PHY 206 SPRING Problem #1 NAME: SIGNATURE: UM ID: Problem #2. Problem #3. Total. Prof. Massimiliano Galeazzi. Midterm #2 March 8, 2006

Similar documents
Chapter 19 The First Law of Thermodynamics

PHY 206 SPRING Problem #1. Problem #2 NAME: SIGNATURE: Problem #3 UM ID: Problem #4. Problem #5. Problem #6. Total. Prof. Massimiliano Galeazzi

THERMODYNAMICS. Chapter Twelve MCQ I

Handout 12: Thermodynamics. Zeroth law of thermodynamics

Irreversible Processes

Handout 12: Thermodynamics. Zeroth law of thermodynamics

Distinguish between an isothermal process and an adiabatic process as applied to an ideal gas (2)

Cyclic Processes. water

Lecture 7, 8 and 9 : Thermodynamic process by: Asst. lect. Karrar Al-Mansoori CONTENTS. 7) Thermodynamic process, path and cycle 2

Downloaded from

Process Nature of Process

Conservation of Energy

Thermodynamic system is classified into the following three systems. (ii) Closed System It exchanges only energy (not matter) with surroundings.

Dr. Gundersen Phy 206 Test 2 March 6, 2013

NOTE: Only CHANGE in internal energy matters

The Laws of Thermodynamics

Irreversible Processes

1. Second Law of Thermodynamics

PHYSICS 214A Midterm Exam February 10, 2009

Survey of Thermodynamic Processes and First and Second Laws

More Thermodynamics. Specific Specific Heats of a Gas Equipartition of Energy Reversible and Irreversible Processes

TB [103 marks] The damping of the system is now increased. Which describes the change in ƒ 0 and the change in A 0?

The laws of Thermodynamics. Work in thermodynamic processes

Version 001 HW 15 Thermodynamics C&J sizemore (21301jtsizemore) 1

The goal of thermodynamics is to understand how heat can be converted to work. Not all the heat energy can be converted to mechanical energy

The First Law of Thermodynamics

1. Second Law of Thermodynamics

Summarizing, Key Point: An irreversible process is either spontaneous (ΔS universe > 0) or does not occur (ΔS universe < 0)

Lecture 2.7 Entropy and the Second law of Thermodynamics During last several lectures we have been talking about different thermodynamic processes.

Lecture 5. PHYC 161 Fall 2016

Chapter 12. The Laws of Thermodynamics

(prev) (top) (next) (Throughout, we will assume the processes involve an ideal gas with constant n.)

Aljalal-Phys March 2004-Ch21-page 1. Chapter 21. Entropy and the Second Law of Thermodynamics

Atkins / Paula Physical Chemistry, 8th Edition. Chapter 3. The Second Law

Chapter 16 Thermodynamics

The Kinetic Theory of Gases

PHYS 1101 Practice problem set 6, Chapter 19: 7, 12, 19, 30, 37, 44, 53, 61, 69

Heat and Thermodynamics. February. 2, Solution of Recitation 2. Consider the first case when air is allowed to expand isothermally.

1985B4. A kilogram sample of a material is initially a solid at a temperature of 20 C. Heat is added to the sample at a constant rate of 100

Chapter 2 Carnot Principle

Heat What is heat? Work = 2. PdV 1

Chapter 20. Heat Engines, Entropy and the Second Law of Thermodynamics. Dr. Armen Kocharian

UNIVESITY OF SWAZILAND FACl.JLTY OF SCIENCE AND ENGINEERING DEPARTMENT OF PHYSICS

The first law of thermodynamics continued

Chapter 19. First Law of Thermodynamics. Dr. Armen Kocharian, 04/04/05

Chapter 19. Heat Engines

Physics 202 Homework 5

Lecture Notes 2014March 13 on Thermodynamics A. First Law: based upon conservation of energy

Q1. A) 53.3 cm/s B) 59.8 cm/s C) 77.5 cm/s D) 35.1 cm/s E) 44.7 cm/s. Ans: 1.6 Q2.

Reversibility. Processes in nature are always irreversible: far from equilibrium

Physics 231. Topic 14: Laws of Thermodynamics. Alex Brown Dec MSU Physics 231 Fall

Physics Fall Mechanics, Thermodynamics, Waves, Fluids. Lecture 32: Heat and Work II. Slide 32-1

CH 15. Zeroth and First Law of Thermodynamics

Reversible Processes. Furthermore, there must be no friction (i.e. mechanical energy loss) or turbulence i.e. it must be infinitely slow.

General Physics I (aka PHYS 2013)

Chapter 20 The Second Law of Thermodynamics

Physics 7B Midterm 1 Problem 1 Solution

Entropy in Macroscopic Systems

Final Review Solutions

Physics 111. Thursday, Dec. 9, 3-5pm and 7-9pm. Announcements. Thursday, December 9, 2004

Thermodynamics: The Laws

C e. Negative. In a clockwise cycle, the work done on the gas is negative. Or for the cycle Qnet = +600 J and U = 0 so W = Q = 600 J

A thermodynamic system is taken from an initial state X along the path XYZX as shown in the PV-diagram.

The First Law of Thermodynamics

MidTerm. Phys224 Spring 2008 Dr. P. Hanlet

Lecture 2 - Thursday, May 11 th, 3pm-6pm

Phys102 First Major- 161 Code: 20 Coordinator: Dr. A. Naqvi Saturday, October 29, 2016 Page: 1

Work, heat and the first law of thermodynamics

Heat Engines and the Second Law of Thermodynamics

Physics 150. Thermodynamics. Chapter 15

CHAPTER - 12 THERMODYNAMICS

First Law of Thermodynamics


Details on the Carnot Cycle

Problem: Calculate the entropy change that results from mixing 54.0 g of water at 280 K with 27.0 g of water at 360 K in a vessel whose walls are

THERMODYNAMICS CONCEPTUAL PROBLEMS

LESSON No. 9 WORK TRANSFER: In thermodynamics the work can be defined as follows:

Entropy & the Second Law of Thermodynamics

Chapter 3 - First Law of Thermodynamics

Worksheet for Exploration 21.1: Engine Efficiency W Q H U

This follows from the Clausius inequality as a consequence of the second law of thermodynamics. Therefore. (for reversible process only) (22.

Lecture Notes Set 4c: Heat engines and the Carnot cycle

Phys102 Term: 103 First Major- July 16, 2011

What is thermodynamics? and what can it do for us?

AP Physics Thermodynamics Wrapup

Chapter 19 The First Law of Thermodynamics

Renewable Energy. Theory: The Ideal Gas Law The equation of state for an ideal gas is written: PV = nrt

U = 4.18 J if we heat 1.0 g of water through 1 C. U = 4.18 J if we cool 1.0 g of water through 1 C.

Topic 3 &10 Review Thermodynamics

Physics 1501 Lecture 37

THERMODYNAMICS b) If the temperatures of two bodies are equal then they are said to be in thermal equilibrium.

dv = adx, where a is the active area of the piston. In equilibrium, the external force F is related to pressure P as

Chapter 12 Thermodynamics

HEAT, WORK, AND THE FIRST LAW OF THERMODYNAMICS

Classical Physics I. PHY131 Lecture 36 Entropy and the Second Law of Thermodynamics. Lecture 36 1

The first law of thermodynamics. U = internal energy. Q = amount of heat energy transfer

Lecture 25 Goals: Chapter 18 Understand the molecular basis for pressure and the idealgas

Physics 121, April 24. Heat and the First Law of Thermodynamics. Department of Physics and Astronomy, University of Rochester

Physics 121, April 24. Heat and the First Law of Thermodynamics. Physics 121. April 24, Physics 121. April 24, Course Information

Q1. A) 21.0 ms B) 63.1 ms C) 31.5 ms D) 45.2 ms E) 73.1 ms. Ans: Q2.

Transcription:

PHY 06 SPRING 006 Prof. Massimiliano Galeazzi Midterm # March 8, 006 NAME: Problem # SIGNAURE: UM ID: Problem # Problem # otal Some useful relations: st lat of thermodynamic: U Q - W Heat in an isobaric process: Q n C P Heat in an isochoric process: Q n C Adiabatic processes: γ p const., const. Work: dw p d Ideal gas equation of state: p n R Engine efficiency: e W Entropy: Wave equation: Speed of propagation of a wave on a string: Q H dq ds y( v µ y( v t

PHY 06 Spring 006 Midterm # March 8, 006 Problem # A quantity of n moles of a diatomic ideal gas is initially enclosed in a volume o and its pressure is p o. he gas expands adiabatically to three times its initial volume, it is then compressed isothermally to the original volume, and finally it is brought back to its initial state at constant volume. All processes are reversible. [NOE: Write the results only in terms of p o, o, and n a) What is the initial temperature of the gas? b) Draw the processes on the p- diagram to the left. c) What is the temperature of the gas during the isothermal compression? d) Calculate the heat transfer for each of the three processes. e) What is the efficiency of an engine operating through this cycle? f) What is the total change in entropy of the system for one cycle? p a) p oo nro, o o o nr γ c) Process ab is adiabatic, therefore const. γ ( ) p oo γ o o, C nr γ nr, with γ P C d) Q ab 0 (adiabatic) c poo poo Q bc Wbc nr ln nr ln ln b nr p oo poo Q ca nc ( a c ) n R γ p oo nr nr poo ln Q e) C Qbc ln e + + Q γ H Qca p oo f) he cycle is fully reversible, therefore S 0 p o o 7 γ ( ) ( ) o R R 7 γ ( )

PHY 06 Spring 006 Midterm # March 8, 006 Problem # One mole of an ideal gas at temperature o fills one third of a cylinder. he gas is kept in there through a thin partition. he rest of the cylinder is empty. he cylinder is also provided with a piston and is in thermal contact with a heat reservoir much bigger than the cylinder (see figure). a) he thin partition is broken and the gas expands freely occupying the whole volume. What are the work W, the heat transfer Q, and the change in internal energy U of the gas? b) Calculate the entropy change S in the gas. c) he gas is then compressed using the piston to its initial volume. he temperature is maintained constant through the contact with the heat reservoir. he reservoir is big enough that its temperature does not change. Calculate the change in entropy of the gas for this process. d) he processed described in a) and c) bring the gas from an initial state back to the same state, therefore representing a cyclical process. What is the change in entropy of the Universe for the full cycle? e) Is your answer to question d) in agreement with the second law of thermodynamic? Explain in no more than one sentence. f) Since the process is cyclical, you could restore the thin partition and start a new cycle again. Could you use this cycle either as thermal engine or a refrigerator? Qualitatively explain what happens to work and heat transfer in the cycle. a) Isothermal (there is the heat reservoir) U 0 ; Free expansion W0; (first law of D) Q0. b) Process is irreversible, use reversible isothermal dq Q W nr ln i S R ln w N f w Also, i S kb ln kb ln NkB ln nr ln R ln wi wi f nr ln dq Q W c) As in the first method in b): i S R ln R ln d) Suniverse S gas + Sreservoir S + S + Sreservoir Sreservoir, f

PHY 06 Spring 006 Midterm # March 8, 006 f nr ln dq Q Qgas W reservoir gas i Sreservoir R ln R ln e) YES, the entropy of the universe is bigger than zero and it is always supposed to be bigger or equal to zero. f) NO, first of all, there is only heat reservoir and we have seen that engines and refrigerators need. Second, the net result of the process is to convert work into heat, therefore this is not a refrigerator, nor an engine. 4

PHY 06 Spring 006 Midterm # March 8, 006 Problem # A very long string is fixed at one end on a wall and on the other end is wrapped around a wheel of negligible mass and then attached to a mass M to provide the tension in the string (see figure). he string has mass m and length L. Assume that the string is long enough to neglect reflections at the wall. A transverse wave is generated on the string. If the origin if the x-axis is placed at the wheel, the wave can be described by the function: ( Asin [ B( Cx y for x>0 and t>0, With A, B, and C constant. [Write your results in terms of M, m, L, A, B, and x o. You may need the trigonometric expression for a generic angle θ: sin θ sinθ cosθ. a) Show that the wave function satisfies the wave equation. b) What is the value of the constant C? c) What is the maximum displacement of the particles in the string? d) What is the maximum transverse speed v y of the particles in the string? e) If you measure the speed and the acceleration of the particles at a point with coordinate x o, at what time t does the string reach the maximum displacement at the point? f) At what time t does it reach the maximum transverse speed at that point of coordinate x o? a) y( y( v t y ABC sin[ B( Cx cos B( Cx ABC sin B Cx t y AB C cos[ B( Cx y ABsin[ B( Cx cos[ B( Cx ABsin B( Cx t y AB cos[ B( Cx t AB C cos[ B( Cx AB cos[ B( Cx v Satisfied for C C v v [ [ ( ) [

PHY 06 Spring 006 Midterm # March 8, 006 b) See a), or remember than wave equation is satisfied for y y( x ± v and considering that y( Asin [ B( Cx Asin [ BC( x C v C c) y is max for sin [ ( Cx ± B y max A y v y t v y is max for sin [ B ( Cx ± v max AB d) ABsin[ B( Cx cos[ B( Cx ABsin[ B( Cx t Cxo B B Cxo t,, t Cxo 4B e) y is max for sin [ B ( Cx ± ( Cx,,,... f) y taking the first value: ( Cx B o B o v is max for sin [ B ( Cx ± ( ),... B o taking the first value: ( Cx 6