (60 pts.) 21 (40 pts.) 22 (20 pts.)

3.20 Exam 1 Fall 2003 SOLUTIONS

CH10007/87. Thermodynamics. Dr Toby Jenkins

AP CHEMISTRY 2007 SCORING GUIDELINES (Form B)

Thermodynamic Third class Dr. Arkan J. Hadi

Chapter 2 First Law Formalism

Lecture Notes 1: Physical Equilibria Vapor Pressure

Answers to Assigned Problems from Chapter 2

10, Physical Chemistry- III (Classical Thermodynamics, Non-Equilibrium Thermodynamics, Surface chemistry, Fast kinetics)

Solution: 1) Energy lost by the hot water: q = m C p ΔT. q = (72.55 g) (4.184 J/g 1 C 1 ) (24.3 C) q = J. 2) Energy gained by the cold water:

AAE THERMOCHEMISTRY BASICS

BCIT Fall Chem Exam #2

Outline of the Course

Answers to assigned problems in Engel and Reid

Chapter 7. Entropy. by Asst.Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn

4) It is a state function because enthalpy(h), entropy(s) and temperature (T) are state functions.

I.G Approach to Equilibrium and Thermodynamic Potentials

10/1/ st Law of Thermodynamics (Law of Conservation of Energy) & Hess s Law. Learning Targets

Chapter 7. Entropy: A Measure of Disorder

Work and heat. Expansion Work. Heat Transactions. Chapter 2 of Atkins: The First Law: Concepts. Sections of Atkins

Exam 2 Solutions. for a gas obeying the equation of state. Z = PV m RT = 1 + BP + CP 2,

Phase Equilibrium: Preliminaries

Thermochemistry. Energy and Chemical Change

Physical Chemistry I FINAL EXAM SOLUTIONS

THERMODYNAMICS. Extensive properties Intensive properties

Pressure Volume Temperature Relationship of Pure Fluids

Hence. The second law describes the direction of energy transfer in spontaneous processes

P(N,V,T) = NRT V. = P(N,V,T) dv

Chpt 19: Chemical. Thermodynamics. Thermodynamics

Disorder and Entropy. Disorder and Entropy

Transcription:

Solutions to Problem Set 6 1. non- ideal gas, 1 mol 20.0 L 300 K 40.0 L 300 K isothermal, reversible Equation of state: (a)b is a constant independent of T Given du = ( U/ T) V dt + ( U/ V) T dv U = U(T,V) 1 dt = 0 Given, isothermal 2 du = ( U/ V) T dv 3 Derived from first and second law of thermodynamics 4 ( p/ T) V = {RT[V -1 + BV -2 ]} = R[V -1 + BV -2 ] = T R[V -1 + BV -2 ] - RT[V -1 + BV -2 ] = 0 and using cross derivatives (see lecture notes Part 4) Apply to this equation of state, where B (has units of volume) is indep of T ΔU = 0 From Eq 1, 2 and 5 6 dh = ( H/ T) p dt + ( H/ p) T dp H = H(T,p) We could do this: ( H/ p) T = - T( V/ T) p + V Derived from first and second law of thermodynamics and using cross derivatives (see lecture notes Part 4) But equation of state is too complex to derive( V/ T) p. Use another route to get ΔH H = U + pv Definition 7 ΔH = ΔU + p f V f -p i V i p = RT/V + BRT/V 2 p i = (0.0820578)(300)[(1/20.0) + (B/20.0 2 )] p f = (0.0820578)(300)[(1/40.0) + (B/40.0 2 )] Δ(pV) = (0.0820578)(300)B(-0.025) ΔH = 0+(0.0820578)(300)B (-0.025) = (8.31451J mol -1 K -1 )(300)B (-0.025) Substituting known values ΔU = q + W First law of thermodynamics 10 q REV = - W REV From Eq 6 and 10 11 5 8 9

W = - p op dv Definition 12 p op = p gas reversible 13 W REV = - [RT/V + BRT/V 2 ]dv = - RT ln(v f /V i ) +2BRT[(1/V f )-(1/V i )] Using equation of state for this gas and integrating 14 W REV = -RT[ ln(2) + 2B (-0.025)] Substituting 15 = - (8.31451J mol -1 K -1 )(300) [ln(2) -0.050B] = - q REV ds = δq REV /T ΔS = q REV /T Second Law of thermodynamics reversible isothermal 16 17 ΔS = + (8.31451J mol -1 K -1 ) From Eq 15 and 17 18 [ln(2) -0.050B] A = U - TS Definition 19 ΔA = ΔU - TΔS For constant T (this problem) 20 ΔA = 0 300 (8.31451J mol -1 K -1 ) From Eq 6, 18 and 20 21 [ln(2) -0.050B] G = H - TS Definition 21 ΔG = ΔH - TΔS For constant T (this problem) 22 ΔG = 300 (8.31451J mol -1 K -1 ) [-0.025B - ln(2) +0.050B] = 300 (8.31451J mol -1 K -1 ) From Eq 9, 18 and 22 23 [0.025B - ln(2) ] (b) B is dependent on T B = a + bt + ct 2 Given 24 Just as in deriving Eq 5 but with the additional term 25 in (db/dt) Where db/dt = b + 2cT ( p/ T) V = {RT[V -1 + BV -2 ]} = R[V -1 + BV -2 ] + RV -2 (db/dt) = RT[V -1 + BV -2 ] - RT[V -1 + BV -2 ] +RTV -2 (db/dt) ( U/ V) T = RTV -2 (db/dt) ΔU = ( U/ V) T dv = RT(dB/dT) V -2 dv = RT(dB/dT)[(1/V f ) (1/V i )] = (8.31451J mol -1 K -1 )(300) (b + 600c) [- 0.025] p = RT/V + BRT/V 2 p i = (0.0820578)(300)[(1/20.0) + (B/20.0 2 )] p f = (0.0820578)(300)[(1/40.0) + (B/40.0 2 )] Δ(pV) = (0.0820578)(300)B(-0.025) From Eq 3 and 25 Note that ΔU 0 when B is dependent on T The equation for Δ(pV) is the same as when B is independent of T 26 27

ΔH = (8.31451J mol -1 K -1 )(300) (b + 600c) (-0.025) + (8.31451J mol -1 K -1 )(300)B (-0.025) The same as in part (a) except that ΔU from Eq 26 and B = a + bt + ct 2 are inserted. 28 = - (8.31451J mol -1 K -1 )(300) 0.025 [a+bt+ct 2 + b + 600c] ΔS = + (8.31451J mol -1 K -1 ) [ln(2) -0.050B] ΔA = ΔU - TΔS ΔA = (8.31451J mol -1 K -1 )(300) (b + 600c) (- 0.025) 300 (8.31451J mol -1 K -1 ) [ln(2) -0.050B] ΔG = ΔH - TΔS ΔG = - (8.31451J mol -1 K -1 )(300) 0.025 [a+bt+ct 2 + b + 600c] 300 (8.31451J mol -1 K -1 ) [ln(2) -0.050B] ΔG = + (8.31451J mol -1 K -1 )(300) [0.025 (a+bt+ct 2 ) - 0.025(b + 600c)- ln(2)] (c) B is a constant independent of temperature and the process is carried out irreversibly ΔU = 0 ΔH = - (8.31451J mol -1 K -1 )(300)(0.025)B ΔS = + (8.31451J mol -1 K -1 ) [ln(2) -0.050B] ΔA = 300 (8.31451J mol -1 K -1 ) [ln(2) -0.050B] ΔG = 300 (8.31451J mol -1 K -1 ) [0.025B - ln(2) ] But q IRREV and W IRREV are not the same as in part (a). q IRREV + W IRREV = 0 q IRREV < TΔS W IRREV > W REV W IRREV > -TΔS The equation is derived the same as in part (a) except that B = a + bt + ct 2 is inserted The same as in part (a) except that ΔU from Eq 26 and B = a + bt + ct 2 are inserted. ΔH from Eq 28 and B = a + bt + ct 2 are inserted. 31 All are exactly the same as in part (a) because these are state functions and do not depend on the path (reversible or otherwise). What do we know about the non-state functions? Because ΔU = 0 here because of Clausius inequality: TΔS > q IRREV Algebraic relation! Remember W is a signed quantity. Negative value means work is done by system on the surroundings. Less work than maximum work appears in the surroundings for irreversible process 29 30

2. non- ideal gas, 2 mol 10 atm 600 K 5 atm 300 K irreversible Equation of state: pv = nrt + nbp B = 0.035 L mol -1 J mol -1 K -1 du = ( U/ T) V dt + ( U/ V) T dv U = U(T,V) 1 Derived from first and second law of thermodynamics 2 and using cross derivatives (see lecture notes Part 4) p = nrt/(v-nb) Given equation of state 3 ( p/ T) V = nr/(v-nb) Apply to this equation of state, where B (has units of volume) is independent of T 4 = nrt/(v-nb) - nrt/(v-nb) = 0 du = C V dt + 0 From Eq 1 and 4 5 General relation derived from U = U(T,V) and the first 6 law of thermodynamics (see lecture notes Part 2) V-nB = nrt/p Rearrange equation of state and differentiate 7 ( V/ T) p = nr/p C p -C V = [p +0](nR/p) = nr From Eq 6, 7 and 4 8 C V = C p nr = n[52.675 From Eq 8 and given C p 9 + 1.561 10-2 T- 8.31451] J mol -1 K -1 C V = 2 mol (44.361+ 1.561 10-2 T) J mol -1 K -1 ΔU = 2 mol Integrating Eq 5 and using given C V 10 (44.361 + 1.561 10-2 T)dT ΔU = 2{44.361(T 2 -T 1 ) + (1/2)1.561 10-2 (T 2 2 -T 2 1 )} = 2{(44.361(300-600) + (1/2)1.561 10-2 (300 2-600 2 )} = 2{-13308.3-2107.3} Integrating Eq 5 and using given C V 11 = - 30830 J dh = ( H/ T) p dt + ( H/ p) T dp H = H(T,p) 12 ( H/ p) T = - T( V/ T) p + V. Derived from first and second law of thermodynamics 13 and using cross derivatives (see lecture notes Part 4) ( V/ T) p = nr/p and V = nrt/p +nb From Eq 7 and Equation of state substituted into Eq 13 ( H/ p) T = nb 14 dh = C p dt + nb dp ΔH = n( 52.675 + 1.561 10-2 T)dT + nbdp From Eq 12 and 14 15

ΔH = 2{52.675(T 2 -T 1 ) + (1/2)1.561 10-2 (T 2 2 -T 2 1 )} +2 0.035 L mol -1 [p 2 -p 1 ] = 2{(52.675(300-600) + (1/2)1.561 10-2 (300 2-600 2 )} +2 0.035[5-10] 8.31451/0.0820578 = 2{-15802.5-2107.3-17.7 } = - 35850 J Integrating and substituting known values We would have obtained the same answer by using H = U+pV ΔH = ΔU + Δ(pV) It happens that dv=0 here ds = ( S/ T) p dt + ( S/ p) T dp S = S(T,p) 12 Derived (see lecture notes Part 4) starting from 13 and use of cross derivatives ds = (C p /T)dT - nrdp/p Using Eq 7 and 2 into Eq 12 14 ΔS = n( 52.675/T + 1.561 10-2 )dt Using Eq 7 and 2 into Eq 12 - nr dp/p 15 ΔS = 2{ Integrating and substituting given values 16 52.675ln(T 2 /T 1 )+1.561 10-2 (T 2 -T 1 ) - 8.31451ln(p 2 /p 1) } ΔS = 2{ 52.675ln(300/600) +1.561 10-2 (300-600) We would have obtained the same answer by using - 8.31451ln(5/10) } ds = ( S/ T) V dt + ( S/ V) T dv ( S/ V) T = ( p/ T) V but it happens that dv=0 here, so ΔS = - 52.12 J K -1 ΔS = C V dt/t = same answer A = U TS ΔA = ΔU - Δ(TS) S abs, f = S abs,i + ΔS Δ(TS) = 300S abs,i - 600[S abs, i + ΔS ] ΔA = - 30830 J - 300S abs, i - 600[S abs, i - 52.12] G = H -TS ΔG = ΔH - Δ(TS) S abs, f = S abs,i + ΔS Δ(TS) = 300S abs,i - 600[S abs, i + ΔS ] ΔG = - 35850 J - 300S abs, i - 600[S abs, i - 52.12] Definition We already have ΔU. Since dt is not zero for this problem, it requires that we know absolute entropy. We have to have either the absolute entropy for the gas at the initial state or at the final state in order to calculate ΔA. If we are given data to calculate S abs, i, then we can do it. Definition We already have ΔH. As from above argument, we have to use the absolute entropy for the gas at the initial state and at the final state in order to calculate ΔG 8

3. non- ideal gas, 2 mol p i =1 atm T i =300 K p f =11 atm T f =300 K V i = a + RT i /p i V f = a + RT f /p f Equation of state: pv = RT +ap a = a(t) du = ( U/ T) V dt + ( U/ V) T dv U = U(T,V) 1 Derived from first and second law of thermodynamics 2 and using cross derivatives (see lecture notes Part 4) pv = RT +ap Given equation of state 3 p = RT/(V-a), a=a(t) ( p/ T) V = R/(V-a) + RT(da/dT)/(V-a) 2 Apply to this equation of state 4 = p + RT 2 (da/dt)/(v-a) 2 - p ( U/ V) T = RT 2 (da/dt)/(v-a) 2 Q.E.D. Using Eq 2 5 dh = ( H/ T) p dt + ( H/ p) T dp H = H(T,p) 6 ( H/ p) T = - T( V/ T) p + V. Derived from first and second law of thermodynamics 7 and using cross derivatives (see lecture notes Part 4) V = RT/p + a From Equation of state (Eq 3) ( V/ T) p = R/p + da/dt 8 ( H/ p) T = - T( V/ T) p + V = -RT/p -T(da/dT) + RT/p + a Substituting Eq 8 into Eq 7 ( H/ p) T = a -T(da/dT) Q.E.D. 9 We will need da/dt in Eq 5 and Eq 9 10 a = 0.02 + 0.005 10-2 (T-300) L mol -1 The given 3 values of a(t) changes linearly with T by 0.005 per 100 K increment (da/dt) = 0.005 10-2 L mol -1 K -1 du = ( U/ T) V dt + ( U/ V) T dv U = U(T,V) 1 du = C V dt + RT 2 (da/dt)/(v-a) 2 dv Substituting Eq 5 into Eq 1 11 In this problem dt = 0 ΔU = 2 mol Integrating Eq 11 RT 2 (da/dt) (V-a) - 2 dv = 2 mol RT 2 (da/dt) (-1)(V-a) -1 f i 12 = - 2 mol RT 2 f (da/dt) p/rt i Using equation of state to express (V-a) -1 = - 2 mol T(da/dT) [p f - p i ] 13 ΔU = - 2 mol (300 K)(0.005 10-2 L substituting values mol -1 K -1 ) [11-1] atm Using conversion factor 8.31451 J / 0.0820578 L atm ΔU = - 0.30 L atm or - 30.4 J 14 dh = ( H/ T) p dt + ( H/ p) T dp H = H(T,p) 6 dh = 0 + {a - T(da/dT)}dp Using dt = 0 and Eq 9 into Eq 6 15

ΔH = 2 mol {0.02 L mol -1 - (300 K)(0.005 10-2 L mol -1 K -1 )} [p f - p i ] ΔH = 2 mol 0.005 L mol -1 [11-1] atm ΔH = +0.10 L atm or +10.1 J W = - p op dv W max = - p gas dv = -2 mol RT (V-a) -1 dv = -2 mol RT ln [(V f - a)/(v i - a)] = -2 mol RT ln (p i /(p f ) W max = - 2 mol (8.31451 J mol -1 K -1 )(300K)ln(1/11) W max = +11963 J dg = Vdp - SdT dt = 0 for this process ΔG = Vdp = 2 mol (RT/p + a)dp ΔG =2 mol { RTln(p f /pi) + a[p f -pi] } = 2 mol {(8.31451 J mol -1 K -1 )(300K) ln(11/1) + 0.02 L mol -1 [11-1] atm 8.31451J /0.0820578 L atm} ΔG = +11963 J + 20.26 J ΔG = +11983 J da = - pdv - SdT dt = 0 for this process ΔA = - pdv = - RTdV/(V-a) = - 2 mol RT ln [(V f - a)/(v i - a)] = - 2 mol RT ln (p i /(p f ) ΔA = - 2 mol (8.31451 J mol -1 K -1 )(300K)ln(1/11) ΔA = +11963 J Integrating Eq 15 Substituting values Maximum work is obtained by carrying out the process reversibly, in which case p op = p gas Integrating Using the equation of state Substituting values one of the fundamental equations of thermodynamics (Lecture notes Part 5) Using the equation of state Integrating one of the fundamental equations of thermodynamics (Lecture notes Part 5) Integrating Using the equation of state Substituting the values Note that ΔA = W max, as it should be 16 17 18 19

4. 0.5 mol CH 4 (g) at 298 K is burned in excess O 2 under adiabatic conditions. T final =? The reaction at 298 K CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) Standard enthalpy of combustion at 298 K = Δ comb H 298 = 2Δ form H 298 (H 2 O(l)) +Δ form H 298 (CO 2 (g)) - Δ form H 298 (CH 4 (g)) - 2Δ form H 298 (O 2 (g)) Δ comb H 298 = 2Δ form H 298 (H 2 O(l) ) +Δ form H 298 (CO 2 (g) ) - Δ form H 298 (CH 4 (g) ) = 2(-285.83)+(-393.51)-(-74.81) kj mol -1 Δ comb H 298 = - 890.36 kj mol -1 Δ form H 298 for the elements in their standard state is defined as zero Using Δ form H 298 given in the provided table for CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) all at 298 K 1 2 (a) at constant pressure ~ 1 bar q p = 0 adiabatic (insulated container and piston) 0.5 mol CH 4 0.5 mol CO 2 (g) 10 mol O 2 [10-1] mol O 2 (g) 0.5(2) mol H 2 O(g) 298 K T f =? dh = ( H/ T) p dt + ( H/ p) T dp H = H(T,p) 3 dh = C p dt + 0 ΔH = q p q p = 0 Given dp = 0 Given adiabatic 4 5 6 ΔH = ½ Δ comb H 298 + 0.5 Tf 298 C p (CO 2 g)dt + 9 Tf 298 C p (O 2 g)dt + 1 373 298 C p (H 2 Oliq)dT + 1 Δ vap H 373 (H 2 O) + 1 Tf 373 C p (H 2 Og)dT = 0 Need to sum up all the contributions to ΔH 7

0.5(- 890.36 10 3 ) + 0.5(37.11)(T f - 298) + 9(29.355)(T f - 298) +1(75.291)(373-298) + 1(40.656 10 3 ) + 1(33.58)(T f - 373) = 0 Substitute values of ΔH in J mol -1 from Eq 2 and into Eq 7 and look up C p in J mol -1 K -1 for CO 2 (g), O 2 (g), H 2 O(liq) and H 2 O(g) in given table T f = 1567 K Solving Eq 8 for T f 9 (b) at constant volume adiabatic (insulated container) with rigid walls 8 initial final 0.5 mol CH 4 0.5 mol CO2(g) 10 mol O 2 [10-0.5] mol O2(g) 0.5(2) mol H 2 O(g) 298 K T f =? du = ( U/ T) V dt + ( U/ V) T dv U = U(T,V) 10 du = ( U/ T) V dt + 0 Given dv = 0 11 ΔU = q V 12 q V = 0 Given adiabatic 13 ΔU = ½ Δ comb U 298 Need to sum up all the contributions to ΔU 14 + 0.5 Tf 298 C V (CO 2 g)dt + 9 Tf 298 C V (O 2 g)dt + 1 373 298 C V (H 2 Oliq)dT + 1 Δ vap U 373 (H 2 O) + 1 Tf 373 C V (H 2 Og)dT = 0 H = U + pv Definition 15 ΔH = ΔU + Δ(pV) 16 Δ comb H 298 = Δ comb U 298 Apply Eq 16 to the reaction CH 4 (g) + 2O 2 (g) 17 +(pv) products - (pv) reactants (pv) products = (298 R) 1 (pv) reactants = (298 R) 3 (pv) products - (pv) reactants = - 2(298 R) = - 2(298)(8.31451) (pv) products - (pv) reactants = - 4955 J mol -1 Δ comb U 298 = - 890.36 10 3 - (-4955) = - 885.40 10 3 J mol -1 CO 2 (g) + 2H 2 O(l) at 298 K Assuming ideal gas behavior for reactants and products, neglecting pv for liquid 19 Substituting the values from Eq 2 & 19 into Eq 17 20 18 21

Δ vap H 373 = Δ vap U 373 + (pv) gas - (pv) liq Δ vap U 373 = Δ vap H 373 - Δ(pV) Δ(pV) = (373R) = 3101 J mol -1 Δ vap U 373 = (40.656 10 3 ) - 3101 Substituting the values = 37.554 10 3 J mol -1 ΔU = ½( - 885.40 10 3 ) + 0.5(28.8)(T f - 298) + 9(21.0)(T f - 298) +1(75.2)(373-298) + 1(37.554 10 3 ) + 1(25.3)(T f - 373) = 0 T f = 2053 K Solving Eq 26 for T f Note that this is a higher final temperature than the final temperature reached in doing the combustion at constant pressure. Apply Eq 16 to vaporization 22 Assuming ideal gas behavior for gas and neglecting pv for liquid Using values for Eq 21 & 25 into Eq 14 and looking up C V values in J mol -1 K -1 for CO 2 (g), O 2 (g), H 2 O(liq) and H 2 O(g) in given table This is because all of the chemical energy released by the reaction at constant volume went into raising the temperature of the products, whereas, some of the chemical energy released by the reaction at constant pressure went into Work against p op = 1 bar. 23 24 25 26 5. The formation of H 2 O(g) at 298 K H 2 (g) + ½O 2 (g) H 2 O(g) given Δ form H 298 = 241.82 kj mol -1 Initial Final 1 mol H 2 (g) (1-x) mol H 2 (g) 10 mol O 2 (g) (10-½x) mol O 2 (g) x mol H 2 O(g) T = 298 T = 2000 K constant pressure set up the problem as shown above Recognize that the reaction is that for the formation of H 2 O(g) dh = ( H/ T) p dt + ( H/ p) T dp H = H(T,p) 1 dh = C p dt + 0 ΔH = q p Given dp = 0 2 3 q p = 0 Assume that all the Δ rxn H is accounted for by the final 4 temperature of 2000 K, i.e., as if H 2 was burned in an insulated cylinder/piston at ~ 1 atm or 1 bar. ΔH = x Δ form H 298 + (1-x) 2000 298 C p (H 2 g)dt + (10-½x) 2000 298 C p (O 2 g)dt + x 2000 298 C p (H 2 Og)dT = 0 Need to sum up all the contributions to ΔH 5

x (- 241.82 10 3 ) + (1-x) (28.824)(2000-298) + (10-½x)(29.355)(2000-298) + x (33.58)(2000-298) = 0 x = 0.47 fraction of original H 2 that had undergone combustion Substitute value of Δ form H 298 in J mol -1 into Eq 5 and look up C p in J mol -1 K -1 for H 2 (g), O 2 (g), and H 2 O(g) in given table Solving Eq 6 for x 7 6 6. f /p of ethane as a function of pressure at 600 K Z = 1.0000 Given at 600 K 1-0.000612 (p/1 bar) + 2.661 10-6 (p/1bar) 2-1.390 10-9 (p/1bar) 3-1.077 10-13 (p/1bar) 4 ln(f /p) = p 0 [(Z-1)/p] T dp which we derived in class (see lecture notes Part 5) 2 (Z-1)/p = - 0.000612 From Eq 1 3 + 2.661 10-6 (p/1 bar) - 1.390 10-9 (p/1bar) 2-1.077 10-13 (p/1bar) 3 ln(f /p) = p 0 [(Z-1)/p] T dp = - 0.000612 (p/1 bar) +½ 2.661 10-6 (p/1bar) 2 - (1/3) 1.390 10-9 (p/1bar) 3 - (1/4) 1.077 10-13 (p/1bar) 4 Substituting Eq 3 into Eq 2 and integrating 4 (f /p) = exp{- 0.000612 (p/1 bar) +½ 2.661 10-6 (p/1bar) 2 - (1/3) 1.390 10-9 (p/1bar) 3 - (1/4) 1.077 10-13 (p/1bar) 4 } at 600 K Rearranging Eq 4 5

7. Distribution of a real gas in a gravity field F = mg F z+dz + df = F z See lecture notes Part 1 for starting this problem Pressure at any height z is determined by the column 1 2 of fluid above that height. Start withforce arising from the weight of fluid in the column at z (F z ) and compare that with the force arising from the weight of the fluid in the column at z+dz. Let ρ be the density at height z The mass of fluid in the column between z and z+dz is ρadz A is the cross sectional area of the column df = (ρadz)g 3 The pressure: p at height z = F z /A p + dp at height z+dz = F z+dz /A A(p+dp) +df = Ap df = -Adp Substituting Eq 4&5 into Eq 2 Rearranging 6 dp = - ρgdz Substituting Eq 6 into Eq 3 7 ρ = M/V where M is the molar mass and V is the molar volume dp = - (M/V) gdz substituting for ρ in Eq 7 8 dp/p = - (M/ZRT) gdz Using Z = pv/rt or V = ZRT/p into Eq 8 9 p p0 (dp/p) = - (M/ZRT) g z 0 dz ln(p/p 0 ) = - (Mg/ZRT) z p = p 0 exp[- (Mg/ZRT) z] (a) if Z > 1, exp[- (Mg/ZRT)z] > exp[- (Mg/RT)z] That is, need to go to larger height to drop from p 0 (at the surface) to the same pressure p it would have been had the gas been ideal. This means the gas distribution is broader when Z > 1. (b) if Z < 1, exp[- (Mg/ZRT)z] < exp[- (Mg/RT)z] That is, don t need to go as high to drop from p 0 to the same pressure p it would have been had the gas been ideal. This means the gas distribution is narrower Z < 1. Integrating between z = 0 and z 10 Comparing with ideal case where Z = 1 If Z = 1+Bp dp/p = - [M/(1+Bp)RT] gdz Substitute this into Eq 9 {(1+Bp)/p} dp = - [Mg/RT] dz Rearranging 4 5 11 12

p0 p (dp/p) + B p0 p dp = - (Mg/RT) 0 z dz ln(p/p 0 ) + B(p-p 0 ) = - (Mg/RT) z Integrating between z = 0 and z to be compared with the ideal case: ln(p/p 0 ) = - (Mg/RT) z 8. Fugacity of a gas Z = 1 + [b-(a/rt)](p/rt) (Z-1)/p = [b-(a/rt)]/rt Given Rearranging 1 2 ln(f /p) = p 0 [(Z-1)/p] T dp which we derived in class (see lecture notes Part 5) 3 ln(f /p) = p 0 {b-(a/rt)]/rt} T dp = [b-(a/rt)]/rt} T p Substituting Eq 2 into Eq 3 Integrating (f /p) = exp[(p/rt)[b-(a/rt)] f = p exp[(p/rt)[b-(a/rt)] 9. The actual chemical potential is Recall that the standard state for a gas corresponds 1 μ(t,p) = μ (T) + RTln (f/1bar) to a fictitious state of an ideal gas at 1 bar and temperature T for which the chemical potential is μ (T) μ(t,p) = μ (T) + Rln (f/1bar) Dividing Eq 1 by T 2 T T (1/n) G = (1/n) G + Rln (f/1bar) Recall that for a substance μ = G/n where n is the 3 T T number of moles (1/n)[ / T(G/T)] p = (1/n)[ / T(G /T)] p + R ln (f/1bar)/ T] p Take the derivative of both sides with respect to T at constant p [ / T(G/T)] p = - H/T 2 The Gibbs-Helmholtz equation (see lecture notes Part 5) - H/T 2 = - H /T 2 +R ln (f/1bar)/ T] p Substitute Eq 5 into Eq 4, where H and H are now molar quantities H = H - R T 2 ln (f/1bar)/ T] p H is H ideal since the standard state at temperature T Q.E.D. corresponds to a fictitious gas behaving ideally at 1 bar, whereas H is the molar enthalpy of the real gas. 4 5 6 7

2024 © sciencedocbox.com Privacy Policy | Terms of Service | Feedback