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Quiz 8 (Ch. 11)Energy required to change the temperature of an object

c - specific heat

Energy required during the phase change

L - latent heat of vaporization or fusion

LmQ

TTTTcmQ

if

±=

−=∆∆=

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Chapter 12

The Laws of Thermodynamics(Part 1)

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OutlineWork in thermodynamic processesThe first law of thermodynamicsFour types of thermal processes, PV diagramsSpecific heat and degrees of freedomHeat enginesSecond law of thermodynamics

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Work in Thermodynamic Processes – Assumptions

Dealing with an ideal gasAssumed to be in thermodynamic equilibrium

Every part of the gas is at the same temperatureEvery part of the gas is at the same pressure

Ideal gas law applies (PV=nkBT)

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Work in a Gas CylinderGas volume V, pressure PA force F is applied to slowly compress the gasPressure P=F/A

The compression is slow enough for all the system to remain essentially in thermal equilibrium

Work W=F∆yWork done on the gas

W = - P ∆V

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Work on a GasWhen the gas is compressed∆V is negative The work done on the gas is positive

When the gas is allowed to expand∆V is positiveThe work done on the gas is negative

When the volume remains constantNo work is done on the gas

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Notes about the Work Equation

W = - P ∆V

The pressure remains constant during the expansion or compression

This is called an isobaric process (P=const)

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

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PV DiagramsUsed when the pressure and volume are known at each step of the processThe magnitude of the work done on a gas that takes it from initial state to final state is equal to the area under the curve on the PV diagram

This is true whether or not the pressure stays constant

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PV Diagrams, cont.The curve on the diagram is called the pathtaken between the initial and final statesThe work done depends on the particular path

Same initial and final states, but different amounts of work are done

a b c d10

Types of Thermal Processes

Isobaric (P=const)Pressure stays constantHorizontal line on the PV diagram

Isovolumetric (V=const)Volume stays constant, work W=0Vertical line on the PV diagram

Isothermal (T=const)Temperature stays the same

Adiabatic (Q=const)No heat is exchanged with the surroundings

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PV diagram: Isothermal and Adiabatic processes

Adiabatic Q=constIsothermal T=constWork can be found from PV diagram as an area under the graph

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Quick Quiz 12.2

Identify the paths A, B, C, and D in Figure as

isobaric, isothermal, isovolumetric, or adiabatic.

For path B, Q = 0.

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First Law of Thermodynamics

Energy conservation lawThe First Law of Thermodynamics tells that the internal energy of a system can be increased by

Adding energy Q to the systemDoing work W on the system

Applicable to all types of processesProvides a connection between microscopic and macroscopic worlds

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First Law, cont.Energy transfers occur

By doing workRequires a macroscopic displacement of an object through the application of a force

By heatOccurs through the random molecular collisions

Both result in a change in the internal energy, ∆U, of the system

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First Law, EquationIf a system undergoes a change from an initial state to a final state, then

∆U = Uf – Ui = Q + W

Q is the energy transferred to the system by heatW is the work done on the system∆U is the change in internal energy

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First Law – Signs Signs of the terms in the equation

QPositive if energy is transferred to the system by heatNegative if energy is transferred out of the system by heat

WPositive if work is done on the systemNegative if work is done by the system

∆UPositive if the temperature increasesNegative if the temperature decreases

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Notes About Work

Positive work increases the internal energy of the systemNegative work decreases the internal energy of the systemThis is consistent with the definition of mechanical work

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Results of ∆UInternal energy is related to microscopic characteristics (KE, PE of atoms)Changes in the internal energy result in changes in the measurable macroscopic variables of the system

These includePressureTemperatureVolume

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∆U in Isovolumetric processAt V=const work W=0, therefore ∆U=QFor ideal gas U=(3/2)nRTIntroduce molar specific heat at constant volume for an ideal gas

Cv = 3/2 R

The change in internal energy can be expressed as

∆U = nCv∆T

n is number of moles20

Molar Specific HeatA gas with a large molar specific heat requires more energy for a given temperature changeThe value depends on the structure of the gas moleculeThe value also depends on the ways the molecule can store energy

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Degrees of Freedom

Each way a gas can store energy is called a degree of freedomEach degree of freedom contributes ½ R to the molar specific heatSee table 12.1 for some Cv values

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Isobaric process

P = constThermal energy is transferred into gas by the heat

Q= nCp∆TCp - molar heat capacity at P=constCp = 5/2 R

For ideal gasCp - Cv = R

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Example: Problem #9n = 1 moleTi = 0oCP = 1 atmVf=4Vi

Tf - ?Won gas -?R=8.31 J/mole K

VPWnRTPV∆−=

=

P=const

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Example: Problem #17A = 0.15 m2

h =20.0 cmP = 6000 Pa∆U = - 8.0 J

Wby gas -?Q -?R=8.31 J/mole K

gason input WQUVPW

+=∆∆=

P=const

h

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Isolated System

An isolated system does not interact with its surroundingsNo energy transfer takes place and no work is doneTherefore, the internal energy of the isolated system remains constant (∆U=const)

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Cyclic Processes

A cyclic process is one in which the process originates and ends at the same state

Uf = 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

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Heat Engine

A heat engine takes in energy by heat and partially converts it to other formsIn general, a heat engine carries some working substance through a cyclic process

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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)

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

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Second Law of Thermodynamics

Constrains the First LawEstablishes which processes actually occurHeat engines are an important application

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Second Law of ThermodynamicsNo heat engine operating in a cycle can absorb energy from a reservoir and use it entirely for the performance of an equal amount of work

Kelvin – Planck statementMeans that Qc cannot equal 0

Some Qc must be expelled to the environment

Means that e must be less than 100%32

Summary of the First and Second Laws

First LawWe cannot get a greater amount of energy out of a cyclic process than we put in

Second LawWe can’t break even

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Thermal 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

1eng h c c

h h h

W Q Q Q

Q Q Q

−≡ = = −e

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Example: Problem #25

Qh = 3Weng

a) e -?b) Qc/Qh -?

h

c

h

eng

QQ

QW

e −== 1

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Heat Pumps and RefrigeratorsHeat engines can run in reverse

Energy is injectedEnergy is extracted from the cold reservoirEnergy is transferred to the hot reservoir

This process means the heat engine is running as a heat pump

A refrigerator is a common type of heat pumpAn air conditioner is another example of a heat pump

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Heat Pump, contThe work is what you pay forThe Qc is the desired benefitThe coefficient of performance (COP) measures the performance of the heat pump running in cooling mode

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Heat Pump, COP

In cooling mode,

The higher the number, the betterA good refrigerator or air conditioner typically has a COP of 5 or 6

cQCOP

W=

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Heat Pump, COP

In heating mode,

The heat pump warms the inside of the house by extracting heat from the colder outside airTypical values are greater than one

HQCOP

W=