Lecture 26: Chapter 16 Thermodynamics Final: Weds. Dec 15, 2010, 7-9 PM

Location: EE129 Material: Chapter 11 - Chapter 16 Bring a number 2 pencil, your calculator and your student ID. You can bring a one-page (two-side) crib sheet.

NOTE: AN EQUATION SHEET WILL BE GIVEN BUT YOU ARE REMINDED THAT IT IS YOUR RESPONSIBILITY TO CREATE WHATEVER TWO-SIDED CRIB SHEET YOU WANT TO BRING TO THIS EXAM.

Physics 218 Fall 2010 Temperature Zeroth Law of Thermodynamics: *Two systems individually in thermal equilibrium with a third system are in thermal equilibrium with each other.

Implications:

*Temperature is an indicator of thermal equilibrium.

*All parts of a system must be in thermal equilibrium if the system is defined by a single temperature.

Physics 218 Fall 2010 First Law of Thermodynamics First Law of Thermodynamics: *The change in internal energy U of a system is equal to the heat flow into the system plus the work done the system. U  Q  W Implications: *U of system tends to increase if heat is absorbed by the system or if positive work is done on the system

*U of system tends to decrease if heat is lost by the system or if negative work is done on the system. Physics 218 Fall 2010 First Law of Thermodynamics

*The change in internal energy U of a system is equal to the heat flow into the system plus the work done the system. U  Q  W

Meaning of + sign Meaning of - Sign

Q Heat flows into the system Heat flows out of the system

W Work done by the system Surroundings do positive work on system

U Decreases Increases

Physics 218 Fall 2010 Thermal Processes Isobaric: Constant Pressure

Physics 218 Fall 2010 If P=constant, W done by Thermal Processes gas is proportional to the volume change Isobaric: Constant Pressure

As the gas expands, it does work on the piston. Force on the piston by the gas is:

F  Po A The piston moves from position xi to xf.

Vi  Axi Vf  Ax f

Work done by the gas:

W  F(x f  xi )  Po A(x f  xi )  Po (Ax f  Axi )  Po (Vf  Vi ) W  PV Physics 218 Fall 2010 Thermal Processes Isobaric: Constant Pressure

1.00 gram of water vaporizes isobarically at atmospheric pressure ( 1.013 x105 Pa). Its volume in the liquid state is 1.00 cm3 and in the vapor state 1671 cm3. What is the work done by expansion? What is the change in internal energy?

W  Po (Vf  Vi )  169J 6 Q  mLv  0.001kg *2.26x10 J / kg  2600J U   Q W 2600 1691692600  2090 2431JJJWQU

Heat is added to vaporize, thus work is done by system (+) in expansion

Physics 218 Fall 2010 Thermal Processes Isochoric: Constant Volume

U  Q

3 U  nRT 2

Physics 218 Fall 2010 Thermal Processes Isothermal: Constant T

For an ideal gas isotherm, P-V relationship PV  NkT  constant For any given isotherm, NkT const P= P  constant the work done by the V V expanding gas is

Because T=0:  Vf   Vf  W=NkT ln   nRT ln  3  Vi   Vi  U  nRT  0 2 Q  W Work goes into heat or heat goes into work

Physics 218 Fall 2010 Thermal Processes Isothermal: Constant T Suppose 0.50 mol of a monoatomic ideal gas is changed from state A to state D by one of the processes shown on the PV diagram of the figure below. Find the total work done by the gas if it follows the constant temperature path A-C followed by the constant pressure path C-D

 Vf   Vf  W=NkT ln   nRT ln   Vi   Vi  Work done by gas is positive. Gas expands, to PV  nRT  constant maintain constant PV temperature, heat flows T=  constant into the system nR

V f   nRTW ln   VP  Vi 

Physics 218 Fall 2010 Thermal Processes Adiabatic: without heat transfer

DEMO H16 Q  0 U  W

When work is done by the system, the internal energy decreases. When work is done on the system, the internal energy increases.

Physics 218 Fall 2010 Thermal Processes: Summary

Physics 218 Fall 2010 Quiz

Rank the three ideal gas expansions according to the amount of work done by the gas, starting with the least.

A. a,b,c B. b,c,a C. c,a,b D. b,a,c E. c,b,a

Physics 218 Fall 2010