Lecture 9 Overview (Ch. 1-3)

Slides:



Advertisements
Similar presentations
Chapter 3 Interactions and Implications. Entropy.
Advertisements

The Kinetic Theory of Gases
Lecture 2, p 1 Thermal reservoir Today: Equipartition First Law of Thermodynamics  Internal energy  Work and heat Lecture 2: Ideal Gases.
Thermodynamics versus Statistical Mechanics
Lecture 2 The First Law of Thermodynamics (Ch.1)
Lecture 6. Entropy of an Ideal Gas (Ch. 3)
First Law of Thermodynamics Physics 102 Professor Lee Carkner Lecture 6 “of each the work shall become manifest, for the day shall declare it, because.
Lecture 8. Thermodynamic Identities (Ch. 3) We have been considering the entropy changes in the processes where two interacting systems exchanged the thermal.
Internal Energy Physics 202 Professor Lee Carkner Lecture 14.
Lecture 2 The First Law of Thermodynamics (Ch.1)
First law of thermodynamics
Internal Energy Physics 202 Professor Lee Carkner Lecture 16.
Absolute Zero Physics 313 Professor Lee Carkner Lecture 15.
Fig The net work done by the system in the process aba is –500 J.
The Kinetic Theory of Gases Chapter 19 Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Equilibrium Processes in Gases Slow (almost unchanging) with respect to the typical relaxation times of a system Equilibrium processes are represented.
Lecture 5: 2 nd and 3 rd Laws of Thermodynamics 2 nd law: The entropy of an isolated system never decreases. Nernst’s Theorem: The entropy of a system.
17.4 State Variables State variables describe the state of a system
Advance Chemical Engineering Thermodynamics
1 Introduction Physics 313 Professor Lee Carkner Lecture 1.
Lecture 8. Systems with a “Limited” Energy Spectrum The definition of T in statistical mechanics is consistent with our intuitive idea of the temperature.
Lecture 6. Entropy of an Ideal Gas (Ch. 3) Find  (U,V,N,...) – the most challenging step S (U,V,N,...) = k B ln  (U,V,N,...) Solve for U = f (T,V,N,...)
Thermal contact Two systems are in thermal (diathermic) contact, if they can exchange energy without performing macroscopic work. This form of energy.
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.
More Thermodynamics1 ► Specific Heats of a Gas ► Equipartition of Energy ► Reversible and Irreversible Processes  Carnot Cycle  Efficiency of Engines.
Lecture 5 – The Second Law (Ch. 2)
1 Chapter 7. Applications of the Second Law. 2 Consider entropy changes in various reversible (!!!) processes We have: (a) Adiabatic process Hence a reversible.
Chapter 4: Applications of the First Law Different types of work: Configuration work: (reversible process) Dissipative work: (irreversible process) Adiabatic.
Lecture 9 Overview (Ch. 1-3) Format of the first midterm: four problems with multiple questions. The Ideal Gas Law, calculation of  W,  Q and dS for.
Entropy Property Relationships Chapter 7b. The T-ds relations Consider an internally reversible process occurring in a closed system.
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.
The Kinetic Theory of Gases
1 Property Relationships Chapter 6. 2 Apply the differential form of the first law for a closed stationary system for an internally reversible process.
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,
Kinetic theory of gases The macroscopic behavior of an ideal gas can be explained by the kinetic theory of gases which involves the application of the.
Compressing a rod: The equation of state for a rod is: Again, if the compression is done quasi-statically (e.g., with a vise), then.
Lecture 7. Thermodynamic Identities (Ch. 3). Diffusive Equilibrium and Chemical Potential Sign “-”: out of equilibrium, the system with the larger  S/
H. Saibi January 20 th,  The internal Energy of an Ideal Gas  Work and the PV Diagram for a Gas  Heat capacities of Gases  Heat capacities of.
Lecture 29: 1st Law of Thermodynamics
Lecture 7 – The Second Law (Ch. 2)
Results of Midterm points # of students GradePoints A> 85 B+B B60-79 C+C C30-54 D<
The Kinetic Theory of Gases
12. Thermodynamics Temperature
Lecture 17 Overview Ch. 4-5 List of topics
Classical Ideal Gas.
Lecture 6. Entropy of an Ideal Gas (Ch. 3)
The First Law of Thermodynamics
Ideal Gas in the Canonical Ensemble
Results of Midterm 1 # of students
The Third Law of Thermodynamics
The Kinetic Theory of Gases
Gibbs’ Paradox.
Mechanics & Thermodynamics
Quasistatic processes The relation of heat and work
Equipartition of energy (….and some problems with
The First Law of Thermodynamics
Back to the 1st law: PV cycles
Recall the Equipartition Theorem: In Ch 6,
The Kinetic Theory of Gases
Chapter 4 ENERGY ANALYSIS OF CLOSED SYSTEMS
Thermodynamics.
The Kinetic Theory of Gases
Classical Ideal Gas.
The Micro/Macro Connection
Heat Flow.
A THERMODYNAMIC SYSTEM
Presentation transcript:

Lecture 9 Overview (Ch. 1-3) Format of the first midterm: three problems with multiple questions. Total: 100 points. The Ideal Gas Law, calculation of W, Q and dS for various ideal gas processes. Einstein solid and two-state paramagnet, multiplicity and entropy, the stat. phys. definition of T, how to get from the multiplicity to the equation of state. Only textbook and cheat-sheets (handwritten!) are allowed. No homeworks and lecture notes. DO NOT forget to bring your calculator!

Problem 1 One mole of a monatomic ideal gas goes through a quasistatic three-stage cycle (1-2, 2-3, 3-1) shown in the Figure. T1 and T2 are given. (a) (10) Calculate the work done by the gas. Is it positive or negative? (b) (20) Using two methods (Sackur-Tetrode eq. and dQ/T), calculate the entropy change for each stage and for the whole cycle, Stotal. Did you get the expected result for Stotal? Explain. (c) (5) What is the heat capacity (in units R) for each stage? T V V1 V2 T1 T2 1 2 3

Problem 1 (cont.) (a) 1 – 2 V  T  P = const (isobaric process) 2 – 3 V = const (isochoric process) 3 – 1 T = const (isothermal process)

Problem 1 (cont.) (b) 1 – 2 2 – 3 3 – 1 Sackur-Tetrode equation: V 3 V  T  P = const (isobaric process) T1 T2 T 2 – 3 V = const (isochoric process) 3 – 1 T = const (isothermal process) as it should be for a quasistatic cyclic process (quasistatic – reversible), because S is a state function.

Problem 1 (cont.) (b) 1 – 2 2 – 3 3 – 1 - for quasi-static processes V V  T  P = const (isobaric process) V1 1 T1 T2 T 2 – 3 V = const (isochoric process) 3 – 1 T = const (isothermal process)

Problem 1 (cont) (c) Let’s express both Q and dT in terms of dV : V 3 V2 2 1 – 2 V  T  P = const (isobaric process) V1 1 T1 T2 T 2 – 3 V = const (isochoric process) 3 – 1 T = const (isothermal process), dT = 0 while Q  0 At home: recall how these results would be modified for diatomic and polyatomic gases.

Problem 2 One mole of a monatomic ideal gas goes through a quasistatic three-stage cycle (1-2, 2-3, 3-1) shown in the Figure. Process 3-1 is adiabatic; P1 , V1 , and V2 are given. (a) (10) For each stage and for the whole cycle, express the work W done on the gas in terms of P1, V1, and V2. Comment on the sign of W. (b) (5) What is the heat capacity (in units R) for each stage? (c) (15) Calculate Q transferred to the gas in the cycle; the same for the reverse cycle; what would be the result if Q were an exact differential? (d) (15) Using the Sackur-Tetrode equation, calculate the entropy change for each stage and for the whole cycle, Stotal. Did you get the expected result for Stotal? Explain. P 1 2 3 V V1 V2 P1

Problem 2 (cont.) (a) P V V1 V2 P1 1 – 2 2 – 3 3 – 1 P = const (isobaric process) 2 – 3 V = const (isochoric process) 3 – 1 adiabatic process

Problem 2 (cont.) (c) P V V1 V2 P1 1 – 2 2 – 3 3 – 1 P = const (isobaric process) 2 – 3 V = const (isochoric process) 3 – 1 adiabatic process For the reverse cycle: If Q were an exact differential, for a cycle Q should be zero.

Problem 2 (cont.) (d) P P1 V1 V2 1 – 2 V 2 – 3 3 – 1 Sackur-Tetrode equation: P (d) 1 2 P1 3 V1 V2 1 – 2 V  T  P = const (isobaric process) V 2 – 3 V = const (isochoric process) 3 – 1 Q = 0 (quasistatic adiabatic = isentropic process) as it should be for a quasistatic cyclic process (quasistatic – reversible), because S is a state function.

we need to find the equation of this process V=V(T) Problem 3 Calculate the heat capacity of one mole of an ideal monatomic gas C(V) in the quasi-static process shown in the Figure. P0 and V0 are given. P 10 Start with the definition: P0 we need to find the equation of this process V=V(T) 20 30 V0 V 40

Problem 3 (cont.) P V P0 V0 T=const isotherm S=const adiabat V0/2 T=const isotherm S=const adiabat V0/2 5V0/8 Problem 3 (cont.) 50 Does it make sense? C/R the line touches an isotherm 2.5 1.5 1/2 5/8 1 V/ V0 the line touches an adiabat

Problem 4 (10) The ESR (electron spin resonance) set-up can detect the minimum difference in the number of “spin-up” and “spin-down” electrons in a two-state paramagnet N-N =1010. The paramagnetic sample is placed at 300K in an external magnetic field B = 1T. The component of the electron’s magnetic moment along B is  B =  9.3x10-24 J/T. Find the minimum total number of electrons in the sample that is required to make this detection possible. - the high-T limit

Problem 5 (a) Consider a system whose multiplicity is described by the equation: where U is the internal energy, V is the volume, N is the number of particles in the system, Nf is the total number of degrees of freedom, f(N) is some function of N. (10) Find the system’s entropy and temperature as functions of U. Are these results in agreement with the equipartition theorem? Does the expression for the entropy makes sense when T  0? (5) Find the heat capacity of the system at fixed volume. (15) Assume that the system is divided into two sub-systems, A and B; sub-system A holds energy UA and volume VA, while the sub-system B holds UB=U-UA and VB=V-VA. Show that for an equilibrium macropartition, the energy per molecule is the same for both sub-systems. (a) - in agreement with the equipartition theorem When T  0, U  0, and S  -  - doesn’t make sense. This means that the expression for  holds in the “classical” limit of high temperatures, it should be modified at low T.

Problem 5 (cont.) (b) (c)