Download Chapter 2: Internal Energy

Winter 2013 Chem 254: Introductory Thermodynamics
Chapter 2: Internal Energy (U), Work (w), Heat (q), Enthalpy (H) ................................................ 13
Heat Capacities ......................................................................................................................... 16
Calculating ΔU, ΔH, w, q in Ideal Gas ........................................................................................ 18
Isothermal Compression ........................................................................................................... 21
Reversible Process (limiting process) ....................................................................................... 22
Isothermal Expansion ............................................................................................................... 22
Chapter 2: Internal Energy (U), Work (w), Heat (q), Enthalpy (H)
Internal Energy (excludes motion and rotation of vessel)
 Look at isolated part of universe
U  U system  U Environment
Total = isolated
First law of thermodynamics:
- Total U for isolated system is constant
- Energy can be exchanged between various components
- Energy forms can be interconverted
Eg. Chemical En  Heat  Work
Utotal  U system  U environement  0
Chapter 2: Internal Energy, Work, Heat and Enthalpy
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Winter 2013 Chem 254: Introductory Thermodynamics
Work
In classical mechanics, move object a distance d with force F in direction of
displacement is work  N m = J
F  mg
d h
w  mgh (kg m s-2 m = N m = J)
w  mgd cos
w  mgd
cos  
h
d
h
 mgh
d
General formula
w   F  dL Line integral
PV work (constant external pressure)
m applies constant force P 
F
A
F
( Ah)  Pext (V1  V2 )
A
 Vinitial ) Joules, or L Bar (1 L Bar = 100 J)
w  mgh  Fh 
w   Pext (V final
Chapter 2: Internal Energy, Work, Heat and Enthalpy
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Winter 2013 Chem 254: Introductory Thermodynamics
More general formula for PV work, P does not need to be constant
Vf
w   Pext dV
Vi
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
w  Q
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
U system  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
qsystem  qenvironment
note: Tsystem  Tenvironment for heat to flow
Isolated system
Chapter 2: Internal Energy, Work, Heat and Enthalpy
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Winter 2013 Chem 254: Introductory Thermodynamics
Touter  Tinner (regulate)
So there is no flow of heat
U system  U environment  0
U inner  0
Beaker + Lab +… = environment (isolated)
U I  0 ; U II  0
U I  U II  0
Chemical Energy
Butane + O2  CO2  H 2
Note : U II  0 even if temperature 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. Heat capacity of water is 4.18 J/g K = 1 calorie
1) Heat capacity is dependent on heat
Eg. 10 oC  11 oC and 80 oC  81 oC, require slightly different energies
2) At least 2 types of heat capacity
a) Keep volume constant CV
b) Keep pressure constant CP
3) Heat capacity is proportional to amount of substance
Molar heat capacities : CP ,m , CV ,m
n moles : CV  nCV ,m , CP  nCP,m
4) General formula
Chapter 2: Internal Energy, Work, Heat and Enthalpy
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Winter 2013 Chem 254: Introductory Thermodynamics
Tf
qV   CV dT
Ti
If CV is constant over temperature range:
qV  CV  dT  CV T T f  CV T f  Ti 
i
Ti
Tf
T
qV  CV (T )
And
qP  CP (T )
Which is larger CP or CV ?
Relation for CP and CV for ideal gas?
V2  V1 ; T2  T1
U  qP  w  qP  Pext (V2  V1 )
U  qP  nR(T2  T1 )
PV  nRT
qP  CP T ; U  qV  CV T
CP T  CV T  nRT
CP  CV  nR
or
CP,m  CV ,m  R
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
U  CV T
Bomb calorimetry
qVsystem  qVsurrounding  CVCalorimeter Tmeasure
 U reaction
Chapter 2: Internal Energy, Work, Heat and Enthalpy
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Winter 2013 Chem 254: Introductory Thermodynamics
qPsystem  qVsurroundings
qPsystem  CVCalorimeter Tmeasure
qPsystem  H reaction
True definition of Enthalpy
H  U  ( PV )
  PV   PV
2 2  PV
1 1 ; for PV
1 1  PV
2 2;
H  U  PV
At constant Pressure
H  U   PV
2 2  PV
1 1
H  U  P  V 
H  qP  w  P  V   qP  P(V )  P(V )
H  qP  CP T
Completely general : U , H are function of state
 specify T ,V , P
U  U (T2 , P2 ,V2 )  U (T1 , P1 ,V1 )
H  H (T2 , P2 ,V2 )  H (T1 , P1 ,V1 )
Change in U , H are the same for both paths
Change in q, w are different for different
paths
Calculating ΔU, ΔH, w, q in Ideal Gas
1) Calculating U , H is easy if T is known
U  U (T )  U  CV dT  CV T ]Tif  CV T f  Ti 
Tf
T
Ti
U  CV T
for any process
H  H (T )  .....
Chapter 2: Internal Energy, Work, Heat and Enthalpy
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Winter 2013 Chem 254: Introductory Thermodynamics
H  CP T
for any process (if CP is constant)
We know CP  CV  nR
Special cases:
Isothermal Process T is constant T  0 ; U  H  0
2) Work: w    Pext dV
- Constant V
PV work only
Vi  V f  w  0 ;
q  qV  U
Vf
- Constant Pext w   Pext  dV   Pext (V f  Vi )
Vi
; q  qP  H
Isothermal reversible process: (Reversible process: delicate, see later)
1
nRT is constant
Pext  nRT
V
V f dV
V
w  nRT 
 nRT  ln V V f
i
Vi V
 Vf 
 nRT  ln V f  ln Vi   nRT ln  
 Vi 
 Vf 
w  nRT ln  
 Vi 
3) Heat
Adiabatic process :
q  0 by definition
U  q  w ; U  w
Adiabatic Reversible Process
nRT
V
V f nRT
U  w   
dV
Vi
V
nRT
nCV ,m dT 
dV
V
dT nR
nCV ,m

dV
T
V
q  0 , U  w ,
Pext 
Chapter 2: Internal Energy, Work, Heat and Enthalpy
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Winter 2013 Chem 254: Introductory Thermodynamics
dT
i T
 Tf
nCv ,m ln 
 Ti
nCV ,m 
f
dV
V

 Vf 
  nR ln  

 Vi 
 nR 
f
i
 Tf 
 Vf 
ln     ln  
R
 Ti 
 Vi 
For adiabatic reversible process:
Cv ,m
P
C
 Tf 
 Vf 
 Tf 
ln     ln   OR P ,m ln     ln  i
P
R
R
 Ti 
 Vi 
 Ti 
 f
Cv ,m

 V f  CP ,m  Pi 
ln  
 OR ln   
P 
V
C
i
V
,
m



 f 
 Tf 
 Vf 
ln     ln  
R
 Ti 
 Vi 
Cv ,m
1)
R
 Tf 
 Vf 
 V f  CV , m
R
ln    
ln    ln  
CV ,m  Vi 
 Ti 
 Vi 
Tf
Ti

Vf
R
CV , m
Vi
 Tf 
 Vf 
ln     ln  
R
 Ti 
 Vi 
Cv ,m
2)
 nRT f
 Tf 
P 
ln     ln 
 i 
 P
R
nRTi 
 Ti 
 f
CV ,m
 Tf
  ln 
 Ti
  Pi
 
  Pf



P
 Tf 
  ln    ln  i
P
 Ti 
 f
P
 CV ,m
  Tf 
 1 ln     ln  i

P
 R
  Ti 
 f



 Pi
 CV ,m  R   T f 
ln


ln





R

  Ti 
 Pf



P
 Tf 
ln     ln  i
P
R
 Ti 
 f



CP , m



Chapter 2: Internal Energy, Work, Heat and Enthalpy
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Winter 2013 Chem 254: Introductory Thermodynamics
Adiabatic Isobaric Process
Constant external pressure AND q  0
Isothermal Compression
Constant external pressure
w   Pf V f  Vi   0
q  w  0 (because U  0 because isothermal)
What is work in 2-step process?
w2  Pint Vint  Vi   Pf V f  Vint 
w2  w1 ; q2  q1
Chapter 2: Internal Energy, Work, Heat and Enthalpy
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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  w4  w3  w2  w1 ; q5  q4  q3  q2  q1
The more steps, the less w and less heat
Reversible Process (limiting process)
Pext  Pgas at each step
nRT
V
Isothermal Reversible Process
Pext 
dV
Vi
Vi V
 Vf 
V
 nRT ln |Vif  nRT ln  
 Vi 
Vf
w    Pext dV  nRT 
Vf
work, q is minimal
Isothermal Expansion
Chapter 2: Internal Energy, Work, Heat and Enthalpy
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Winter 2013 Chem 254: Introductory Thermodynamics
w1   Pf V f  Vi   0 ;
q1  w1  0
w2  w1 ; q2  q1
w3  w2  w1 ; q3  q2  q1
More processes more work ( w ), more heat ( q )
w5  w4  w3  w2  w1 ; q5  q4  q3  q2  q1
Limiting Expansion Work
ansion
wlcompression
 wlexp
imit
imit
Vf
Vf
Vi
Vi
wlimit    Pext dV   nRT
Vf
dV
 nRT ln
V
Vi
Chapter 2: Internal Energy, Work, Heat and Enthalpy
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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 mass)
Follows arrows in reverse: add mass, piston 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
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