Chem131_Chapter2

7/28/2019 Chem131_Chapter2

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CHAPTER 2

First Law of Thermodynamics


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Two types of quantities

1) State functions: Depend only on the thermodynamic state of the system not on howthe system got there (e.g., P, T, V, ...)

FIRST LAW OF THERMODYNAMICS

Thermodynamics is based on laws that have never been observed to be violated


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2) Process-dependent quantities: Depend on the way in which the system changesits thermodynamic state



surroundings

system

state a

Pa, Ta, Va


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surroundings

system

state b

Pb, Tb, Vb

Pa, Ta, Va

system

state a


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surroundings

system

state a

system

state bPa, Ta, Va

Pb, Tb, Vb


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Two types of processes

1) Reversible: System and surroundings are infinitesimally close to equilibrium throughout

the process.

2) Irreversible: System and/or surroundings not in equilibrium except at beginning and atthe end (sometimes in between).



Natural processes: irreversible


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Work (mechanical): Involves moving an object against a force (e.g., gravity, friction, piston)

Work may be

Positive: work done on the system

Negative: work done by the system

P-V WORK

Work (w) depends on the specific path

surroundings

system

state a

system

state b


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w = Fdr = PextAn

r =

PextAdr

w = PextdV

Expansion: dr > 0 dV > 0 w < 0

w = F dr

dr

Force = gravityF = PextAn

|n| = 1

A = area

System

Fn

A

w > 0dV < 0Compression: dr < 0

P-V WORK


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Force = gravity

w = PextdV

System

F drn

A

surroundings

system

state a

system

state b

w =

f

Vi

Pext(T, V)dV

Line integral

P-V WORK


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Examples

n = 1 mole of gas

w = P2(V2 V1)

P1, V1, TP2, V2, T

T1 = T2 = T

w =

f

Vi

Pext(T, V)dV

V2V1

P1

P2

Pext

V

w < 0 work is done by the system

P-V WORK


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Examples

n = 1 mole of gas

P1, V1, TP2, V2, T

T1 = T2 = T

w =

f

Vi

Pext(T, V)dV

V2V1

P1

P2

Pext

V


w = P1(V2 V1)

P-V WORK


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Examples

n = 1 mole of gas

P1, V1, TP2, V2, T

T1 = T2 = T

w =

f

Vi

Pext(T, V)dV

V2V1

P1

P2

Pext

V


Pa

Va

w = Pa(Va V1) P2(V2 Va)

P-V WORK


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Examples

n = 1 mole of gas

P1, V1, TP2, V2, T

T1 = T2 = T

w =

f

Vi

Pext(T, V)dV

V2V1

P1

P2

Pext

V

P-V WORK


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Examples

n = 1 mole of gas

P1, V1, TP2, V2, T

T1 = T2 = T

w =

f

Vi

Pext(T, V)dV

V2V1

P1

P2

Pext

V

Reversible processThe system is always infinitesimally close to equilibrium

= ext

P-V WORK


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Examples

n = 1 mole of gas

T1 = T2 = T

w = f

ViPext(T, V)dV

V2V1

P1

P2

Pext

V

Reversible processThe system is always infinitesimally close to equilibrium: P = Pext

P =RT

V

Ideal gas

w = RT

V2

V1

dV

Vw = RT ln

V2

V1

Maximum amount of work that can be done by (on) a system at constant T

P-V WORK


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Heat is the entity that flows across a boundary between two systems or between asystem and its surroundings as a result of a temperature difference between them

Molecular scale: kinetics energy transferred via motion and/or collisions and/or radiation

Convention: Heat flows from higher T to lower T

A process that involves no heat flow is adiabatic

Observation

When an infinitesimal amount of heat flows at constant pressure into a system thetemperature of the system increases according to

q= CPdT

HEAT

CP(T

) : Heat capacity at constant pressure (depends on the system)


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Heat may be transferred reversibly or irreversibly

HEAT

qP

=

T2

T1 CP

(T)dTq=

CPdT

CP =

q

T

P

Heat capacity at constant pressure:

qP > 0 heat flows into the system Convention:

IMPORTANT(1) A system doesnt have a q orw. They arent state functions

(2) q and w may both lead to a temperature change

(3) q and w have same units


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It is based on experimental observations and can be stated in various ways

qcycle + wcycle = 0

In general qand ware different fromzero and depend on the specific path

(q+ w) = 0


2

1

a

b

1

(q+ w) +

1

2

(q+ w) = 0

path a path b


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2

1

(q+ w) + 1

2

(q+ w) = 0

path a path b

2

1

(q+ w) = 2

1

(q+ w)

path a path b

dU= q+ w

U = internal energy (extensive state function)

q+ w = exact differential !!!


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INTERNAL ENERGY

Properties

dU = 0 The internal energy of a system that undergoes

a cyclic process does not change

The change in internal energy of a

system whose state change from 1 to 2depends only on the initial and finalstate and is independent of the way inwhich the change occurs

U =21

dU= U2 U1

independent of the path


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(2) Thermodynamics doesnt really care about the absolute value of U but

only about the difference U between two states the zero can be set anywhere (be consistent!!!)

(1) Knowing U means knowing the function not just one valueU(n,T,V) or U(n,T,P) for single component systemsU(n,T,P,x1,x2,...,xN-1) for multicomponent systems

Important

INTERNAL ENERGY


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dV = 0Constant volume process:

HEAT CAPACITY AT CONSTANT V

dU= q+ w

dU= q Pext dV

U= q

Change in internal energy with an infinitesimal changein temperature in a constant volume process

Cv =

U(n,V,T)

T

V,n

Heat capacity at constant V

Heat absorbed with an infinitesimal change intemperature in a constant volume process


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HEAT CAPACITY AT CONSTANT V

Cv =

U(n,V,T)T

V,n

Heat absorbed with an infinitesimal change in

temperature in a constant volume process

Molar capacity: Divide by the number of moles

CV,m(T) =

Um(V, T)

T

V

Um =U

n

Change in internal energy with an infinitesimal changein temperature in a constant volume process

O S O G


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Changes in U(V,T) associated with a change in the thermodynamic state

z

x

y

=1

xz y

z

x

y

x

y

z

y

z

x

= 1

RELATIONS INVOLVING U

(single component, homogeneous system for now)

Recall:

z = z(x, y) dz = zxy

dx+zyx

dy



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U= U(V, T) dU=U

T

VdT+

UV

TdV

dU = CVdT+U

V

T

dVCV= U

T

V

U

V

T

=

U

T

V

T

V

U

U

VT

V

TU

T

UV

= 1



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dU=U

T

VdT+

UV

TdV


U= U(V, T)

dU = CVdT+U

V

T

dVCV= U

T

V

U

VT

= U

TV

T

VU

U

VT

= CVT

VU

J = 0 for ideal gas

U

V

T

= CV JJ = Joule coefficient



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dU=U

T

V

dT+U

V

T

dV

dU= CV dT CV J dV


U(V,T) convenient because:

(1) dU can be related to measurable quantities (CV, J)

(2) dU = dq when dV = 0, no work (T,P can change but there is no compression or expansion)

ENTHALPY


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Many experiments involve constant P

ENTHALPY

Enthalpy: H= U+ PV

dH= dU + PdV + V dP

= q + w + PdV + V dP

= q PextdV + P dV + V dP

= q + (P Pext)dV + V dP

ENTHALPY


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CP includes heat adsorbed to cause temperature rise and system expansion

CV does not involve system expansion

ENTHALPY

dH= q

Cp =

H

T

p

Heat capacity at constant P

CV =

U

T

V

Heat capacity at constant V

At constant pressure: P = Pext

CP > CV

dH= q + (P Pext)dV + V dP

ENTHALPY


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ENTHALPY

dH= HT

P

dT+HP

T

dP

dH= CPdT+

H

P

T

dP

Following the same procedure as for U(T,V)

dH= CPdT CP T

PH

dP

dH= CPdT CP JT dP

JT =

T

P

H

Joule-Thomsoncoefficient

for ideal gasJT = 0

SUMMARY


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SUMMARY

Changes associated with the equation of state as well as with the first law ofthermodynamics that can be expressed in terms of measurable quantities

(T, P) =1

V

V

T

P,n

(T, P) = 1

V

V

P

T,n

Expansion coefficient Isothermal compressibility

CV =

U

T

V

=

q

dT

V

p =

H

T

p

=

q

dT

p

Heat capacity at constant V Heat capacity at constant P

J =

T

V

U

= 1

CV

U

V

T

Joule coefficient Joule-Thomson coefficient

JT =

T

P

H

= 1

CP

H

P

T

SUMMARY


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SUMMARY

Relation between heat capacities

CV =

U

T

V

=

q

dT

V

Cp =

H

T

p

=

q

dT

p

Heat capacity at constant V Heat capacity at constant P

CPCV =

HT

P

UT

V

CP CV = (U+ PV)T

P

UT

V

CP CV =

U

T

P

+ P

V

T

P

U

T

V

SUMMARY


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SUMMARY


dU = UT

V

dT+UV

T

dVBut,

At constant P, dUP =

U

T

V

dTP +

U

V

T

dVP

U

T

P

=

U

T

V

+

U

V

T

V

T

P

dividing by dTP

CPCV =

UT

P

+ PV

T

P

UT

V

SUMMARY


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SUMMARY


CP CV =

U

T

P

+ P

V

T

P

U

dT

V

UTP

= UTV

+ UV

T

VTP

Using

CP CV =

U

V

T

+ P

V

T

P

SUMMARY


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SUMMARY

dU= CV dT CV J dV

dH= CPdT CP JT dP

JOULE EXPERIMENT


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J =T

VU

=

1

CVU

VT

JOULE EXPERIMENT

dU= CV dT+

U

V

T

dV

dU= CV dT CV J dV

adiabaticwalls

plug

vacuum

T1, V1

adiabaticwalls

T2, V2

q = 0adiabatic walls

vacuumPext = 0 w = 0

U = 0

T

V

U

JOULE THOMSON EXPERIMENT


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JOULE-THOMSON EXPERIMENT

dH= CPdT+

H

P

T

dP

dH= CPdTCP

T

P

H

dP

JT= T

PH

P1 P2

P1, V1, T1adiabatic

walls

adiabaticwalls P1 > P2

porous

plug



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dH= CPdT+

H

P

T

dP

dH= CPdTCP

T

P

H

dP

JT= T

PH

P1 P2

adiabaticwalls




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dH= CPdT+

H

P

T

dP

dH= CPdTCP

T

P

H

dP

JT= T

PH

P1 P2

adiabaticwalls


q = 0adiabatic walls

P2, V2, T2

w =PdVreversible process w =

P1(0

V1)

P2(V2

0)



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q = 0

w = 1 1 2 22 1 = 1 1 2 2

first law

1 + 1 1 = 2 + 2 2

H1 = H2

constant enthalpy!!!



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dH= CPdT+

H

P

T

dP

dH= CPdTCP

T

P

H

dP

JT= T

PH

P1 P2

adiabaticwalls


P2, V2, T2

H = 0 T

PH



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JT = TP

H

= 1

CPH

PT

H

PT

> 0

H

P

T

< 0

Low T / High P:

High T / Low P:

JT < 0

JT > 0



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JT = TP

H

= 1

CPH

PT

H

PT

> 0

H

P

T

< 0

Low T / High P:

High T / Low P:

ISENTHALPIC CURVES

It is possible to use the J-T experimentto liquefy gas (see book)

PERFECT GAS


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PERFECT GAS

V= nRT

U

V

T

= 0

U = U(T)

dU = CVdT

CV = CV(T)

H= U+ PV= U+ nRT

H= H(T)

CP = CP(T) =dH

dT

PERFECT GAS


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PERFECT GAS

CPCV =

UV

T

+ PV

TP

V= nRT

U

V

T

= 0

V

T

P

=

nR

P

P V = nR

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Chem131_Chapter2