To Date…. Ideal Monatomic Gas Cv = 3/2R Cp = Cv + R = 5/2 R

Slides:

Advertisements

Similar presentations

Derivation of thermodynamic equations

Advertisements

Lecture 1: Energy Reading: Zumdahl 9.1 Outline –Energy: Kinetic and Potential –System vs. Surroundings –Heat, Work, and Energy.

Lecture 7: Thermo and Entropy Reading: Zumdahl 10.2, 10.3 Outline –Isothermal processes –Isothermal gas expansion and work –Reversible Processes.

Ideal Monatomic Gas C v = 3/2R C p = C v + R = 5/2 R Polyatomic Gas C v > 3/2R C p > 5/2 R All Ideal Gases  E = q + w w = -P ext  V (for now)  E = nC.

Chapter 18: Heat,Work,the First Law of Thermodynamics

* Reading Assignments:

First Law of Thermodynamics

1 UCT PHY1025F: Heat and Properties of Matter Physics 1025F Heat & Properties of Matter Dr. Steve Peterson THERMODYNAMICS.

Chem Ch 19/#1 Today’s To Do List l Start Chapter 19: 1st Law P-V work State Functions 1st Law Adiabatic Processes.

Internal Energy Physics 202 Professor Lee Carkner Lecture 14.

Lecture 2: Enthalpy Reading: Zumdahl 9.2, 9.3 Outline –Definition of Enthalpy (  H) –Definition of Heat Capacity (C v and C p ) –Calculating  E and.

Lect# 7: Thermodynamics and Entropy Reading: Zumdahl 10.2, 10.3 Outline: Isothermal processes (∆T = 0) Isothermal gas expansion and work(w) Reversible.

Lecture 1: Energy Reading: Zumdahl 9.1 Outline –Energy: Kinetic and Potential –System vs. Surroundings –Heat, Work, and Energy.

Physics 2 Chapter 19 problems Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.

How many people would MISS the thermo midterm if we move it to MONDAY, Feb 15? Note: it’s only okay to miss a midterm with an excused absence, e.g. : a.

Internal Energy Physics 202 Professor Lee Carkner Lecture 16.

Thermodynamics The study of the transformation of energy from one form into another. Chemical Thermodynamics The energy changes taking place during a chemical.

Specific Heat Thermodynamics Professor Lee Carkner Lecture 8.

The First Law of Thermodynamics

Thermodynamics.

Lecture 2: Heat Capacities/State functions Reading: Zumdahl 9.3 Outline –Definition of Heat Capacity (C v and C p ) –Calculating  E and  H using C v.

Lecture 3: Hess’ Law Reading: Zumdahl 9.5 Outline –Definition of Hess’ Law –Using Hess’ Law (examples)

Lecture 6: Thermo and Entropy

Lecture 2: Enthalpy Reading: Zumdahl 9.2, 9.3 Outline –Definition of Enthalpy (  H) –Definition of Heat Capacity (C v and C p ) –Calculating  E and.

Lecture 1: Energy and Enthalpy Reading: Zumdahl 9.1 and 9.2 Outline –Energy: Kinetic and Potential –System vs. Surroundings –Heat, Work, and Energy –Enthalpy.

Knight: Chapter 17 Work, Heat, & the 1st Law of Thermodynamics

Fig The net work done by the system in the process aba is –500 J.

Unit 9: Gases Ideal Gas Law. After today you will be able to… Explain what an ideal gas is Calculate an unknown pressure, temperature, volume, or amount.

Physical Chemistry I (TKK-2246) 13/14 Semester 2 Instructor: Rama Oktavian Office Hr.: M.13-15, Tu , W , Th.

The Thermodynamic Behavior of Gases. Variables and Constants.

Thermodynamics. Thermodynamic Process in which energy is transferred as heat and work.

31.1 Thermodynamics of Mixing of Ideal Solutions For the process where solute and solvent are mixed to form an ideal solution at constant temperature and.

MSEG 803 Equilibria in Material Systems 2: First Law of TD Prof. Juejun (JJ) Hu

Thermodynamic Quantities Defined Internal Energy = U = the sum of all the energy held by the molecules: * the PE stored in their chemical bonds, attractions.

CHAPTER 16 : THERMODYNAMICS

COMBINED AND IDEAL GAS LAWS. COMBINED GAS LAW  Do variables remain constant for gases???  Temperature, pressure, and volume are CONSTANTLY changing.

17.4 State Variables State variables describe the state of a system

Advance Chemical Engineering Thermodynamics

Ideal Gas Law (Equation):

P203/4c17:1 Chapter 17: The First Law of Thermodynamics Thermodynamic Systems Interact with surroundings Heat exchange Q = heat added to the system(watch.

Chapter 19. “A theory is the more impressive the greater the simplicity of its premises, the more different kinds of things it relates, and the more extended.

Chemical Thermodynamics 2013/ th Lecture: Manipulations of the 1 st Law and Adiabatic Changes Valentim M B Nunes, UD de Engenharia.

when system is subdivided? Intensive variables: T, P Extensive variables: V, E, H, heat capacity C.

Chapter 4: Applications of the First Law Different types of work: Configuration work: (reversible process) Dissipative work: (irreversible process) Adiabatic.

Gas Laws Chemistry Mrs. Coyle. Factors Affecting Gas Pressure Number of Moles (Amount of gas) –As the number of particles increases, the number of collisions.

Ch15 Thermodynamics Zeroth Law of Thermodynamics If two systems are in thermal equilibrium with a third system, then they are in thermal equilibrium with.

Heat & The First Law of Thermodynamics

Lecture 4 – The First Law (Ch. 1) Monday January 14 th Finish previous class: functions of state Reversible work Enthalpy and specific heat Adiabatic processes.

Chemical Equilibrium By Doba Jackson, Ph.D.. Outline of Chpt 5 Gibbs Energy and Helmholtz Energy Gibbs energy of a reaction mixture (Chemical Potential)

Thermodynamics Internal energy of a system can be increased either by adding energy to the system or by doing work on the system Remember internal energy.

IV. Kinetic theory (continued – see previous lecture) 5. Heat capacitance a) Monoatomic gas } b) Equipartition principle. Degrees of freedom Diatomic gas,

1 Work and Heat Readings: Chapter Internal Energy -Initial kinetic energy is lost due to friction. -This is not completely true, the initial kinetic.

B2 Thermodynamics Ideal gas Law Review PV=nRT P = pressure in Pa V = volume in m3 n = # of moles T= temperature in Kelvin R = 8.31 J K -1 mol -1 m = mass.

Gas Processes. Gas Process The thermodynamic state of a gas is defined by pressure, volume, and temperature. A “gas process” describes how gas gets from.

Dr.Salwa Al Saleh Lecture 8 Mechanical Work Reversible Processes Free Expansion.

Q18. First Law of Thermodynamics. 1.A quantity of an ideal gas is compressed to half its initial volume. The process may be adiabatic, isothermal or isobaric.

The First Law of Thermodynamics Ideal Gas Processes

Lecture 29: 1st Law of Thermodynamics

Physics 101: Lecture 27, Pg 1 Physics 101: Lecture 27 Thermodynamics l Today’s lecture will cover Textbook Chapter Check your grades in grade.

On expanding isothermally from 2L to 4L, an ideal gas does 6J of work, as the pressure drops from 2 atm to 1 atm. By how much must it expand to do an additional.

Lecture #20: Thermodynamics AP Physics B. Work done by a gas Suppose you had a piston filled with a specific amount of gas. As you add heat, the temperature.

Application of Environment Spatial Information System HW – Perfect Gas Minkasheva Alena Thermal Fluid Engineering Lab. Department of Mechanical Engineering.

11.1 1st Law of Thermodynamics A law is a statement which summarizes our experiences. Among the most fundamental laws (there are no known exceptions to.

The Kinetic Theory of Gases

ERT 108/3 PHYSICAL CHEMISTRY EXERCISES FIRST LAW OF THERMODYNAMICS

Physical Chemistry I (TKK-2246)

The Kinetic Theory of Gases

Physics 101: Lecture 27 Thermodynamics

Prof. Marlon Flores Sacedon

Internal Energy and the First Law of Thermodynamics

Presentation transcript:

To Date…. Ideal Monatomic Gas Cv = 3/2R Cp = Cv + R = 5/2 R Polyatomic Gas Cv > 3/2R Cp > 5/2 R All Ideal Gases • DE = q + w • w = -PextDV (for now) • DE = nCvDT = qV • DH = nCpDT = qP • If DT = 0, then DE = 0 and q = -w

Lecture 3: State Functions Reading: Zumdahl 9.3, 9.4 Outline Example of Thermo. Pathways State Functions (9.4…for laboratory)

Thermodynamic Pathways: an Example Example 9.2. We take 2.00 mol of an ideal monatomic gas undergo the followng: Initial (State A): PA = 2.00 atm, VA = 10.0 L Final (State B): PB = 1.00 atm, VB = 30.0 L • We’ll do this two ways: Path 1: Expansion then Cooling Path 2: Cooling then Expansion

Thermodynamic Jargon When doing pathways, we usually keep one variable constant. The language used to indicate what is held constant is: Isobaric: Constant Pressure Isothermal: Constant Temperature Isochoric: Constant Volume

Thermodynamic Path: A series of manipulations of a system that takes the system from an initial state to a final state isochoric isobaric

Pathway 1 Step 1. Constant pressure expansion (P = 2 atm) from 10.0 l to 30.0 l. PDV = (2.00 atm)(30.0 L - 10.0 L) = 40.0 L.atm = (40.0 L atm)(101.3 J/L.atm) = 4.0 x 103 J = -w And DT = PDV/nR = 4.05 x 103 J/(2 mol)(8.314 J/mol.K) = 243.6 K

Pathway 1 (cont.) Step 1 is isobaric (constant P); therefore, q1 = qP = nCpDT = (2mol)(5/2R)(243.6 K) = 1.0 x 104 J = DH1 And DE1 = nCvDT = (2mol)(3/2R)(243.6 K) = 6.0 x 103 J (check: DE1 = q1 + w1 = (1.0 x 104 J) -(4.0 x 103 J) = 6.0 x 103 J )

Pathway 1 (cont.) Step 2: Isochoric (const. V) cooling until pressure is reduced from 2.00 atm to 1.00 atm. First, calculate DT: Now, DT = DPV/nR (note: P changes, not V) = (-1.00 atm)(30.0 L)/ (2 mol)(.0821 L atm/mol K) = -182.7 K

Pathway 1 (cont.) q2 = qv = nCvDT = (2 mol)(3/2R)(-182.7 K) = - 4.6 x 103 J and DE2 = nCvDT = -4.6 x 103 J and DH2 = nCpDT = -7.6 x 103 J Finally w2 = 0 (isochoric…no V change)

Pathway 1 (end) Thermodynamic totals for this pathway are the sum of values for step 1 and step 2 q = q1 + q2 = 5.5 x 103 J w = w1 + w2 = -4.0 x 103 J DE = DE1 + DE2 = 1.5 x 103 J DH = DH1 + DH2 = 2.5 x 103 J

Next Pathway • Now we will do the same calculations for the green path.

Pathway 2 Step 1: Isochoric cooling from P = 2.00 atm to P = 1.00 atm. First, calculate DT: DT = DPV/nR = (-1.00 atm)(10.0 L)/ (2 mol)R = -60.9 K

Pathway 2 (cont.) Then, calculate the rest for Step 1: q1 = qv = nCvDT = (2 mol)(3/2 R)(-60.9 K) = -1.5 x 103 J = DE1 DH1 = nCPDT = (2 mol)(5/2 R)(-60.9 K) = -2.5 x 103 J w1 = 0 (constant volume)

Pathway 2 (cont.) Step 2: Isobaric (constant P) expansion at 1.0 atm from 10.0 l to 30.0 l. DT = PDV/nR = (1 atm)(20.0 L)/(2 mol)R = 121.8 K

Pathway 2 (cont.) Then, calculate the rest: q2 = qp = nCPDT = (2 mol)(5/2 R)(121.8 K) = 5.1 x 103 J = DH2 DE2 = nCvDT = (2 mol)(3/2 R)(121.8 K) = 3.1 x 103 J w1 = -PDV = -20 l.atm = -2.0 x 103 J

Pathway 2 (end) Thermodynamic totals for this pathway are again the sum of values for step 1 and step 2: q = q1 + q2 = 3.6 x 103 J w = w1 + w2 = -2.0 x 103 J DE = DE1 + DE2 = 1.5 x 103 J DH = DH1 + DH2 = 2.5 x 103 J

Comparison of Path 1 and 2 Note: Energy and Enthalpy are the same, Pathway 1 q = 5.5 x 103 J w = -4.1 x 103 J DE = 1.5 x 103 J DH = 2.5 x 103 J Pathway 2 q = 3.6 x 103 J w = -2.0 x 103 J DE = 1.5 x 103 J DH = 2.5 x 103 J Note: Energy and Enthalpy are the same, but heat and work are not the same!

State Functions A State Function is a function in which the value only depends on the initial and final state….NOT on the pathway taken. In this example, start in Seattle, end in Chicago, but you take different paths to get from one place to the other.

Thermodynamic State Functions Thermodynamic State Functions: Thermodynamic properties that are dependent on the state of the system only. (Example: DE and DH) Other variables will be dependent on pathway (Example: q and w). These are NOT state functions. The pathway from one state to the other must be defined.