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Winter 2013 Chem 254: Introductory

Chapter 2: Internal (U), (w), (q), (H) ...... 13 Heat Capacities ...... 16 Calculating ΔU, ΔH, w, q in Ideal ...... 18 Isothermal Compression ...... 21 Reversible Process (limiting process) ...... 22 Isothermal Expansion ...... 22

Chapter 2: (U), Work (w), Heat (q), Enthalpy (H)

Internal Energy (excludes and rotation of vessel)

 Look at isolated part of

UUU Total = isolated

First law of thermodynamics: - Total U for is constant - Energy can be exchanged between various components - Energy forms can be interconverted Eg. Chemical En Heat  Work

UUUtotal   system   environement  0

Chapter 2: Internal Energy, Work, Heat and Enthalpy 13

Winter 2013 Chem 254: Introductory Thermodynamics

Work In classical , move object a d with F in direction of is work  N m = J

F mg dh w mgh (kg m s-2 m = N m = J)

h w mgd cos cos  d h w mgd mgh d

General formula

w F dL Line

PV work (constant external )

F m applies constant force P  A F w mgh  Fh ()() Ah  P V  V A ext 12

w  Pext() V final  V initial , or L Bar (1 L Bar = 100 J)

Chapter 2: Internal Energy, Work, Heat and Enthalpy 14

Winter 2013 Chem 254: Introductory Thermodynamics

More general formula for PV work, P does not need to be constant V w f P dV V ext i Sign Convention : Work done on the system raises internal energy of system ( w  0) Work done by the system lowers the internal energy ( w  0)

Other forms of work: - electrical work wQ  Q is charge in coulombs  difference in potential (in Volts or J/C) Run a current over  Q I t I is current (in Amps or C/s) w I t

Important: Work is associated with a process, with change. Work is transitory. You cannot say that a system contains that amount of energy or heat

Heat: associated with a process going from State 1  State 2

Usystem  q  w q is heat; w is the work

Heat is exchanged between system and environment

q  0: system loses energy q  0 : system gains energy qq system environment

note: TTsystem environment for heat to flow

Isolated system

Chapter 2: Internal Energy, Work, Heat and Enthalpy 15

Winter 2013 Chem 254: Introductory Thermodynamics

TTouter inner (regulate) So there is no flow of heat

UUsystem   environment  0

Uinner 0

Beaker + Lab +… = environment (isolated) U 0; U 0 I II UU    0 I II Butane + O CO H 2 2 2

Note : UII 0even if increases! Why? Chemical energy of butane is converted to heat.

Heat Capacities

The amount of energy (heat) required to raise the temperature of 1 gram of substance by 1 oC. of is 4.18 J/g K = 1

1) Heat capacity is dependent on heat Eg. 10 oC  11 oC and 80 oC 81 oC, require slightly different

2) At least 2 types of heat capacity

a) Keep constant CV

b) Keep pressure constant CP

3) Heat capacity is proportional to

Molar heat capacities : CPm, ,CVm,

n moles : CV nC V, m , CP nC P, m

4) General formula Chapter 2: Internal Energy, Work, Heat and Enthalpy 16

Winter 2013 Chem 254: Introductory Thermodynamics

T q f C dT VVT i

If CV is constant over temperature range: T f Tf qV C V dT  C V T  C V T f  T i T Ti  i

qVV C() T

And qPP C() T

Which is larger CP or CV ? Relation for and for ?

VV ; TT 2121

U  qP  w  q P  P ext () V21  V PV nRT

U  qP  nR() T21  T qPP C T ; U  qVV  C  T

CPV T  C  T  nR  T

CPV C nR or CCRP,, m V m

Therefore CP is larger than CV . At constant P , the system also does PV work when raising T . (analysis for ideal gas)

No work because V is constant

U  qV  w  qV

UCT V 

Bomb

qsystem  q surrounding   C  T V V V measure U reaction

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Winter 2013 Chem 254: Introductory Thermodynamics

system surroundings qqPV system Calorimeter qP  C V  T measure system qHP reaction

True definition of Enthalpy

H   U  () PV PV  PV2 2  PV 1 1 ; for PV1 1 PV 2 2 ; H U PV

At constant Pressure

H   U  PV2 2  PV 1 1  HUPV     

H  qPP  w  P  V  q  P()()  V  P  V

H  qPP  C  T

Completely general :UH, are function of state specify TVP,,

UUTPVUTPV (,,)(,,)2 2 2  1 1 1 HHTPVHTPV (,,)(,,)  2 2 2 1 1 1

Change in UH, are the same for both paths Change in qw, are different for different paths

Calculating ΔU, ΔH, w, q in Ideal Gas

1) Calculating UH, is easy if T is known T UUT()]   Uf CdTCT Tf  CT  T T V V Ti V f i  i

UCT V  for any process

HHT( ) .....

Chapter 2: Internal Energy, Work, Heat and Enthalpy 18

Winter 2013 Chem 254: Introductory Thermodynamics

HCT P  for any process (if CP is constant)

We know CPV C nR

Special cases: T is constant T 0 ; UH    0

2) Work: w P dV PV work only  ext

- Constant V Vif V  w  0 ; q qV   U

Vf - Constant Pext w  P dV   P() V  V ; q q   H extV ext f i P i

Isothermal reversible process: (Reversible process: delicate, see later) 1 P nRT nRT is constant ext V

Vf dV V w  nRT   nRTln V  f V Vi i V

Vf  nRTln Vfi  ln V   nRT ln Vi

Vf w nRT ln  Vi

3) Heat

Adiabatic process : q  0 by definition U  q  w; Uw

Adiabatic Reversible Process nRT q  0 , Uw, P  ext V

Vf nRT U  w   dV V i V nRT nC dT dV Vm, V dT nR nC dV Vm, TV

Chapter 2: Internal Energy, Work, Heat and Enthalpy 19

Winter 2013 Chem 254: Introductory Thermodynamics

ffdT dV nC nR Vm, iiTV

TVff    nCvm, ln  nR ln   TVii   

Cvm, TVff    ln  ln   RTVii    For adiabatic reversible process: C TV    C T P V C P vm, lnff ln OR Pm, lnf  ln i OR lnf  Pm, ln i         RTVii    RTPif VCPi V, m f

1)

R C TVVf R  f   f  Vm, ln   ln    ln   TCVVi  V, m  i   i  R

TVCVm, ff TVii

2)

C T nRT P Vm, lnff  ln  i   R Ti P f nRT i T P ln f i  TPif T P  lnf  ln i   TPif C T P Vm, 1 lnf   ln i   RTPif  CRVm,  Tf Pi ln ln RTP if C T P Pm, lnf  ln i   RTPif

Chapter 2: Internal Energy, Work, Heat and Enthalpy 20

Winter 2013 Chem 254: Introductory Thermodynamics

Adiabatic Constant external pressure AND q  0

Isothermal Compression

Constant external pressure w  Pf V f  V i   0 qw   0 (because U 0 because isothermal)

What is work in 2-step process?

w2 P int V int  Vi  P f V f  V int 

ww21 ; qq21

Chapter 2: Internal Energy, Work, Heat and Enthalpy 21

Winter 2013 Chem 254: Introductory Thermodynamics

Conclusion: w and q depend on details of process, not only on initial and final state.

Repeat for 3 step, 4…. 

w5 w 4  w 3  w 2  w 1 ; q5 q 4  q 3  q 2  q 1 The more steps, the less w and less heat

Reversible Process (limiting process)

PPext gas at each step nRT P  ext V Isothermal Reversible Process

VVffdV w  P dV   nRT VVext iiV V  nRTln |Vf   nRT ln f work, q is minimal Vi  Vi

Isothermal Expansion

Chapter 2: Internal Energy, Work, Heat and Enthalpy 22

Winter 2013 Chem 254: Introductory Thermodynamics

w1   Pf V f  V i   0 ;

qw11   0

ww ; qq 21 21

w3 w 2 w 1 ; q3 q 2 q 1

More processes more work ( w ), more heat ( q )

w5 w 4  w 3  w 2  w 1 ; q5 q 4  q 3  q 2  q 1

Limiting Expansion Work compressionexp ansion wwlimit limit

VVffdV V w  P dV   nRT   nRT ln f limit VVext ii VVi

Chapter 2: Internal Energy, Work, Heat and Enthalpy 23

Winter 2013 Chem 254: Introductory Thermodynamics

Grains of sand : I can run process either way The thermodynamic work is the same both ways for reversible process

Irreversible Process (Big chunks of )

Follows arrows in reverse: add mass, rises? ; removes mass, piston lowers?

This is absurd, hence: Why do irreversible processes run in one way and not another? What is special about irreversible?

Chapter 2: Internal Energy, Work, Heat and Enthalpy 24