a) Use the following equation from the lecture notes: = ( J K 1 mol 1) ( ) 10 L


 Bertram Mosley
 1 years ago
 Views:
Transcription
1 hermodynamics: Examples for chapter One mole of nitrogen gas is allowed to expand from 0.5 to 10 L reversible and isothermal process at 300 K. Calculate the change in molar entropy using a the ideal gas law and b the van der Waals equation with a = atm L 2 mol 2, b = L mol 1. a Use the following equation from the lecture notes: S = nr ln V2 V 1 = J K 1 mol 1 10 L ln 0.5 L = J K 1 mol 1 b Start with the following equation see lecture notes: S he equation of state gives: S = P V P = nr V nb an2 P V 2 V = nr V 2 V nb S = nr V nb V 1 or S = R ln Plugging in the numerical values gives: V2 b V 1 b = nr V nb dv = nr ln V2 nb V 1 nb S = J K 1 mol 1 10 L 1 mol L mol 1 ln 0.5 mol 1 mol L mol 1 = J K 1 mol 1 1
2 2. Derive the relation for C P C V for a gas that follows the van der Waals equation. Based on the lecture notes we have: C P C V = V α 2 κ with α = 1 V V P and κ = 1 V V P And the van der Waals equation can be written in the following forms: P = R V b ā V or = 1 P 2 R V Pb + ā V ab V 2 Next α and κ must be calculated. Since the van der Waals equation of state cannot be written for the molar volume easily, one must use the reciprocal identity: V P = 1 / V P α 1 = V V P = V P R a R V + 2ab P V = 2 R V 3 R a R V + 2ab R V 2 It is convenient to eliminate pressure P from this equation by using the van der Waals equation: α 1 = Next we calculate κ: V V b 2a R V + 2ab R V 2 = α = V V b 1 V V V 2a b R V b 2 2 2a V b R V 2
3 κ 1 = V P V = V R V b 2 2a V 2 Plugging both α and κ into the expression for C P C V and simplifying, we get: C P C V = R 1 2a V b 2 R V 3 3. Estimate the change in molar entropy of liquid benzene at 25 C and atmospheric pressure when the pressure is raised to 1000 bar? he coefficient of thermal expansion α is K 1, the density is g cm 3, and the molar mass is g mol 1. Based on the lecture notes we have: S P = P Inserting the values into this equation we get: = V α S V α P S g mol m cm K Pa g cm 3 = J K 1 mol 1 4. Derive the expression for U/ the internal pressure for a gas following the virial equation with Z = 1 + B/V. Consider only PV  work and a reversible process. Note that B = B i.e. it depends on temperature. 3
4 Recall the first law of thermodynamics: du = dq + dw. For PV  work dw = PdV and by the second law: dq = ds. hus: du = ds PdV and dividing both sides by dv and imposing constant temperature: U = he equation of state gives: S P P = V P V R = 1 + B V P = R V + BR V 2 P Furthermore, differentiation of P with respect to gives: P V = R V + BR B R V + 2 V 2 By inserting these results into the expression derived above: U = R V + BR B V + 2 V R V 2 R V BR V 2 = B V R 2 V 2 5. Consider the process of freezing water at 10 C for which the enthalpy change is 5619 J mol 1 and entropy change is J K 1 mol 1. What is the Gibbs energy of freezing water at constant temperature of 10 C? he Gibbs energy can be written in terms of enthalpy: G = H S. At constant temperature we have: G = H S = 5619 J mol K J K 1 mol 1 = J mol 1. 4
5 6. a Integrate the GibbsHelmholtz to obtain an expression for G 2 at temperature 2 in terms of G 1 at 1, assuming that H is independent of temperature. b Obtain an expression for G 2 using a more accurate approximation that H = H C P, where 1 is an arbitrary reference temperature. Assume that C P is temperature independent. Hint: Integrate one side with respect to variable G/ and the other with respect to. a Integrate both sides of the GibbsHelmholtz equation see lecture notes: H 2 = G/ 2 H 2 d = G 2 / 2 d G/ 1 G 1 / 1 1 H 1 = G 2 / 2 G 1 / G 2 = H 2 + G 1 2 / b Assume temperature dependency for H = H C P : G 2 / 2 G 1 / 1 d G/ = 2 1 H 1 2 d C P G 2 G 1 1 = H 1 1 C P ln d 2 C P G 2 = G H C P 2 ln
6 7. An ideal gas is allowed to expand reversibly and isothermally 25 C from a pressure of 1 bar to a pressure of 0.1 bar. a What is the change in molar Gibbs energy? b What would be the change in molar Gibbs energy if the process occurred irreversibly? a Use the following equation see lecture notes: G 2 = R ln P2 P bar = J K 1 mol K ln 1 bar = kj mol 1 b Gibbs energy is a state function and depends only on the endpoints. hus the answer is the same as in a. 8. Helium is compressed isothermally and reversibly at 100 C from a pressure of 2 to 10 bar. Calculate a q per mole, b w per mole, c Ḡ, d Ā, e H, f Ū, g S, assuming that helium is an ideal gas. a and b Ideal gas: U = 0 and q = w. he integrated P V work is: w = R ln P2 P 1 10 bar = J K 1 mol K ln 2 bar = 4993 J mol 1 and hence q = 4993 J mol 1 c he following equation in the lecture notes yields: Ḡ = R ln P2 10 bar = J K 1 mol K ln P 1 2 bar = 4993 J mol 1 6
7 d A = U S and at constant : A = U S. U = 0 and A = S. Also by definition: G = U + PV S = U S = S. So Ḡ = Ā in this case and Ā = 4993 J mol 1. e H = U + PV = U = 0. f As already stated U = 0. g S = G/ = J K 1 mol oluene molecular weight g mol 1 is vaporized at constant external pressure and its boiling point, 111 C constant temperature. he heat of vaporization at this temperature is J g 1. For the vaporization of toluene, calculate a w per mole, b q per mole, c H, d Ū, e Ḡ, and f S. In part a assume that the volume of the liquid is negligible and that toluene vapor behaves according to the ideal gas law this assumption is not needed elsewhere. Note also that evaporation at the boiling point is a reversible process. a w = P V = PV fin V init = R 0 = J K 1 mol K = 3193 J mol 1. he final volume was obtained from the ideal gas law PV = nr. b and c For vaporization process we have q = H = J g g mol 1 = kj mol 1. d Ū = q+w = J mol J mol 1 = kj mol 1. Note that if ideal gas behavior was assumed here, U would have been zero as it would depend only on temperature. e and f G = U +PV S and hence G = U +V P +P V S S. At constant temperature and pressure P = = 0 and therefore G = U + P V S. From part a P V = 3193 J mol 1 and we need to calculate S. S = q rev / = kj mol 1 /384 K = 86.8 J K 1 mol 1. And finally G = 0 by combining all the required terms. 7
8 10. Calculate mix G and mix S for the formation of a quantity of air containing 1 mol of gas by mixing nitrogen and oxygen at K. Air may be taken to be 80% nitrogen and 20% oxygen in mol %. Assume ideal gas behavior. In the lecture notes, the following realtions were given: mix G = R n 1 lny 1 + n 2 lny 2 mix S = R n 1 lny 1 + n 2 lny 2 mix H = mix V = 0 n 1 = 0.80 mol of N 2 y 1 = 0.80 and n 2 = 0.20 mol of O 2 y 2 = By inserting the values in above equations, we get: mix G = 1239 J mol 1 and mix S = J K 1 mol 1. 8
= T T V V T = V. By using the relation given in the problem, we can write this as: ( P + T ( P/ T)V ) = T
hermodynamics: Examples for chapter 3. 1. Show that C / = 0 for a an ideal gas, b a van der Waals gas and c a gas following P = nr. Assume that the following result nb holds: U = P P Hint: In b and c,
More informationProblem Set 3 Solutions
Chemistry 360 Dr Jean M Standard Problem Set 3 Solutions 1 (a) One mole of an ideal gas at 98 K is expanded reversibly and isothermally from 10 L to 10 L Determine the amount of work in Joules We start
More informationProblem Set 2 Solutions
Chemistry 360 Dr. Jean M. Standard roblem Set Solutions 1. The atmospheric surface pressure on Venus is 90 bar. The atmosphere near the surface is composed of 96% carbon dioxide and 4% other gases. Given
More informationThermodynamics Answers to Tutorial # 1
Thermodynamics Answers to Tutorial # 1 1. (I) Work done in free expansion is Zero as P ex = 0 (II) Irreversible expansion against constant external pressure w = P ex (V 2 V 1 ) V 2 = nrt P 2 V 1 = nrt
More informationvap H = RT 1T 2 = 30.850 kj mol 1 100 kpa = 341 K
Thermodynamics: Examples for chapter 6. 1. The boiling point of hexane at 1 atm is 68.7 C. What is the boiling point at 1 bar? The vapor pressure of hexane at 49.6 C is 53.32 kpa. Assume that the vapor
More informationI 2 (g) 2I(g) 0.12 g 254 g mol 1 = mol
hermodynamics: Examples for chapter 1. 1. When 0.12 g of solid iodine was vaporized at 1670 K, the resulting gas displaced 20.2 cm 3 of dry air at 298 K and 99.9 kpa pressure. How many iodine molecules
More informationEntropy Changes & Processes
Entropy Changes & Processes Chapter 4 of Atkins: he Second Law: he Concepts Section 4.3, 7th edition; 3.3, 8th edition Entropy of Phase ransition at the ransition emperature Expansion of the Perfect Gas
More informationProblem Set 4 Solutions
Chemistry 360 Dr Jean M Standard Problem Set 4 Solutions 1 Two moles of an ideal gas are compressed isothermally and reversibly at 98 K from 1 atm to 00 atm Calculate q, w, ΔU, and ΔH For an isothermal
More informationTHERMODYNAMIC PROPERTIES AND CALCULATION. Academic Resource Center
THERMODYNAMIC PROPERTIES AND CALCULATION Academic Resource Center THERMODYNAMIC PROPERTIES A quantity which is either an attribute of an entire system or is a function of position which is continuous and
More informationApplied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur
Applied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture  7 Ideal Gas Laws, Different Processes Let us continue
More informationChem 338 Homework Set #2 solutions September 12, 2001 From Atkins: 2.8, 2.15, 2.16, 2.17, 2.18, 2.21, 2.23, 2.26
Chem 8 Homework Set # solutions September 1, 001 From Atkins:.8,.15,.16,.17,.18,.1,.,.6.8) A sample of methane of mass 4.50 g occupies 1.7 L at 10 K. (a) Calculate the work done when the gas expands isothermally
More informationFinal Exam CHM 3410, Dr. Mebel, Fall 2005
Final Exam CHM 3410, Dr. Mebel, Fall 2005 1. At 31.2 C, pure propane and nbutane have vapor pressures of 1200 and 200 Torr, respectively. (a) Calculate the mole fraction of propane in the liquid mixture
More informationAnswer, Key Homework 6 David McIntyre 1
Answer, Key Homework 6 David McIntyre 1 This printout should have 0 questions, check that it is complete. Multiplechoice questions may continue on the next column or page: find all choices before making
More information2.5(a) Enthalpy. Chapter 2. The First Law. P.27
2.5(a) Enthalpy Chapter 2. The First Law. P.27 Justification 2.1 The relation H = q p For a general infinitesimal change in the state of the system, U changes to U + du, p changes to p + dp, and V changes
More informationPROPERTIES OF PURE SUBSTANCES
Thermodynamics: An Engineering Approach Seventh Edition in SI Units Yunus A. Cengel, Michael A. Boles McGrawHill, 2011 Chapter 3 PROPERTIES OF PURE SUBSTANCES Mehmet Kanoglu University of Gaziantep Copyright
More informationExpansion and Compression of a Gas
Physics 6B  Winter 2011 Homework 4 Solutions Expansion and Compression of a Gas In an adiabatic process, there is no heat transferred to or from the system i.e. dq = 0. The first law of thermodynamics
More informationLesson 5 Review of fundamental principles Thermodynamics : Part II
Lesson 5 Review of fundamental principles Thermodynamics : Part II Version ME, IIT Kharagpur .The specific objectives are to:. State principles of evaluating thermodynamic properties of pure substances
More informationc. Applying the first law of thermodynamics from Equation 15.1, we find that c h c h.
Week 11 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution
More informationChapter 8 Maxwell relations and measurable properties
Chapter 8 Maxwell relations and measurable properties 8.1 Maxwell relations Other thermodynamic potentials emerging from Legendre transforms allow us to switch independent variables and give rise to alternate
More informationNext, solid silicon is separated from other solid impurities by treatment with hydrogen chloride at 350 C to form gaseous trichlorosilane (SiCl 3 H):
University Chemistry Quiz 5 2014/12/25 1. (5%) What is the coordination number of each sphere in (a) a simple cubic cell, (b) a bodycentered cubic cell, and (c) a facecentered cubic cell? Assume the
More informationEntropy and The Second Law of Thermodynamics
The Second Law of Thermodynamics (SL) Entropy and The Second Law of Thermodynamics Explain and manipulate the second law State and illustrate by example the second law of thermodynamics Write both the
More informationBasic problems for ideal gases and problems to the first law of thermodynamics and cycles
Basic problems for ideal gases and problems to the first law of thermodynamics and cycles SHORT SUMMARY OF THE FORMS: Equation of ideal gases: Number of moles: Universal gas constant: General gas equation
More informationGibbs Free Energy and Chemical Potential. NC State University
Chemistry 433 Lecture 14 Gibbs Free Energy and Chemical Potential NC State University The internal energy expressed in terms of its natural variables We can use the combination of the first and second
More informationProblem # 2 Determine the kinds of intermolecular forces present in each element or compound:
Chapter 11 Homework solutions Problem # 2 Determine the kinds of intermolecular forces present in each element or compound: A. Kr B. NCl 3 C. SiH 4 D. HF SOLUTION: Kr is a single atom, hence it can have
More informationExergy: the quality of energy N. Woudstra
Exergy: the quality of energy N. Woudstra Introduction Characteristic for our society is a massive consumption of goods and energy. Continuation of this way of life in the long term is only possible if
More informationProblem Set 1 3.20 MIT Professor Gerbrand Ceder Fall 2003
LEVEL 1 PROBLEMS Problem Set 1 3.0 MIT Professor Gerbrand Ceder Fall 003 Problem 1.1 The internal energy per kg for a certain gas is given by U = 0. 17 T + C where U is in kj/kg, T is in Kelvin, and C
More informationFinal Exam Review Questions PHY Final Chapters
Final Exam Review Questions PHY 2425  Final Chapters Section: 17 1 Topic: Thermal Equilibrium and Temperature Type: Numerical 12 A temperature of 14ºF is equivalent to A) 10ºC B) 7.77ºC C) 25.5ºC D) 26.7ºC
More information2.1. PURE SUBSTANCE:
Pure substance Phase change processes Use of thermodynamic tables Diagrams v, Pv Ideal gas equation 2.1. PURE SUBSANCE: Substance that has a fixed chemical composition throughout is called a PURE SUBSANCE.
More informationChemical Process calculation III
Chapter 7 Ideal and Real Gases Gas, Liquid, and Solid Chemical Process calculation III Gas: a substance in a form like air, relatively low in density and viscosity Liquid: a substance that flows freely
More informationProblems of Chapter 2
Section 2.1 Heat and Internal Energy Problems of Chapter 2 1 On his honeymoon James Joule traveled from England to Switzerland. He attempted to verify his idea of the interconvertibility of mechanical
More informationAssigned questions for Lecture 14 are listed below. The questions occur in the following editions of Physical Chemistry by P.W.
Assigned questions for Lecture 14 are listed below. The questions occur in the following editions of Physical Chemistry by P.W. Atkins: 10th edition 9th edition 8th edition Note: The letter P in front
More informationThe final numerical answer given is correct but the math shown does not give that answer.
Note added to Homework set 7: The solution to Problem 16 has an error in it. The specific heat of water is listed as c 1 J/g K but should be c 4.186 J/g K The final numerical answer given is correct but
More informationChapter 4 The Properties of Gases
Chapter 4 The Properties of Gases Significant Figure Convention At least one extra significant figure is displayed in all intermediate calculations. The final answer is expressed with the correct number
More informationH C p = T Enthalpy can be obtained from the Legendre transform of the internal energy as follows, U + T = P T (RT )
roblem Set 2 Solutions 3.20 MI rofessor Gerbrand Ceder Fall 2003 roblem 1.1 1 S(U, ) : ds du + d 1 ariables here are U and and intensive variables are and. o go to 1 as a natural variables take the Legendre
More informationStatistical Mechanics, Kinetic Theory Ideal Gas. 8.01t Nov 22, 2004
Statistical Mechanics, Kinetic Theory Ideal Gas 8.01t Nov 22, 2004 Statistical Mechanics and Thermodynamics Thermodynamics Old & Fundamental Understanding of Heat (I.e. Steam) Engines Part of Physics Einstein
More informationTHE KINETIC THEORY OF GASES
Chapter 19: THE KINETIC THEORY OF GASES 1. Evidence that a gas consists mostly of empty space is the fact that: A. the density of a gas becomes much greater when it is liquefied B. gases exert pressure
More informationEXERCISES. 16. What is the ionic strength in a solution containing NaCl in c=0.14 mol/dm 3 concentration and Na 3 PO 4 in 0.21 mol/dm 3 concentration?
EXERISES 1. The standard enthalpy of reaction is 512 kj/mol and the standard entropy of reaction is 1.60 kj/(k mol) for the denaturalization of a certain protein. Determine the temperature range where
More informationExtensive variables: volume, mass, energy Intensive variables: pressure, temperature, density
Properties of Gases If a gas is sufficiently dilute it obeys the ideal gas law The ideal gas law can also be written PV = nrt A molar quantity is indicated by the bar across the top. The ideal gas law
More informationGases. States of Matter. Molecular Arrangement Solid Small Small Ordered Liquid Unity Unity Local Order Gas High Large Chaotic (random)
Gases States of Matter States of Matter Kinetic E (motion) Potential E(interaction) Distance Between (size) Molecular Arrangement Solid Small Small Ordered Liquid Unity Unity Local Order Gas High Large
More informationName: SOLUTIONS. Physics 240, Exam #1 Sept (4:155:35)
Name: SOLUTIONS Physics 240, Exam #1 Sept. 24 2015 (4:155:35) Instructions: Complete all questions as best you can and show all of your work. If you just write down an answer without explaining how you
More informationChemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten. Chapter 10 Gases
Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 10 Gases A Gas Has neither a definite volume nor shape. Uniformly fills any container.
More informationPhysics Courseware Physics I Ideal Gases
Physics Courseware Physics I Ideal Gases Problem 1. How much mass of helium is contained in a 0.0 L cylinder at a pressure of.0 atm and a temperature of.0 C? [The atomic mass of helium is 4 amu] PV (
More informationEntropy. Objectives. MAE 320  Chapter 7. Definition of Entropy. Definition of Entropy. Definition of Entropy. Definition of Entropy + Δ
MAE 320  Chapter 7 Entropy Objectives Defe a new property called entropy to quantify the secondlaw effects. Establish the crease of entropy prciple. Calculate the entropy changes that take place durg
More informationThe First Law of Thermodynamics
The First Law of Thermodynamics (FL) The First Law of Thermodynamics Explain and manipulate the first law Write the integral and differential forms of the first law Describe the physical meaning of each
More informationReading. Spontaneity. Monday, January 30 CHEM 102H T. Hughbanks
Thermo Notes #3 Entropy and 2nd Law of Thermodynamics Monday, January 30 CHEM 102H T. Hughbanks Reading You should reading Chapter 7. Some of this material is quite challenging, be sure to read this material
More informationThe first law: transformation of energy into heat and work. Chemical reactions can be used to provide heat and for doing work.
The first law: transformation of energy into heat and work Chemical reactions can be used to provide heat and for doing work. Compare fuel value of different compounds. What drives these reactions to proceed
More informationLecture 6: Intro to Entropy
Lecture 6: Intro to Entropy Reading: Zumdahl 0., 0., 0.3 Outline: Why enthalpy isn t enough. Statistical interpretation of entropy Boltzmann s Formula Heat engine leads to Entropy as a state function Problems:Z0.4,
More informationExam 4  PHYS 101. Name: Class: Date: Multiple Choice Identify the choice that best completes the statement or answers the question.
Name: Class: Date: Exam 4  PHYS 101 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. A steel tape measure is marked such that it gives accurate measurements
More informationThermodynamics of Mixing
Thermodynamics of Mixing Dependence of Gibbs energy on mixture composition is G = n A µ A + n B µ B and at constant T and p, systems tend towards a lower Gibbs energy The simplest example of mixing: What
More informationTHE CLAUSIUS INEQUALITY
Part IV Entropy In Part III, we introduced the second law of thermodynamics and applied it to cycles and cyclic devices. In this part, we apply the second law to processes. he first law of thermodynamics
More informationSecond Law of Thermodynamics
Thermodynamics T8 Second Law of Thermodynamics Learning Goal: To understand the implications of the second law of thermodynamics. The second law of thermodynamics explains the direction in which the thermodynamic
More informationChapter 3: Evaluating Properties
Chapter 3: Evaluating Properties Lava flowing into the Pacific Ocean in Hawaii. Photo courtesy of Mike Benson. Property Relations in Engineering hermodynamics ENGINEERING CONEX o apply the energy balance
More information(b) 1. Look up c p for air in Table A.6. c p = 1004 J/kg K 2. Use equation (1) and given and looked up values to find s 2 s 1.
Problem 1 Given: Air cooled where: T 1 = 858K, P 1 = P = 4.5 MPa gage, T = 15 o C = 88K Find: (a) Show process on a Ts diagram (b) Calculate change in specific entropy if air is an ideal gas (c) Evaluate
More informationFrom the equation (186) of textbook, we have the Maxwell distribution of speeds as: v 2 e 1 mv 2. 2 kt. 2 4kT m. v p = and f(2vp) as: f(v p ) = 1
Solution Set 9 1 MaxwellBoltzmann distribution From the equation 186 of textbook, we have the Maxwell distribution of speeds as: fv 4πN In example 185, the most probable speed v p is given as m v e
More informationPhysics 2101 Section 3 April 26th: Chap. 18 : Chap Ann n ce n e t nnt : Exam #4, April Exam #4,
Physics 2101 Section 3 April 26 th : Chap. 181919 Announcements: n nt Exam #4, April 28 th (Ch. 13.618.8) 18.8) Final Exam: May 11 th (Tuesday), 7:30 AM Make up Final: May 15 th (Saturday) 7:30 AM Class
More informationSystem. System, Boundary and surroundings: Nature of heat and work: Sign convention of heat: Unit7 Thermodynamics
Unit7 Thermodynamics Introduction: The term Thermo means heat and dynamics means flow or movement.. So thermodynamics is concerned with the flow of heat. The different forms of the energy are interconvertible
More informationChapter 3. Real gases. Chapter : 1 : Slide 1
Chapter 3 Real gases Chapter11111 1 : 1 : Slide 1 1 Real Gases Perfect gas: only contribution to energy is KE of molecules Real gases: Molecules interact if they are close enough, have a potential energy
More informationLecture 14 Chapter 19 Ideal Gas Law and Kinetic Theory of Gases. Chapter 20 Entropy and the Second Law of Thermodynamics
Lecture 14 Chapter 19 Ideal Gas Law and Kinetic Theory of Gases Now we to look at temperature, pressure, and internal energy in terms of the motion of molecules and atoms? Relate to the 1st Law of Thermodynamics
More informationPhys222 W11 Quiz 1: Chapters 1921 Keys. Name:
Name:. In order for two objects to have the same temperature, they must a. be in thermal equilibrium.
More informationEntropy and the Second Law of Thermodynamics. The Adiabatic Expansion of Gases
Lecture 7 Entropy and the Second Law of Thermodynamics 15/08/07 The Adiabatic Expansion of Gases In an adiabatic process no heat is transferred, Q=0 = C P / C V is assumed to be constant during this process
More informationThermal Properties of Matter
Chapter 18 Thermal Properties of Matter PowerPoint Lectures for University Physics, Thirteenth Edition Hugh D. Young and Roger A. Freedman Lectures by Wayne Anderson Goals for Chapter 18 To relate the
More informationPhys 2101 Gabriela González
Phys 2101 Gabriela González If volume is constant ( isochoric process), W=0, and Tp 1 is constant. pv=nrt If temperature is constant ( isothermal process), ΔE int =0, and pv is constant. If pressure is
More information2. A process that releases heat into the surroundings is called. A process that can be reversed by an infinitesimal change in a parameter
Sample quiz and test questions Chapter 2. I. Terms and short answers 1. A system that can exchange neither matter nor energy with its surroundings is called isolated 2. A process that releases heat into
More informationThe Relationships Between. Internal Energy, Heat, Enthalpy, and Calorimetry
The Relationships Between Internal Energy, Heat, Enthalpy, and Calorimetry Recap of Last Class Last class, we began our discussion about energy changes that accompany chemical reactions Chapter 5 discusses:
More informationEntropy in Chemistry
LECURE 5 Can we predict: If a reaction or process can occur? How will it proceed? How fast will it go? Entropy in Chemistry hermodynamics tells us nothing about rates tells everything about the rest So
More informationThe First Law of Thermodynamics: Closed Systems. Heat Transfer
The First Law of Thermodynamics: Closed Systems The first law of thermodynamics can be simply stated as follows: during an interaction between a system and its surroundings, the amount of energy gained
More informationSo T decreases. 1. Does the temperature increase or decrease? For 1 mole of the vdw N2 gas:
1. One mole of Nitrogen (N2) has been compressed at T0=273 K to the volume V0=1liter. The gas goes through the free expansion process (Q = 0, W = 0), in which the pressure drops down to the atmospheric
More informationGive all answers in MKS units: energy in Joules, pressure in Pascals, volume in m 3, etc. Only work the number of problems required. Chose wisely.
Chemistry 45/456 0 July, 007 Midterm Examination Professor G. Drobny Universal gas constant=r=8.3j/molek=0.08latm/molek Joule=J= Ntm=kgm /s 0J= Latm. Pa=J/m 3 =N/m. atm=.0x0 5 Pa=.0 bar L=03 m 3.
More informationChemistry Problem Set 4 Answers Professor: C. E. Loader
roblem Set 4 Answers rofessor: C. E. Loader. At 5.0 ºC, the vapor pressure of ice is 3.03 mm and that of liquid water is 3.63 mm. Calculate G for the change of one mole of liquid water at 5.0 ºC to solid
More informationPSS 17.1: The Bermuda Triangle
Assignment 6 Consider 6.0 g of helium at 40_C in the form of a cube 40 cm. on each side. Suppose 2000 J of energy are transferred to this gas. (i) Determine the final pressure if the process is at constant
More informationThe Gas Laws. The effect of adding gas. 4 things. Pressure and the number of molecules are directly related. Page 1
The Gas Laws Describe HOW gases behave. Can be predicted by the theory. The Kinetic Theory Amount of change can be calculated with mathematical equations. The effect of adding gas. When we blow up a balloon
More informationThermodynamics 2nd year physics A. M. Steane 2000, revised 2004, 2006
Thermodynamics 2nd year physics A. M. Steane 2000, revised 2004, 2006 We will base our tutorials around Adkins, Equilibrium Thermodynamics, 2nd ed (McGrawHill). Zemansky, Heat and Thermodynamics is good
More informationChapter 18 Temperature, Heat, and the First Law of Thermodynamics. Problems: 8, 11, 13, 17, 21, 27, 29, 37, 39, 41, 47, 51, 57
Chapter 18 Temperature, Heat, and the First Law of Thermodynamics Problems: 8, 11, 13, 17, 21, 27, 29, 37, 39, 41, 47, 51, 57 Thermodynamics study and application of thermal energy temperature quantity
More informationIntroduction to the Ideal Gas Law
Course PHYSICS260 Assignment 5 Consider ten grams of nitrogen gas at an initial pressure of 6.0 atm and at room temperature. It undergoes an isobaric expansion resulting in a quadrupling of its volume.
More informationCHAPTER 12. Gases and the KineticMolecular Theory
CHAPTER 12 Gases and the KineticMolecular Theory 1 Gases vs. Liquids & Solids Gases Weak interactions between molecules Molecules move rapidly Fast diffusion rates Low densities Easy to compress Liquids
More informationTutorial 6 GASES. PRESSURE: atmospheres or mm Hg; 1 atm = 760 mm Hg. STP: Standard Temperature and Pressure: 273 K and 1 atm (or 760 mm Hg)
T41 Tutorial 6 GASES Before working with gases some definitions are needed: PRESSURE: atmospheres or mm Hg; 1 atm = 760 mm Hg TEMPERATURE: Kelvin, K, which is o C + 273 STP: Standard Temperature and Pressure:
More informationCourse 2 Mathematical Tools and Unit Conversion Used in Thermodynamic Problem Solving
Course Mathematical Tools and Unit Conversion Used in Thermodynamic Problem Solving 1 Basic Algebra Computations 1st degree equations  =0 Collect numerical values on one side and unknown to the otherside
More informationName Class Date. In the space provided, write the letter of the term or phrase that best completes each statement or best answers each question.
Assessment Chapter Test A Chapter: States of Matter In the space provided, write the letter of the term or phrase that best completes each statement or best answers each question. 1. The kineticmolecular
More informationPage 1. Set of values that describe the current condition of a system, usually in equilibrium. What is a state?
What is a state? Set of values that describe the current condition of a system, usually in equilibrium Is this physics different than what we have learned? Why do we learn it? How do we make the connection
More informationExam 4 Practice Problems false false
Exam 4 Practice Problems 1 1. Which of the following statements is false? a. Condensed states have much higher densities than gases. b. Molecules are very far apart in gases and closer together in liquids
More information1 MOLECULAR WEIGHT DETERMINATION BY THE VICTOR MEYER METHOD
1 MOLECULAR WEIGHT DETERMINATION BY THE VICTOR MEYER METHOD OBJECTIVE: To determine the molecular weight of an unknown liquid using the Victor Meyer apparatus. INTRODUCTION: The Victor Meyer method consists
More information= 1.038 atm. 760 mm Hg. = 0.989 atm. d. 767 torr = 767 mm Hg. = 1.01 atm
Chapter 13 Gases 1. Solids and liquids have essentially fixed volumes and are not able to be compressed easily. Gases have volumes that depend on their conditions, and can be compressed or expanded by
More information1 CHAPTER 10 THE JOULE AND JOULETHOMSON EXPERIMENTS
AER 0 E JOLE AND JOLEOMON EXERIMEN 0 Introduction Equation 84, γ constant, tells us how to calculate the drop in temperature if a gas expands adiabatically and reversibly; it is expanding against an external
More informationLecture Notes: Gas Laws and Kinetic Molecular Theory (KMT).
CHEM110 Week 9 Notes (Gas Laws) Page 1 of 7 Lecture Notes: Gas Laws and Kinetic Molecular Theory (KMT). Gases Are mostly empty space Occupy containers uniformly and completely Expand infinitely Diffuse
More informationCHAPTER 20. Q h = 100/0.2 J = 500 J Q c = J = 400 J
CHAPTER 20 1* Where does the energy come from in an internalcombustion engine? In a steam engine? Internal combustion engine: From the heat of combustion (see Problems 19106 to 19109). Steam engine:
More informationChapter 19 Thermodynamics
19.1 Introduction Chapter 19 Thermodynamics We can express the fundamental laws of the universe which correspond to the two fundamental laws of the mechanical theory of heat in the following simple form.
More informationEnergy in Thermal Processes: The First Law of Thermodynamics
Energy in Thermal Processes: The First Law of Thermodynamics 1. An insulated container half full of room temperature water is shaken vigorously for two minutes. What happens to the temperature of the water?
More informationSimple Mixtures. Atkins 7th: Sections ; Atkins 8th: The Properties of Solutions. Liquid Mixtures
The Properties of Solutions Simple Mixtures Atkins 7th: Sections 7.47.5; Atkins 8th: 5.45.5 Liquid Mixtures Colligative Properties Boiling point elevation Freezing point depression Solubility Osmosis
More informationThe Gas, Liquid, and Solid Phase
The Gas, Liquid, and Solid Phase When are interparticle forces important? Ron Robertson Kinetic Theory A. Principles Matter is composed of particles in constant, random, motion Particles collide elastically
More informationLecture 4: 09.16.05 Temperature, heat, and entropy
3.012 Fundamentals of Materials Science Fall 2005 Lecture 4: 09.16.05 Temperature, heat, and entropy Today: LAST TIME...2 State functions...2 Path dependent variables: heat and work...2 DEFINING TEMPERATURE...4
More informationChapter 19. Chemical Thermodynamics. The reverse reaction (two eggs leaping into your hand with their shells back intact) is not spontaneous.
Chapter 19. Chemical Thermodynamics SOURCE: Chemistry the Central Science: Prentice hall I. Spontaneous Processes Thermodynamics is concerned with the question: will a reaction occur? First Law of Thermodynamics:
More information31 Copyright Richard M. Felder, Lisa G. Bullard, and Michael D. Dickey (2014)
EQUATIONS OF STATE FOR GASES Questions A gas enters a reactor at a rate of 255 SCMH. What does that mean? An orifice meter mounted in a process gas line indicates a flow rate of 24 ft 3 /min. The gas temperature
More informationBomb Calorimetry. Example 4. Energy and Enthalpy
Bomb Calorimetry constant volume often used for combustion reactions heat released by reaction is absorbed by calorimeter contents need heat capacity of calorimeter q cal = q rxn = q bomb + q water Example
More informationFUNDAMENTALS OF ENGINEERING THERMODYNAMICS
FUNDAMENTALS OF ENGINEERING THERMODYNAMICS System: Quantity of matter (constant mass) or region in space (constant volume) chosen for study. Closed system: Can exchange energy but not mass; mass is constant
More informationCHEMISTRY. Matter and Change. Section 13.1 Section 13.2 Section 13.3. The Gas Laws The Ideal Gas Law Gas Stoichiometry
CHEMISTRY Matter and Change 13 Table Of Contents Chapter 13: Gases Section 13.1 Section 13.2 Section 13.3 The Gas Laws The Ideal Gas Law Gas Stoichiometry State the relationships among pressure, temperature,
More informationVaporLiquid Equilibria
31 Introduction to Chemical Engineering Calculations Lecture 7. VaporLiquid Equilibria Vapor and Gas Vapor A substance that is below its critical temperature. Gas A substance that is above its critical
More informationChapter 15: Thermodynamics
Chapter 15: Thermodynamics The First Law of Thermodynamics Thermodynamic Processes (isobaric, isochoric, isothermal, adiabatic) Reversible and Irreversible Processes Heat Engines Refrigerators and Heat
More informationAP CHEMISTRY 2006 SCORING GUIDELINES
AP CHEMISTRY 2006 SCORING GUIDELINES Question 3 3. Answer the following questions that relate to the analysis of chemical compounds. (a) A compound containing the elements C, H, N, and O is analyzed. When
More information= 800 kg/m 3 (note that old units cancel out) 4.184 J 1000 g = 4184 J/kg o C
Units and Dimensions Basic properties such as length, mass, time and temperature that can be measured are called dimensions. Any quantity that can be measured has a value and a unit associated with it.
More informationAn introduction to thermodynamics applied to Organic Rankine Cycles
An introduction to thermodynamics applied to Organic Rankine Cycles By : Sylvain Quoilin PhD Student at the University of Liège November 2008 1 Definition of a few thermodynamic variables 1.1 Main thermodynamics
More information