ECE3030 Physical Foundations of Computer Engineering, Fall 2015 Homework 1 Solution to Problems 1-4

Similar documents
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

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

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

CHEMISTRY GAS LAW S WORKSHEET

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

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

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

THE IDEAL GAS LAW AND KINETIC THEORY

CHAPTER 8: CHEMICAL COMPOSITION

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

CHAPTER 12. Gases and the Kinetic-Molecular Theory

Type: Single Date: Kinetic Theory of Gases. Homework: Read (14.1), Do CONCEPT Q. # (1), Do PROBLEMS # (2, 3, 5) Ch. 14

Lecture Notes: Gas Laws and Kinetic Molecular Theory (KMT).

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

Temperature. Number of moles. Constant Terms. Pressure. Answers Additional Questions 12.1

CHEMISTRY 113 EXAM 4(A)

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

Stoichiometry Exploring a Student-Friendly Method of Problem Solving

Chapter 4 Practice Quiz

atm = 760 torr = 760 mm Hg = kpa = psi. = atm. = atm. = 107 kpa 760 torr 1 atm 760 mm Hg = 790.

Unit 3 Notepack Chapter 7 Chemical Quantities Qualifier for Test

Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 1. A chemical equation. (C-4.4)

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

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

Mole Notes.notebook. October 29, 2014

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

2. The percent yield is the maximum amount of product that can be produced from the given amount of limiting reactant.

Chapter 8: Gases and Gas Laws.

Review - After School Matter Name: Review - After School Matter Tuesday, April 29, 2008

EXPERIMENT 13: THE IDEAL GAS LAW AND THE MOLECULAR WEIGHT OF GASES

CHEMISTRY. Matter and Change. Section 13.1 Section 13.2 Section The Gas Laws The Ideal Gas Law Gas Stoichiometry

Chemistry 13: States of Matter

Gas Laws. vacuum. 760 mm. air pressure. mercury

CLASSICAL CONCEPT REVIEW 8

Name Date Class STOICHIOMETRY. SECTION 12.1 THE ARITHMETIC OF EQUATIONS (pages )

CHEM 120 Online Chapter 7

(a) graph Y versus X (b) graph Y versus 1/X

Other Stoich Calculations A. mole mass (mass mole) calculations. GIVEN mol A x CE mol B. PT g A CE mol A MOLE MASS :

Module 5: Combustion Technology. Lecture 33: Combustion air calculation

Exam 4 Practice Problems false false

Name Class Date. In the space provided, write the letter of the term or phrase that best completes each statement or best answers each question.

Kinetic Theory of Gases. 6.1 Properties of Gases 6.2 Gas Pressure. Properties That Describe a Gas. Gas Pressure. Learning Check.

Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature

2014 Spring CHEM101 Ch1-2 Review Worksheet Modified by Dr. Cheng-Yu Lai,

CHEMICAL EQUILIBRIUM (ICE METHOD)

Gas Laws. Heat and Temperature

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

Kinetic Theory of Gases

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

THE HUMIDITY/MOISTURE HANDBOOK

The Mole. Chapter 10. Dimensional Analysis. The Mole. How much mass is in one atom of carbon-12? Molar Mass of Atoms 3/1/2015

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

Test 5 Review questions. 1. As ice cools from 273 K to 263 K, the average kinetic energy of its molecules will

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

Element of same atomic number, but different atomic mass o Example: Hydrogen

The Mole Notes. There are many ways to or measure things. In Chemistry we also have special ways to count and measure things, one of which is the.

Composition of the Atmosphere. Outline Atmospheric Composition Nitrogen and Oxygen Lightning Homework

(1) The size of a gas particle is negligible as compared to the volume of the container in which the gas is placed.

Name Date Class CHEMICAL QUANTITIES. SECTION 10.1 THE MOLE: A MEASUREMENT OF MATTER (pages )

Chemical Calculations: Formula Masses, Moles, and Chemical Equations

Name Date Class CHEMICAL QUANTITIES. SECTION 10.1 THE MOLE: A MEASUREMENT OF MATTER (pages )

LECTURE I-UNITS OF CONCENTRATION

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

Page 2. Base your answers to questions 7 through 9 on this phase diagram

Chemical Composition. Introductory Chemistry: A Foundation FOURTH EDITION. Atomic Masses. Atomic Masses. Atomic Masses. Chapter 8

Calculating Atoms, Ions, or Molecules Using Moles

Chemistry 110 Lecture Unit 5 Chapter 11-GASES

Chapter 12 Kinetic Theory of Gases: Equipartition of Energy and Ideal Gas Law

Stoichiometry. 1. The total number of moles represented by 20 grams of calcium carbonate is (1) 1; (2) 2; (3) 0.1; (4) 0.2.

A dozen. Molar Mass. Mass of atoms

Physics Notes Class 11 CHAPTER 2 UNITS AND MEASUREMENTS

Mole Calculations Multiple Choice Review PSI Chemistry

Chem 1A Exam 2 Review Problems

What s in a Mole? Molar Mass

Chapter 10. Can You draw the Lewis structure for a given covalently bonded molecule?

1. What is the molecular formula of a compound with the empirical formula PO and a gram-molecular mass of 284 grams?

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

Multiple Choice questions (one answer correct)

Chapter 13 Gases. Review Skills

602X ,000,000,000, 000,000,000, X Pre- AP Chemistry Chemical Quan44es: The Mole. Diatomic Elements

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)

Chapter 29: Kinetic Theory of Gases: Equipartition of Energy and the Ideal Gas Law

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

19 The Kinetic Theory of Gases

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

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

2 The Structure of Atoms

CHEMISTRY STANDARDS BASED RUBRIC ATOMIC STRUCTURE AND BONDING

Chem 1100 Chapter Three Study Guide Answers Outline I. Molar Mass and Moles A. Calculations of Molar Masses

We know from the information given that we have an equal mass of each compound, but no real numbers to plug in and find moles. So what can we do?

Geol 116 The Planet Class 7-1 Feb 28, Exercise 1, Calculate the escape velocities of the nine planets in the solar system

Unit 2: Quantities in Chemistry

Description of the Mole Concept:

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

How much does a single atom weigh? Different elements weigh different amounts related to what makes them unique.

10 The Mole. Section 10.1 Measuring Matter

THE KINETIC THEORY OF GASES

Lecture 5, The Mole. What is a mole?

ANSWER KEY : BUILD AN ATOM PART I: ATOM SCREEN Build an Atom simulation ( an atom )

U3-LM2B-WS Molar Mass and Conversions

Transcription:

ECE3030 Physical Foundations of Computer Engineering, Fall 2015 Homework 1 to Problems 1-4 As a prerequisite to this course, you should understand basic physical quantities such as the universal gas constant (R), Avagadro s number (N A), and Boltzmann s constant (k); in addition, these have been explained both from an intuitive perspective as well as a mathematical perspective. Rather than request repetition of some of the derivations explained in class, this homework will ask for some calculations using the equations as well as some written explanations intutitively explaining what the equations show. Please also utilize the following standard measurement equivalents: 1 atmosphere (atm) = 1.01325 x 10 5 Pascals (P) 0 degrees Celsius ( C) = 273.15 degrees Kelvin (K) 1 unified atomic mass unit (u) = 1 gram per mole (g/mol) 1) (15) Determine the volume of 5 mol of an ideal gas at a temperature of 0 C and a pressure of 1 atm. NOTE1: 0 C and 1 atm are standard temperature and pressure, commonly abbreviated as STP. NOTE2: please give your answer in liters (l ). NOTE3: YOU ARE REQUIRED TO BOX YOUR FINAL ANSWER. According to the ideal gas law: = Equation 1 Where V is volume, is number of moles (amount of the substance), is the ideal/universal gas constant ( = 8.314 1 1 ), is the pressure of the gass. From Equation 1 = 5.000 8.314 1 1 273.2 1.013 10 = 0.1121 10 10 1 = 112.1 1 = 0.1121 = Note 1: = evidences the consistency of the units. Note 2: 1 = 10, 1 = 1 = 10 = 112.1

2) (20) A sample of air with a mass of 100.00 grams (g), collected at sea level, is analyzed and found to consist of the following gases: nitrogen (N2) = 75.52 g oxygen (O 2) = 23.15 g argon (Ar) = 1.28 g carbon dioxide (CO 2) = 0.05 g plus trace amounts of neon, helium, methane and other gases. You look up the elements in the periodic table and find the following: 14.0067 u for nitrogen (N) 15.994 u for oxygen (O) 39.948 u for argon (Ar) 12.011 u for carbon (C) where we assume an average number of neutrons in each atom. The molecular mass is simply the addition of the unified atomic mass unit (u) value for each atom in the molecule, which yield a grams per mole number. Therefore, you calculate the number of moles of each gas type as follows: n(n 2) = 75.52 g / 2(14.0067 g/mol) = 2.6962 mol n(o 2) = 23.15 g / 2(15.994 g/mol) = 0.7234 mol n(ar) = 1.28 g / (39.948 g/mol) = 0.0320 mol n(co2) = 0.05 g / (44.01 g/mol) = 0.0011 mol with the result that the total number of moles in the 100 g sample is 3.4527 mol. Using the ideal gas law and assuming a temperature of 0 C and a pressure of 1 atm, find the volume of the sample. NOTE: YOU ARE REQUIRED TO BOX YOUR FINAL ANSWER. We use Equation 1 and follow a similar approach for = 3.4527 = 3.4527 8.314 1 1 273.2 1.013 10 = 0.07742 10 10 1 = 77.42 1 = 0.07742 = = 77.42

3) (10) In the previous example, the number of moles of carbon dioxide in the sample was found to be 0.0011 mol. Explain how this number was found including how the number of grams per mole was calculated. Use of clear and unambiguous language is required for full credit for this problem; vague and incomplete sentences, including improper grammar, will result in at best some limited partial credit (and possibly zero points especially if the writing is not legible). Taking 15.994 u for oxygen (O), and 12.011 u for carbon (C), we calculate the amount of CO2 in terms of unified atomic mass units as 2(15.994 u) + 12.011 u = 43.999 u. Since 1 u is also equivalent to 1g/mol, then for CO2 43.999 u = 43.999 g/mol. n(co 2) = 0.05 g / (44.01 g/mol) = 0.0011 mol

4) (15) Using (i) the expression = < > for the average (or root-mean-square) velocity of a molecule and (ii) the equation relating the kinetic energy of a molecule to Boltzmann s constant and temperature, give an equation for in terms of Boltzmann s constant (k), temperature (T) and molecular mass (µ). Show how the units of the right hand side work out to be equivalent to the left hand side. For full credit, show all steps clearly and unambiguously. NOTE1: the final correct answer alone will earn zero points; you are required to start with < > on the left hand side of an equation and show the mathematical steps to result in on the left hand side and a function of k, T and µ on the right hand side. You are further required to show how the units of the right hand side reduce to the units of the left hand side (which are m/s where m = meters and s = seconds). NOTE2: the following physical identies may be helpful (i) Joules = Newtons * meters and (ii) Newtons = (kg*m)/s 2. NOTE3: molecular mass (µ) is measured in kg. Method 1 In classical mechanics, the kinetic energy of a particle with mass µ is defined as = 1 2 Equation 2 Due to the statistical nature of each particle in a gas, the average values rather the instantaneous ones are taken in Equation 2. = 1 2 = 1 2 Equation 3 On the other hand, from lecture we related this expression to the Boltzmann constant k and temperature T as 1 2 = = 3 2 Equation 4 Where the factor 3 accounts for the 3 spatial degrees of freedom (x, y and z direction). From Equation 4, = 3 Equation 5 Applying the definition definition, we obtain = = 3 Equation 6

Method 2 Starting from and using the average of the usual definition of kinetic energy (see Equation 3) we obtain = 2 Equation 7 Taking the expresion derived in the lecture (see Equation 4) = 2 = 2 3 2 = 3 Equation 8 Since it is provided that = < >, we obtain the same expression as in Equation 6. Cheking units = 3 Focusing on the units and omitting the numbers in Equation 6 = / = / = /