CHAPTER 27 SPECIFIC HEAT CAPACITIES OF GASES

Size: px
Start display at page:

Download "CHAPTER 27 SPECIFIC HEAT CAPACITIES OF GASES"

Transcription

1 HATE 7 SEIFI HEAT AAITIES OF GASES. N mole, W g/mol, m/s K.E. of the vessel Internal energy of the gas (/) mv (/) J n r(t) 8. T T k. 8.. m g, t.7 cal/g- J. J/al. dq du + dw Now, (for a rigid body) So, dw. So dq du. Q msdt cal Joule..., w or piston kg., A of piston cm o kpa, g m/s, x cm. mg dw pdv oadx A ndt dt n. J. n dq npdt np n ( ) n.. J.. H. al/g, H. al/g M g/ Mol, 8. 7 erg/mol- We know, al/g So, difference of molar specific heats M M al/g Now, J J 8. 7 erg/mol- J. 7 erg/cal.. 7.6, n mole, T K (a) Keeping the pressure constant, dq du + dw, T K, 7/6, m mole, dq du + dw n dt du + dt du npdt dt 7 dt dt 6 dt dt 7 6 DT dt 7dT dt 6 dt J. (b) Kipping olume constant, dv n dt dt J (c) Adiabetically dq, du dw n 8. T T T T J

2 6. m.8 g, cm L T k, a nt or n N T 8. atm. T Q Now, v.6 9. n dt p + v Specific Heat apacities of Gases Q n p dt.87.8 al cm, cm a, Q J (a) Q du + dw du + ( 6 ) du + du J (b) n 8. [ U nt for monoatomic] n (c) du n v dt v dndtu p v (d) v. (roved above) 8. Q Amt of heat given.8. Work done Q, Q W + U for monoatomic gas U Q Q Q n T Q nt nt Again Q n pdt Where > Molar heat capacity at const. pressure. nt ndt nt 9. K K T K T T K U dv K. dq du + dw mcdt dt + pdv msdt dt+ ms + K + K,, b dv dt b dt + dt b + r, b + b 7. df K. onsidering two gases, in Gas() we have,, p (Sp. Heat at const. ), v (Sp. Heat at const. ), n (No. of moles) p & p v v

3 v v v ( ) v & p Specific Heat apacities of Gases In Gas() we have,, p (Sp. Heat at const. ), v (Sp. Heat at const. ), n (No. of moles) p & p v v v v ( ) v v Given n : n : du nv dt & du nv dt nvdt v v r &p v () p So, [from () & ()] v ( ) () & p. p. p. v. v. n n mol (n + n ) dt n v dt + n v dt nv nv.. n n + + p.. n mole, J/mol-k, (a) Temp at A T a, a a nt a 6 T a a a k. n Similarly temperatures at point b k at it is 8 k and at D it is k. (b) For ab process, dq npdt [since ab is isobaric] T b T a ( ) 9 J For bc, dq du + dw [dq, Isochorie process] n dq du n v dt T c T a () J (c) Heat liberated in cd n p dt n T d T c Heat liberated in da n v dt T a T d J ( ) 7 J Ka Ka a d Ta Td cm Tc c b Tb cm 7.

4 . (a) For a, b is constant So, T T For b,c is constant So, T T 6 T T T T 6 6 k 9 k Specific Heat apacities of Gases (b) Work done Area enclosed under the graph cc kpa 6 J J (c) Q Supplied n v dt Now, n considering at pt. b T v Q bc dt a, b. 6 dt T 6.67 Q supplied to be n p dt [ p ] 6.9 (.67) dt.9 T (d) Q U + w Now, U Q w Heat supplied Work done (.9 +.9) 9.8. In Joly s differential steam calorimeter v ml m ( ) m Mass of steam condensed.9 g, L al/g. J/g m Mass of gas present g,,.9. v.89.9 J/g-K ( ) 6.. Since it is an adiabatic process, So const. (a) (b) T onst. T T Given L, L,..96. T T 7.. a, cc, T k (a).....? (b) T T (). T (). T. k k 7.7 cm c cm a b Ka Ka 7.

5 (c) Work done by the gas in the process m T T T( ) W T T 6.. [ ].7 J (, ). 8.., T 9 k, atm, p atm We know for adiabatic process, T T or (). (9). ().. T (). (9). T..78 T. T (.78) /.. K 9. Ka a, cm 6 m, T k,. (a) Suddenly compressed to cm (). (). (). 8 Ka Specific Heat apacities of Gases T T (). T ().- T 6 K (b) Even if the container is slowly compressed the walls are adiabatic so heat transferred is. Thus the values remain, 8 Ka, T 6 K.. Given (Initial ressure), (Initial olume) (a) (i) Isothermal compression, or, (ii) Adiabatic ompression or o + (b) (i) Adiabatic compression or (ii) Isothermal compression or. Initial pressure Initial olume (a) Isothermally to pressure + Adiabetically to pressure ( ) ( ) + (+)/ Final olume (+)/ 7.

6 Specific Heat apacities of Gases (b) Adiabetically to pressure to ( + ) () Isothermal to pressure / (+)/ Final olume (+)/. nt Given Ka a, cm 6 m, T k (a) n T 8. (b) ( ) moles. 6.6 J/mole (c) Given Ka a,? cm 6 m,. cm 6 m, T k, T? Since the process is adiabatic Hence ( 6 ) ( 6 ) a 78 Ka (d) Q W + U or W U [U, in adiabatic] n dt ( ) J J (e) U n dt J. A B For A, the process is isothermal A A A A A For B, the process is adiabatic, A A A A A ( B ) A ( B ) B B B B B For,, the process is isobaric T T T T T Final pressures are equal. T. B. p A B. A : B : :. : : :. Initial ressure Initial olume Final ressure Final olume Given,, Isothermal workdone nt Ln 7.6

7 Specific Heat apacities of Gases Adiabatic workdone Given that workdone in both cases is same. Hence nt Ln ( )ln nt nt nt ( ) ln ( ) ln We know T const. in adiabatic rocess. T T, or T ( ) T () ( ) Or, T - T or T T (ii) From (i) & (ii) ( ) ln T T T ( ) ln T T T nt (i) [ ].., T k, Lv l (a) The process is adiabatic as it is sudden, ( ). (). / (b) Ka n a W [T T ] T (). T (.). T.. T nt n n w [T T ] K. T ( in m ) (. ). (c) Internal Energy, dq, du dw ( 8.8)J 8.8 J 8 J. (d) Final Temp.. k k. (e) The pressure is kept constant. The process is isobaric. 8.8 J 8 J. Work done ndt ( ) Final Temp K (.). J. Initial Temp (f) Initial volume Final volume L T T T Work done in isothermal ntln ln / T ln.9 (g) Net work done W A + W B + W J. L. 7.7

8 Specific Heat apacities of Gases 6. Given. We know fro adiabatic process T onst. So, T T (eq) As, it is an adiabatic process and all the other conditions are same. Hence the above equation can be applied. So, T. T.. T T. T. T So, T :T : 7. cm,. J/mol-k, T k, 7 cm (a) No. of moles of gas in each vessel, T 8. 7 (b) Heat is supplied to the gas but dv dq du n dt.8. dt dt for (A).8. For (B) dt.8. 7 A.8. A B A [For container A] T T A. 7. cm of Hg..8. B 7 [For ontainer B] B.8. B A cm of Hg. T T Mercury moves by a distance B A.. m. 8. mhe. g,.67, g/mol, mh? 8/mol. Since it is an adiabatic surrounding He dq n dt. dt. (.67 ) dt (i) m H n dt dt m. dt [Where m is the rqd. Mass of H ] Since equal amount of heat is given to both and T is same in both. Equating (i) & (ii) we get. m dt dt m g.67 A He / / T T / / T T : B H 9. Initial pressure, Initial Temperature T Initial olume (a) For the diathermic vessel the temperature inside remains constant A B, Temperature T o For adiabatic vessel the temperature does not remains constant. The process is adiabatic T T T T ( ) T T T T 7.8

9 p ( ) (b) When the values are opened, the temperature remains T through out nt +, nt ( n n )T [Total value after the expt + ] nt nt. For an adiabatic process, v onst. There will be a common pressure when the equilibrium is reached Hence For left For ight () ( ) () ( ) () Equating for both left & right or () ( ) / / Similarly / / / / / / / For right () / / / Specific Heat apacities of Gases For left.() (b) Since the whole process takes place in adiabatic surroundings. The separator is adiabatic. Hence heat given to the gas in the left part Zero. (c) From () Final pressure Again from () / / / ( ) or y / / /. A cm m, M. g. kg, atm pascal, L cm. m. L 8 cm.8 m,. atm The process is adiabatic / / () () (AL). (AL).. log log.9. / / T / / T.9 m / v m/s 7.9

10 Specific Heat apacities of Gases. 8 m/s, T, oh.89 kg/m, r 8. J/mol-k, At ST, a, We know sound Again, o J/mol-k.8 (8) (8).8 Again, or J/mol-k. g kg, cm 6 m cal/mol-ki. J/mol-k J/mol-k 8. ( ) (8.) 8. Since the condition is ST, atm pa m/s. Given o.7 g/cm.7 kg/m,. a, 8. J/mol-k,. KHz. Node separation in a Kundt tube 6 cm, cm m So, - 6 m/s We know, Speed of sound o (6)..7 (6) But J/mol-k Again So, J/mol-k. Hz, T Hz,. cm 6.6 m 6.6 (66 ) m/s m [v nt t M ot m T (66 ) m v J/mol-k, J/mol-k. o m T ] (66 ) 8. (66 )

Answer, Key Homework 6 David McIntyre 1

Answer, Key Homework 6 David McIntyre 1 Answer, Key Homework 6 David McIntyre 1 This print-out should have 0 questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making

More information

HEAT UNIT 1.1 KINETIC THEORY OF GASES. 1.1.1 Introduction. 1.1.2 Postulates of Kinetic Theory of Gases

HEAT UNIT 1.1 KINETIC THEORY OF GASES. 1.1.1 Introduction. 1.1.2 Postulates of Kinetic Theory of Gases UNIT HEAT. KINETIC THEORY OF GASES.. Introduction Molecules have a diameter of the order of Å and the distance between them in a gas is 0 Å while the interaction distance in solids is very small. R. Clausius

More information

Problem Set 1 3.20 MIT Professor Gerbrand Ceder Fall 2003

Problem 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 information

Problem Set 3 Solutions

Problem 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 information

The final numerical answer given is correct but the math shown does not give that answer.

The 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 information

a) Use the following equation from the lecture notes: = ( 8.314 J K 1 mol 1) ( ) 10 L

a) Use the following equation from the lecture notes: = ( 8.314 J K 1 mol 1) ( ) 10 L hermodynamics: Examples for chapter 4. 1. 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

More information

Problem Set 4 Solutions

Problem 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 information

Gases. Macroscopic Properties. Petrucci, Harwood and Herring: Chapter 6

Gases. Macroscopic Properties. Petrucci, Harwood and Herring: Chapter 6 Gases Petrucci, Harwood and Herring: Chapter 6 CHEM 1000A 3.0 Gases 1 We will be looking at Macroscopic and Microscopic properties: Macroscopic Properties of bulk gases Observable Pressure, volume, mass,

More information

Heat and Work. First Law of Thermodynamics 9.1. Heat is a form of energy. Calorimetry. Work. First Law of Thermodynamics.

Heat and Work. First Law of Thermodynamics 9.1. Heat is a form of energy. Calorimetry. Work. First Law of Thermodynamics. Heat and First Law of Thermodynamics 9. Heat Heat and Thermodynamic rocesses Thermodynamics is the science of heat and work Heat is a form of energy Calorimetry Mechanical equivalent of heat Mechanical

More information

7. 1.00 atm = 760 torr = 760 mm Hg = 101.325 kpa = 14.70 psi. = 0.446 atm. = 0.993 atm. = 107 kpa 760 torr 1 atm 760 mm Hg = 790.

7. 1.00 atm = 760 torr = 760 mm Hg = 101.325 kpa = 14.70 psi. = 0.446 atm. = 0.993 atm. = 107 kpa 760 torr 1 atm 760 mm Hg = 790. CHATER 3. The atmosphere is a homogeneous mixture (a solution) of gases.. Solids and liquids have essentially fixed volumes and are not able to be compressed easily. have volumes that depend on their conditions,

More information

THE KINETIC THEORY OF GASES

THE 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 information

Chapter 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 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 information

Phys222 W11 Quiz 1: Chapters 19-21 Keys. Name:

Phys222 W11 Quiz 1: Chapters 19-21 Keys. Name: Name:. In order for two objects to have the same temperature, they must a. be in thermal equilibrium.

More information

Gas Laws. The kinetic theory of matter states that particles which make up all types of matter are in constant motion.

Gas Laws. The kinetic theory of matter states that particles which make up all types of matter are in constant motion. Name Period Gas Laws Kinetic energy is the energy of motion of molecules. Gas state of matter made up of tiny particles (atoms or molecules). Each atom or molecule is very far from other atoms or molecules.

More information

= 1.038 atm. 760 mm Hg. = 0.989 atm. d. 767 torr = 767 mm Hg. = 1.01 atm

= 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 information

ES-7A Thermodynamics HW 1: 2-30, 32, 52, 75, 121, 125; 3-18, 24, 29, 88 Spring 2003 Page 1 of 6

ES-7A Thermodynamics HW 1: 2-30, 32, 52, 75, 121, 125; 3-18, 24, 29, 88 Spring 2003 Page 1 of 6 Spring 2003 Page 1 of 6 2-30 Steam Tables Given: Property table for H 2 O Find: Complete the table. T ( C) P (kpa) h (kj/kg) x phase description a) 120.23 200 2046.03 0.7 saturated mixture b) 140 361.3

More information

= 800 kg/m 3 (note that old units cancel out) 4.184 J 1000 g = 4184 J/kg o C

= 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 information

18 Q0 a speed of 45.0 m/s away from a moving car. If the car is 8 Q0 moving towards the ambulance with a speed of 15.0 m/s, what Q0 frequency does a

18 Q0 a speed of 45.0 m/s away from a moving car. If the car is 8 Q0 moving towards the ambulance with a speed of 15.0 m/s, what Q0 frequency does a First Major T-042 1 A transverse sinusoidal wave is traveling on a string with a 17 speed of 300 m/s. If the wave has a frequency of 100 Hz, what 9 is the phase difference between two particles on the

More information

vap H = RT 1T 2 = 30.850 kj mol 1 100 kpa = 341 K

vap 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 information

CHAPTER 12. Gases and the Kinetic-Molecular Theory

CHAPTER 12. Gases and the Kinetic-Molecular Theory CHAPTER 12 Gases and the Kinetic-Molecular Theory 1 Gases vs. Liquids & Solids Gases Weak interactions between molecules Molecules move rapidly Fast diffusion rates Low densities Easy to compress Liquids

More information

Chapter 10 Temperature and Heat

Chapter 10 Temperature and Heat Chapter 10 Temperature and Heat What are temperature and heat? Are they the same? What causes heat? What Is Temperature? How do we measure temperature? What are we actually measuring? Temperature and Its

More information

Lecture 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. 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 information

CHEMISTRY GAS LAW S WORKSHEET

CHEMISTRY GAS LAW S WORKSHEET Boyle s Law Charles Law Guy-Lassac's Law Combined Gas Law For a given mass of gas at constant temperature, the volume of a gas varies inversely with pressure PV = k The volume of a fixed mass of gas is

More information

Thermochemistry. r2 d:\files\courses\1110-20\99heat&thermorans.doc. Ron Robertson

Thermochemistry. r2 d:\files\courses\1110-20\99heat&thermorans.doc. Ron Robertson Thermochemistry r2 d:\files\courses\1110-20\99heat&thermorans.doc Ron Robertson I. What is Energy? A. Energy is a property of matter that allows work to be done B. Potential and Kinetic Potential energy

More information

Chapter 6 Thermodynamics: The First Law

Chapter 6 Thermodynamics: The First Law Key Concepts 6.1 Systems Chapter 6 Thermodynamics: The First Law Systems, States, and Energy (Sections 6.1 6.8) thermodynamics, statistical thermodynamics, system, surroundings, open system, closed system,

More information

1.4.6-1.4.8 Gas Laws. Heat and Temperature

1.4.6-1.4.8 Gas Laws. Heat and Temperature 1.4.6-1.4.8 Gas Laws Heat and Temperature Often the concepts of heat and temperature are thought to be the same, but they are not. Perhaps the reason the two are incorrectly thought to be the same is because

More information

Thermodynamics. Chapter 13 Phase Diagrams. NC State University

Thermodynamics. Chapter 13 Phase Diagrams. NC State University Thermodynamics Chapter 13 Phase Diagrams NC State University Pressure (atm) Definition of a phase diagram A phase diagram is a representation of the states of matter, solid, liquid, or gas as a function

More information

explain your reasoning

explain your reasoning I. A mechanical device shakes a ball-spring system vertically at its natural frequency. The ball is attached to a string, sending a harmonic wave in the positive x-direction. +x a) The ball, of mass M,

More information

CHAPTER 14 THE CLAUSIUS-CLAPEYRON EQUATION

CHAPTER 14 THE CLAUSIUS-CLAPEYRON EQUATION CHAPTER 4 THE CAUIU-CAPEYRON EQUATION Before starting this chapter, it would probably be a good idea to re-read ections 9. and 9.3 of Chapter 9. The Clausius-Clapeyron equation relates the latent heat

More information

State Newton's second law of motion for a particle, defining carefully each term used.

State Newton's second law of motion for a particle, defining carefully each term used. 5 Question 1. [Marks 20] An unmarked police car P is, travelling at the legal speed limit, v P, on a straight section of highway. At time t = 0, the police car is overtaken by a car C, which is speeding

More information

The Gas Laws. Our Atmosphere. Pressure = Units of Pressure. Barometer. Chapter 10

The Gas Laws. Our Atmosphere. Pressure = Units of Pressure. Barometer. Chapter 10 Our Atmosphere The Gas Laws 99% N 2 and O 2 78% N 2 80 70 Nitrogen Chapter 10 21% O 2 1% CO 2 and the Noble Gases 60 50 40 Oxygen 30 20 10 0 Gas Carbon dioxide and Noble Gases Pressure Pressure = Force

More information

Gases. States of Matter. Molecular Arrangement Solid Small Small Ordered Liquid Unity Unity Local Order Gas High Large Chaotic (random)

Gases. 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 information

39th International Physics Olympiad - Hanoi - Vietnam - 2008. Theoretical Problem No. 3

39th International Physics Olympiad - Hanoi - Vietnam - 2008. Theoretical Problem No. 3 CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal

More information

momentum change per impact The average rate of change of momentum = Time interval between successive impacts 2m x 2l / x m x m x 2 / l P = l 2 P = l 3

momentum change per impact The average rate of change of momentum = Time interval between successive impacts 2m x 2l / x m x m x 2 / l P = l 2 P = l 3 Kinetic Molecular Theory This explains the Ideal Gas Pressure olume and Temperature behavior It s based on following ideas:. Any ordinary sized or macroscopic sample of gas contains large number of molecules.

More information

Thermodynamics AP Physics B. Multiple Choice Questions

Thermodynamics AP Physics B. Multiple Choice Questions Thermodynamics AP Physics B Name Multiple Choice Questions 1. What is the name of the following statement: When two systems are in thermal equilibrium with a third system, then they are in thermal equilibrium

More information

The First Law of Thermodynamics

The First Law of Thermodynamics Thermodynamics The First Law of Thermodynamics Thermodynamic Processes (isobaric, isochoric, isothermal, adiabatic) Reversible and Irreversible Processes Heat Engines Refrigerators and Heat Pumps The Carnot

More information

Energy Matters Heat. Changes of State

Energy Matters Heat. Changes of State Energy Matters Heat Changes of State Fusion If we supply heat to a lid, such as a piece of copper, the energy supplied is given to the molecules. These start to vibrate more rapidly and with larger vibrations

More information

States of Matter CHAPTER 10 REVIEW SECTION 1. Name Date Class. Answer the following questions in the space provided.

States of Matter CHAPTER 10 REVIEW SECTION 1. Name Date Class. Answer the following questions in the space provided. CHAPTER 10 REVIEW States of Matter SECTION 1 SHORT ANSWER Answer the following questions in the space provided. 1. Identify whether the descriptions below describe an ideal gas or a real gas. ideal gas

More information

KINETIC THEORY OF GASES AND THERMODYNAMICS SECTION I Kinetic theory of gases

KINETIC THEORY OF GASES AND THERMODYNAMICS SECTION I Kinetic theory of gases KINETIC THEORY OF GASES AND THERMODYNAMICS SECTION I Kinetic theory of gases Some important terms in kinetic theory of gases Macroscopic quantities: Physical quantities like pressure, temperature, volume,

More information

Statistical Mechanics, Kinetic Theory Ideal Gas. 8.01t Nov 22, 2004

Statistical 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 information

Chem 338 Homework Set #5 solutions October 10, 2001 From Atkins: 5.2, 5.9, 5.12, 5.13, 5.15, 5.17, 5.21

Chem 338 Homework Set #5 solutions October 10, 2001 From Atkins: 5.2, 5.9, 5.12, 5.13, 5.15, 5.17, 5.21 Chem 8 Homework Set #5 solutions October 10, 2001 From Atkins: 5.2, 5.9, 5.12, 5.1, 5.15, 5.17, 5.21 5.2) The density of rhombic sulfur is 2.070 g cm - and that of monoclinic sulfur is 1.957 g cm -. Can

More information

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Chapter 10 MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) A gas at a pressure of 10.0 Pa exerts a force of N on an area of 5.5 m2. A) 1.8 B) 0.55

More information

State Newton's second law of motion for a particle, defining carefully each term used.

State Newton's second law of motion for a particle, defining carefully each term used. 5 Question 1. [Marks 28] An unmarked police car P is, travelling at the legal speed limit, v P, on a straight section of highway. At time t = 0, the police car is overtaken by a car C, which is speeding

More information

AP Chemistry 2009 Free-Response Questions Form B

AP Chemistry 2009 Free-Response Questions Form B AP Chemistry 009 Free-Response Questions Form B The College Board The College Board is a not-for-profit membership association whose mission is to connect students to college success and opportunity. Founded

More information

PHYS-2010: General Physics I Course Lecture Notes Section XIII

PHYS-2010: General Physics I Course Lecture Notes Section XIII PHYS-2010: General Physics I Course Lecture Notes Section XIII Dr. Donald G. Luttermoser East Tennessee State University Edition 2.5 Abstract These class notes are designed for use of the instructor and

More information

10.7 Kinetic Molecular Theory. 10.7 Kinetic Molecular Theory. Kinetic Molecular Theory. Kinetic Molecular Theory. Kinetic Molecular Theory

10.7 Kinetic Molecular Theory. 10.7 Kinetic Molecular Theory. Kinetic Molecular Theory. Kinetic Molecular Theory. Kinetic Molecular Theory Week lectures--tentative 0.7 Kinetic-Molecular Theory 40 Application to the Gas Laws 0.8 Molecular Effusion and Diffusion 43 Graham's Law of Effusion Diffusion and Mean Free Path 0.9 Real Gases: Deviations

More information

The 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. 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 information

Unit 5 Practice Test. Name: Class: Date: Multiple Choice Identify the choice that best completes the statement or answers the question.

Unit 5 Practice Test. Name: Class: Date: Multiple Choice Identify the choice that best completes the statement or answers the question. Name: Class: Date: Unit 5 Practice Test Multiple Choice Identify the choice that best completes the statement or answers the question. 1) The internal energy of a system is always increased by. A) adding

More information

EXPERIMENT 15: Ideal Gas Law: Molecular Weight of a Vapor

EXPERIMENT 15: Ideal Gas Law: Molecular Weight of a Vapor EXPERIMENT 15: Ideal Gas Law: Molecular Weight of a Vapor Purpose: In this experiment you will use the ideal gas law to calculate the molecular weight of a volatile liquid compound by measuring the mass,

More information

EXERCISES. 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?

EXERCISES. 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 information

Chapter 18 Homework Answers

Chapter 18 Homework Answers Chapter 18 Homework Answers 18.22. 18.24. 18.26. a. Since G RT lnk, as long as the temperature remains constant, the value of G also remains constant. b. In this case, G G + RT lnq. Since the reaction

More information

Define the notations you are using properly. Present your arguments in details. Good luck!

Define the notations you are using properly. Present your arguments in details. Good luck! Umeå Universitet, Fysik Vitaly Bychkov Prov i fysik, Thermodynamics, 0-0-4, kl 9.00-5.00 jälpmedel: Students may use any book(s) including the textbook Thermal physics. Minor notes in the books are also

More information

Boyles Law. At constant temperature the volume occupied by a fixed amount of gas is inversely proportional to the pressure on the gas 1 P = P

Boyles Law. At constant temperature the volume occupied by a fixed amount of gas is inversely proportional to the pressure on the gas 1 P = P Boyles Law At constant temperature the volume occupied by a fixed amount of gas is inversely proportional to the pressure on the gas 1 or k 1 Boyles Law Example ressure olume Initial 2.00 atm 100 cm 3

More information

01 The Nature of Fluids

01 The Nature of Fluids 01 The Nature of Fluids WRI 1/17 01 The Nature of Fluids (Water Resources I) Dave Morgan Prepared using Lyx, and the Beamer class in L A TEX 2ε, on September 12, 2007 Recommended Text 01 The Nature of

More information

Molar Mass of Butane

Molar Mass of Butane Cautions Butane is toxic and flammable. No OPEN Flames should be used in this experiment. Purpose The purpose of this experiment is to determine the molar mass of butane using Dalton s Law of Partial Pressures

More information

AP Chemistry 2009 Free-Response Questions

AP Chemistry 2009 Free-Response Questions AP Chemistry 009 Free-Response Questions The College Board The College Board is a not-for-profit membership association whose mission is to connect students to college success and opportunity. Founded

More information

CHEM 105 HOUR EXAM III 28-OCT-99. = -163 kj/mole determine H f 0 for Ni(CO) 4 (g) = -260 kj/mole determine H f 0 for Cr(CO) 6 (g)

CHEM 105 HOUR EXAM III 28-OCT-99. = -163 kj/mole determine H f 0 for Ni(CO) 4 (g) = -260 kj/mole determine H f 0 for Cr(CO) 6 (g) CHEM 15 HOUR EXAM III 28-OCT-99 NAME (please print) 1. a. given: Ni (s) + 4 CO (g) = Ni(CO) 4 (g) H Rxn = -163 k/mole determine H f for Ni(CO) 4 (g) b. given: Cr (s) + 6 CO (g) = Cr(CO) 6 (g) H Rxn = -26

More information

Determination of the enthalpy of combustion using a bomb calorimeter TEC

Determination of the enthalpy of combustion using a bomb calorimeter TEC Determination of the enthalpy of TEC Related concepts First law of thermodynamics, Hess s law of constant heat summation, enthalpy of combustion, enthalpy of formation, heat capacity. Principle The bomb

More information

Chapter 28 Fluid Dynamics

Chapter 28 Fluid Dynamics Chapter 28 Fluid Dynamics 28.1 Ideal Fluids... 1 28.2 Velocity Vector Field... 1 28.3 Mass Continuity Equation... 3 28.4 Bernoulli s Principle... 4 28.5 Worked Examples: Bernoulli s Equation... 7 Example

More information

Chapter 1 Classical Thermodynamics: The First Law. 1.2 The first law of thermodynamics. 1.3 Real and ideal gases: a review

Chapter 1 Classical Thermodynamics: The First Law. 1.2 The first law of thermodynamics. 1.3 Real and ideal gases: a review Chapter 1 Classical Thermodynamics: The First Law 1.1 Introduction 1.2 The first law of thermodynamics 1.3 Real and ideal gases: a review 1.4 First law for cycles 1.5 Reversible processes 1.6 Work 1.7

More information

AS1 MOLES. oxygen molecules have the formula O 2 the relative mass will be 2 x 16 = 32 so the molar mass will be 32g mol -1

AS1 MOLES. oxygen molecules have the formula O 2 the relative mass will be 2 x 16 = 32 so the molar mass will be 32g mol -1 Moles 1 MOLES The mole the standard unit of amount of a substance the number of particles in a mole is known as Avogadro s constant (L) Avogadro s constant has a value of 6.023 x 10 23 mol -1. Example

More information

Just another. Fuel Cell Formulary. by Dr. Alexander Kabza

Just another. Fuel Cell Formulary. by Dr. Alexander Kabza Just another by Dr. Alexander Kabza This version is from July 16, 2015 and optimized for Smartphones. Latest version see www.kabza.de. Don t hesitate to send me any comments or failures or ideas by email!

More information

Introduction to the Ideal Gas Law

Introduction 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 information

F321 MOLES. Example If 1 atom has a mass of 1.241 x 10-23 g 1 mole of atoms will have a mass of 1.241 x 10-23 g x 6.02 x 10 23 = 7.

F321 MOLES. Example If 1 atom has a mass of 1.241 x 10-23 g 1 mole of atoms will have a mass of 1.241 x 10-23 g x 6.02 x 10 23 = 7. Moles 1 MOLES The mole the standard unit of amount of a substance (mol) the number of particles in a mole is known as Avogadro s constant (N A ) Avogadro s constant has a value of 6.02 x 10 23 mol -1.

More information

Thermodynamics worked examples

Thermodynamics worked examples An Introduction to Mechanical Engineering Part hermodynamics worked examles. What is the absolute ressure, in SI units, of a fluid at a gauge ressure of. bar if atmosheric ressure is.0 bar? Absolute ressure

More information

CLASSICAL CONCEPT REVIEW 8

CLASSICAL CONCEPT REVIEW 8 CLASSICAL CONCEPT REVIEW 8 Kinetic Theory Information concerning the initial motions of each of the atoms of macroscopic systems is not accessible, nor do we have the computational capability even with

More information

Design of heat exchangers

Design of heat exchangers Design of heat exchangers Exchanger Design Methodology The problem of heat exchanger design is complex and multidisciplinary. The major design considerations for a new heat exchanger include: process/design

More information

ME 201 Thermodynamics

ME 201 Thermodynamics ME 0 Thermodynamics Second Law Practice Problems. Ideally, which fluid can do more work: air at 600 psia and 600 F or steam at 600 psia and 600 F The maximum work a substance can do is given by its availablity.

More information

13.3 Factors Affecting Solubility Solute-Solvent Interactions Pressure Effects Temperature Effects

13.3 Factors Affecting Solubility Solute-Solvent Interactions Pressure Effects Temperature Effects Week 3 Sections 13.3-13.5 13.3 Factors Affecting Solubility Solute-Solvent Interactions Pressure Effects Temperature Effects 13.4 Ways of Expressing Concentration Mass Percentage, ppm, and ppb Mole Fraction,

More information

Introductory Laboratory EST - Experiment 9 Calorimetry - 1 - Introductory Laboratory Energy Science and Technology. Experiment 9. Physical Chemistry

Introductory Laboratory EST - Experiment 9 Calorimetry - 1 - Introductory Laboratory Energy Science and Technology. Experiment 9. Physical Chemistry Introductory Laboratory EST - Experiment 9 Calorimetry - 1 - Introductory Laboratory Energy Science and Technology Experiment 9 Physical Chemistry Calorimetry Abstract The heat of combustion of Naphthalene

More information

Worksheet #17. 2. How much heat is released when 143 g of ice is cooled from 14 C to 75 C, if the specific heat capacity of ice is 2.087 J/(g C).

Worksheet #17. 2. How much heat is released when 143 g of ice is cooled from 14 C to 75 C, if the specific heat capacity of ice is 2.087 J/(g C). Worksheet #17 Calculating Heat 1. How much heat is needed to bring 12.0 g of water from 28.3 C to 43.87 C, if the specific heat capacity of water is 4.184 /(g? 2. How much heat is released when 143 g of

More information

Availability. Second Law Analysis of Systems. Reading Problems 10.1 10.4 10.59, 10.65, 10.66, 10.67 10.69, 10.75, 10.81, 10.

Availability. Second Law Analysis of Systems. Reading Problems 10.1 10.4 10.59, 10.65, 10.66, 10.67 10.69, 10.75, 10.81, 10. Availability Readg Problems 10.1 10.4 10.59, 10.65, 10.66, 10.67 10.69, 10.75, 10.81, 10.88 Second Law Analysis of Systems AVAILABILITY: the theoretical maximum amount of reversible work that can be obtaed

More information

Chem 1A Exam 2 Review Problems

Chem 1A Exam 2 Review Problems Chem 1A Exam 2 Review Problems 1. At 0.967 atm, the height of mercury in a barometer is 0.735 m. If the mercury were replaced with water, what height of water (in meters) would be supported at this pressure?

More information

Technical Thermodynamics

Technical Thermodynamics Technical Thermodynamics Chapter 2: Basic ideas and some definitions Prof. Dr.-Ing. habil. Egon Hassel University of Rostock, Germany Faculty of Mechanical Engineering and Ship Building Institute of Technical

More information

AP Chemistry 2010 Free-Response Questions Form B

AP Chemistry 2010 Free-Response Questions Form B AP Chemistry 010 Free-Response Questions Form B The College Board The College Board is a not-for-profit membership association whose mission is to connect students to college success and opportunity. Founded

More information

Chemistry 212 VAPOR PRESSURE OF WATER LEARNING OBJECTIVES

Chemistry 212 VAPOR PRESSURE OF WATER LEARNING OBJECTIVES Chemistry 212 VAPOR PRESSURE OF WATER LEARNING OBJECTIVES The learning objectives of this experiment are to explore the relationship between the temperature and vapor pressure of water. determine the molar

More information

THE IDEAL GAS LAW AND KINETIC THEORY

THE IDEAL GAS LAW AND KINETIC THEORY Chapter 14 he Ideal Gas Law and Kinetic heory Chapter 14 HE IDEAL GAS LAW AND KINEIC HEORY REIEW Kinetic molecular theory involves the study of matter, particularly gases, as very small particles in constant

More information

Module 5: Combustion Technology. Lecture 34: Calculation of calorific value of fuels

Module 5: Combustion Technology. Lecture 34: Calculation of calorific value of fuels 1 P age Module 5: Combustion Technology Lecture 34: Calculation of calorific value of fuels 2 P age Keywords : Gross calorific value, Net calorific value, enthalpy change, bomb calorimeter 5.3 Calculation

More information

AP Physics Course 1 Summer Assignment. Teachers: Mr. Finn, Mrs. Kelly, Mr. Simowitz, Mr. Slesinski

AP Physics Course 1 Summer Assignment. Teachers: Mr. Finn, Mrs. Kelly, Mr. Simowitz, Mr. Slesinski AP Physics Course 1 Summer Assignment Teachers: Mr. Finn, Mrs. Kelly, Mr. Simowitz, Mr. Slesinski On the following pages, there are six sections that use the basic skills that will be used throughout the

More information

CHEM 36 General Chemistry EXAM #1 February 13, 2002

CHEM 36 General Chemistry EXAM #1 February 13, 2002 CHEM 36 General Chemistry EXAM #1 February 13, 2002 Name: Serkey, Anne INSTRUCTIONS: Read through the entire exam before you begin. Answer all of the questions. For questions involving calculations, show

More information

www.mathsbox.org.uk Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx Acceleration Velocity (v) Displacement x

www.mathsbox.org.uk Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx Acceleration Velocity (v) Displacement x Mechanics 2 : Revision Notes 1. Kinematics and variable acceleration Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx differentiate a = dv = d2 x dt dt dt 2 Acceleration Velocity

More information

Heat Exchangers - Introduction

Heat Exchangers - Introduction Heat Exchangers - Introduction Concentric Pipe Heat Exchange T h1 T c1 T c2 T h1 Energy Balance on Cold Stream (differential) dq C = wc p C dt C = C C dt C Energy Balance on Hot Stream (differential) dq

More information

5. Which temperature is equal to +20 K? 1) 253ºC 2) 293ºC 3) 253 C 4) 293 C

5. Which temperature is equal to +20 K? 1) 253ºC 2) 293ºC 3) 253 C 4) 293 C 1. The average kinetic energy of water molecules increases when 1) H 2 O(s) changes to H 2 O( ) at 0ºC 3) H 2 O( ) at 10ºC changes to H 2 O( ) at 20ºC 2) H 2 O( ) changes to H 2 O(s) at 0ºC 4) H 2 O( )

More information

Final Exam CHM 3410, Dr. Mebel, Fall 2005

Final Exam CHM 3410, Dr. Mebel, Fall 2005 Final Exam CHM 3410, Dr. Mebel, Fall 2005 1. At -31.2 C, pure propane and n-butane have vapor pressures of 1200 and 200 Torr, respectively. (a) Calculate the mole fraction of propane in the liquid mixture

More information

Gibbs Free Energy and Chemical Potential. NC State University

Gibbs 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 information

Stirling heat engine Internal combustion engine (Otto cycle) Diesel engine Steam engine (Rankine cycle) Kitchen Refrigerator

Stirling heat engine Internal combustion engine (Otto cycle) Diesel engine Steam engine (Rankine cycle) Kitchen Refrigerator Lecture. Real eat Engines and refrigerators (Ch. ) Stirling heat engine Internal combustion engine (Otto cycle) Diesel engine Steam engine (Rankine cycle) Kitchen Refrigerator Carnot Cycle - is not very

More information

AP Chemistry 2008 Free-Response Questions

AP Chemistry 2008 Free-Response Questions AP Chemistry 008 Free-Response Questions The College Board: Connecting Students to College Success The College Board is a not-for-profit membership association whose mission is to connect students to college

More information

Unit 3: States of Matter Practice Exam

Unit 3: States of Matter Practice Exam Page 1 Unit 3: States of Matter Practice Exam Multiple Choice. Identify the choice that best completes the statement or answers the question. 1. Two gases with unequal masses are injected into opposite

More information

Homework 9. Problems: 12.31, 12.32, 14.4, 14.21

Homework 9. Problems: 12.31, 12.32, 14.4, 14.21 Homework 9 Problems: 1.31, 1.3, 14.4, 14.1 Problem 1.31 Assume that if the shear stress exceeds about 4 10 N/m steel ruptures. Determine the shearing force necessary (a) to shear a steel bolt 1.00 cm in

More information

Chapter 13. Properties of Solutions

Chapter 13. Properties of Solutions Sample Exercise 13.1 (p. 534) By the process illustrated below, water vapor reacts with excess solid sodium sulfate to form the hydrated form of the salt. The chemical reaction is Na 2 SO 4(s) + 10 H 2

More information

EXPERIMENT 9 Evaluation of the Universal Gas Constant, R

EXPERIMENT 9 Evaluation of the Universal Gas Constant, R Outcomes EXPERIMENT 9 Evaluation of the Universal Gas Constant, R After completing this experiment, the student should be able to: 1. Determine universal gas constant using reaction of an acid with a metal.

More information

Standard Free Energies of Formation at 298 K. Average Bond Dissociation Energies at 298 K

Standard Free Energies of Formation at 298 K. Average Bond Dissociation Energies at 298 K 1 Thermodynamics There always seems to be at least one free response question that involves thermodynamics. These types of question also show up in the multiple choice questions. G, S, and H. Know what

More information

Enthalpy of Combustion via Calorimetry

Enthalpy of Combustion via Calorimetry Enthalpy of Combustion via Calorimetry Introduction This experiment measures the enthalpy change when a system consisting of a known amount of a substance in the presence of excess oxygen is quantitatively

More information

Chemistry 110 Lecture Unit 5 Chapter 11-GASES

Chemistry 110 Lecture Unit 5 Chapter 11-GASES Chemistry 110 Lecture Unit 5 Chapter 11-GASES I. PROPERITIES OF GASES A. Gases have an indefinite shape. B. Gases have a low density C. Gases are very compressible D. Gases exert pressure equally in all

More information

ESSAY. Write your answer in the space provided or on a separate sheet of paper.

ESSAY. Write your answer in the space provided or on a separate sheet of paper. Test 1 General Chemistry CH116 Summer, 2012 University of Massachusetts, Boston Name ESSAY. Write your answer in the space provided or on a separate sheet of paper. 1) Sodium hydride reacts with excess

More information

A. Kinetic Molecular Theory (KMT) = the idea that particles of matter are always in motion and that this motion has consequences.

A. Kinetic Molecular Theory (KMT) = the idea that particles of matter are always in motion and that this motion has consequences. I. MOLECULES IN MOTION: A. Kinetic Molecular Theory (KMT) = the idea that particles of matter are always in motion and that this motion has consequences. 1) theory developed in the late 19 th century to

More information

Gases and Kinetic-Molecular Theory: Chapter 12. Chapter Outline. Chapter Outline

Gases and Kinetic-Molecular Theory: Chapter 12. Chapter Outline. Chapter Outline Gases and Kinetic-Molecular heory: Chapter Chapter Outline Comparison of Solids, Liquids, and Gases Composition of the Atmosphere and Some Common Properties of Gases Pressure Boyle s Law: he Volume-Pressure

More information

Temperature Measure of KE At the same temperature, heavier molecules have less speed Absolute Zero -273 o C 0 K

Temperature Measure of KE At the same temperature, heavier molecules have less speed Absolute Zero -273 o C 0 K Temperature Measure of KE At the same temperature, heavier molecules have less speed Absolute Zero -273 o C 0 K Kinetic Molecular Theory of Gases 1. Large number of atoms/molecules in random motion 2.

More information

Carbon Dioxide and an Argon + Nitrogen Mixture. Measurement of C p /C v for Argon, Nitrogen, Stephen Lucas 05/11/10

Carbon Dioxide and an Argon + Nitrogen Mixture. Measurement of C p /C v for Argon, Nitrogen, Stephen Lucas 05/11/10 Carbon Dioxide and an Argon + Nitrogen Mixture Measurement of C p /C v for Argon, Nitrogen, Stephen Lucas 05/11/10 Measurement of C p /C v for Argon, Nitrogen, Carbon Dioxide and an Argon + Nitrogen Mixture

More information