fisika dasar - Minggu 13 - Thermodynamics

04/11/23 Physics 103, Spring 2004, U.Wisconsin 1

Lecture 22Lecture 22

The Laws of Thermodynamics


Work in Thermodynamic Processes Work in Thermodynamic Processes – State Variables– State Variables

State of a systemDescription of the system in terms of state

variables

» Pressure

» Volume

» Temperature (Internal Energy)A macroscopic state of an isolated system can be

specified only if the system is in internal thermal equilibrium


Work in Thermodynamic ProcessesWork in Thermodynamic Processes

Work is an important energy transfer mechanism in thermodynamic systems

Heat is another energy transfer mechanism

QuickTime™ and a Animation decompressor are needed to see this picture.


Work in a Gas CylinderWork in a Gas Cylinder

The gas is contained in a cylinder with a moveable piston

The gas occupies a volume V and exerts pressure P on the walls of the cylinder and on the piston


Work in a Gas Cylinder, cont.Work in a Gas Cylinder, cont.

A force is applied to slowly compress the gas The compression is slow

enough for all the system to remain essentially in thermal equilibrium

W = - P V This is the work done on

the gas


More about Work on a Gas CylinderMore about Work on a Gas Cylinder

When the gas is compressedV is negative The work done on the gas is positive

When the gas is allowed to expandV is positiveThe work done on the gas is negative

When the volume remains constantNo work is done on the gas


Notes about the Work EquationNotes about the Work Equation

The pressure remains constant during the expansion or compressionThis is called an isobaric process

If the pressure changes, the average pressure may be used to estimate the work done


PV DiagramsPV Diagrams

Used when the pressure and volume are known at each step of the process

The work done on a gas that takes it from some initial state to some final state is the negative of the area under the curve on the PV diagramThis is true whether or

not the pressure stays constant


Lecture 22,Lecture 22, Preflight 1Preflight 1

49%

23%

28%

0% 10% 20% 30% 40% 50%

Shown in the picture below are the pressure versus volume graphs for two thermal processes, in each case moving a system from state A to state B along the straight line shown. In which case is the work done by the system the biggest?

1. Case 1

2. Case 2

3. Same

A

B4

2

3 9 V(m3)

Case 1

A

B

4

2

3 9 V(m3)

P(atm)

Case 2

P(atm)

correct

Net Work = area under P-V curve

Area the same in both cases!


Follow up QuestionFollow up Question

A curve on a PV diagram depicts the path taken between the initial (Pi, Vi) and final (Pf, Vf) states

Does the work done depends on the particular path taken if a system moves between the same two states?YesNo


Other ProcessesOther Processes

IsovolumetricVolume stays constantVertical line on the PV diagram

IsothermalTemperature stays the same (PV = constant)

AdiabaticNo heat is exchanged with the surroundings


Processes for Transferring EnergyProcesses for Transferring Energy

By doing work Requires a macroscopic displacement of the point of application of

a force By heat

Occurs by random molecular collisions Results of both

Change in internal energy of the system Generally accompanied by measurable macroscopic variables

» Pressure

» Temperature

» Volume


First Law of ThermodynamicsFirst Law of Thermodynamics Quantities of interest

Q - Heat

» Positive if energy is transferred to the systemW - Work

» Positive if done on the systemU - Internal energy

» Positive if the temperature increases The relationship among U, W, and Q can be expressed as

U = Uf – Ui = Q + W This means that the change in internal energy of a system is

equal to the sum of the energy transferred across the system boundary by heat and the energy transferred by work



46%

21%

33%

0% 10% 20% 30% 40% 50%

Shown in the picture below are the pressure versus volume graphs for two thermal processes, in each case moving a system from state A to state B along the straight line shown. In which case is the change in internal energy of the system the biggest?

1. Case 1

2. Case 2

3. Same

A

B4

2

3 9 V(m3)

Case 1

A

B

4

2

3 9 V(m3)

P(atm)

Case 2

P(atm)correct

T increase higher in case 1 than in case 2Therefore, U increases more in case 1 than in case 2

P1AV1A=2x3=6, P1BV1B=4x9=36, 6 to 36

P2AV2A=4x3=12, P2BV2B=2x9=18, 12 to 18


Follow up QuestionFollow up Question

Shown in the picture below are the pressure versus volume graphs for two thermal processes, in each case moving a system from state A to state B along the straight line shown. In which case is the heat added to the system the biggest?

1. Case 1

2. Case 2

3. Same

A

B4

2

3 9 V(m3)

Case 1

A

B

4

2

3 9 V(m3)

P(atm)

Case 2

P(atm)correct

Q1 = U1 - W

Q2 = U2 - W

W is the same for both, but U1 > U2


Applications of the First Law –Applications of the First Law –Isolated SystemIsolated System

An isolated system does not interact with its surroundings No energy transfer takes place and no work is done Therefore, the internal energy of the isolated system

remains constant


Applications of the First Law –Applications of the First Law –Cyclic ProcessesCyclic Processes

A cyclic process is one in which the process originates and ends at the same stateUf = Ui and Q = -W

The net work done per cycle by the gas is equal to the area enclosed by the path representing the process on a PV diagram


Cyclic Process in a PV DiagramCyclic Process in a PV Diagram

This is an ideal monatomic gas confined in a cylinder by a moveable piston

A to B is an isovolumetric process which increases the pressure

B to C is an isothermal expansion and lowers the pressure

C to A is an isobaric compression

The gas returns to its original state at point A


Applications of the First Law –Applications of the First Law –Isothermal ProcessesIsothermal Processes

Isothermal means constant temperature

The cylinder and gas are in thermal contact with a large source of energy

Allow the energy to transfer into the gas (by heat)

The gas expands and pressure falls to maintain a constant temperature

The work done is the negative of the heat added


Applications of the First Law – Applications of the First Law – Adiabatic ProcessAdiabatic Process

Energy transferred by heat is zero The work done is equal to the change in the internal energy

of the system One way to accomplish a process with no heat exchange is

to have it happen very quickly In an adiabatic expansion, the work done is negative and

the internal energy decreases


Applications of the First Law – Applications of the First Law – Isovolumetric ProcessIsovolumetric Process

No change in volume, therefore no work is done The energy added to the system goes into increasing the

internal energy of the systemTemperature will increase


Additional Notes About the First LawAdditional Notes About the First Law

The First Law is a general equation of Conservation of Energy

There is no practical, macroscopic, distinction between the results of energy transfer by heat and by work

Q and W are related to the properties of state for a system


The First Law and Human The First Law and Human MetabolismMetabolism

The First Law can be applied to living organisms The internal energy stored in humans goes into other forms

needed by the organs and into work and heat The metabolic rate (U / T) is directly proportional to the

rate of oxygen consumption by volumeBasal metabolic rate (to maintain and run organs, etc.) is

about 80 W


Various Metabolic RatesVarious Metabolic Rates

Fig. T12.1, p. 369

Slide 11


Aerobic FitnessAerobic Fitness

One way to measure a person’s physical fitness is their maximum capacity to use or consume oxygen

Fig. T12.2, p. 370

Slide 12


Efficiency of the Human BodyEfficiency of the Human Body

Efficiency is the ratio of the mechanical power supplied to the metabolic rate or total power input

F ig . T 12 .3 , p . 3 7 0

S lide 1 3


Lecture 22,Lecture 22, Preflight 3 & 4Preflight 3 & 4

64%

36%

0% 20% 40% 60% 80%

Consider a hypothetical device that takes 1000 J of heat from a hot reservoir at 300K, ejects 200 J of heat to a cold reservoir at 100K, and produces 800 J of work.

Does this device violate the first law of thermodynamics ?

1. Yes

2. No

This device doesn't violate the first law of thermodynamics because no energy is being created nor destroyed. All the energy is conserved.

correct

Pretty SureNot Quite SureJust Guessing


Heat EngineHeat Engine

A heat engine is a device that converts internal energy to other useful forms, such as electrical or mechanical energy

A heat engine carries some working substance through a cyclical process

QuickTime™ and a Animation decompressor are needed to see this picture.


Heat Engine, cont.Heat Engine, cont.

Energy is transferred from a source at a high temperature (Qh)

Work is done by the engine (Weng)

Energy is expelled to a source at a lower temperature (Qc)


Heat Engine, cont.Heat Engine, cont.

Since it is a cyclical process, U = 0 Its initial and final

internal energies are the same

Therefore, Qnet = Weng The work done by the

engine equals the net energy absorbed by the engine

The work is equal to the area enclosed by the curve of the PV diagram


Thermal Efficiency of a Heat EngineThermal Efficiency of a Heat Engine

Thermal efficiency is defined as the ratio of the work done by the engine to the energy absorbed at the higher temperature

e = 1 (100% efficiency) only if Qc = 0No energy expelled to cold reservoir

h

c

h

ch

h

eng

Q

Q1

Q

QQ

Q

W

e


Second Law of ThermodynamicsSecond Law of Thermodynamics

It is impossible to construct a heat engine that, operating in a cycle, produces no other effect than the absorption of energy from a reservoir and the performance of an equal amount of workKelvin – Planck statementMeans that Qc cannot equal 0

» Some Qc must be expelled to the environmentMeans that e cannot equal 100%


Lecture 22,Lecture 22, Preflight 5Preflight 5Consider a hypothetical device that takes 1000 J of heat from a hot reservoir at 300K, ejects 200 J of heat to a cold reservoir at 100K, and produces 800 J of work.

What is the efficiency of this engine ?

1. 80 %

2. 20%

3. 25%

correct

W (800) = Qhot (1000) - Qcold (200) Efficiency = W/Qhot = 800/1000 = 80%

15%

33%

52%

0% 20% 40% 60%



64%

36%

0% 20% 40% 60% 80%

Consider a hypothetical device that takes 1000 J of heat from a hot reservoir at 300K, ejects 200 J of heat to a cold reservoir at 100K, and produces 800 J of work.

Does this device violate the second law of thermodynamics ?

1. Yes

2. No

This device doesn't violate the second law of thermodynamics.

correct

Pretty SureNot Quite SureJust Guessing

fisika dasar - Minggu 13 - Thermodynamics

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