THERMODYNAMICS

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THERMODYNAMICS
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• Energy
• Kinetic-motion
• Potential position
• Thermal temperature
• Chemical or Nuclear- composition
• Electrical- generator
• Mechanical- Motor
• Also magnetic, surface, strain, etc.

Produced
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• The object of thermodynamics is to seek out
  logically the relations between kinds of energy
  and their diverse manifestations.
• 1st Law Energy cannot be created or destroyed.
  It is conserved.
• Energies can be converted to one another, as long
  as the total quantity of E is the same.
• Any E ? Thermal E possible
• But only mechanical E ? Potential E is possible!

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Steel ball
Case 1 Case 2
Free fall
Etc.
mgh
Losing EP Gaining EK
Losing EK Gaining EP
Hits the ground Assume elastic collision
Same ball
water
T2 gt T1
Losing EP Gaining EK
mgh
T2
T1
The ball rests at the bottom of the beaker
Converted to thermal energy
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• In reality the steel ball remains in the beaker
  and does not jump out of the water. But this fact
  cannot be predicted by the 1st Law!
• ?Every real process has a sequence which is
  recognized as NATURAL the opposite sequence is
  UNNATURAL.
• 2nd Law is concerned with the direction of
  natural processes.
• In combination with 1st Law, it enables us to
  predict the natural direction of any process, and
  as a result to predict the equilibrium situation.

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• Zeroth Law
• Law of Thermal Equilibrium
• When in contact if they dont influence each
  other, the wall is insulating or adiabatic.
• If they influence each other, the wall is
  thermally conducting (diatermic) and the systems
  are in thermal contact. The systems are in
  thermal equilibrium.

p1
p1
p2
p2
V1
V2
V1
V2
In thermal contact
Isolated system
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• Zeroth Law Two systems which are both in thermal
  equilibrium with a third system are in thermal
  equilibirium with each other.
• 1.Systems in thermal equilibrium with each other
  have the same temperature
• 2.Systems not in thermal equilibrium with each
  other have different temperatures.

pB
pC
pA
B
C
A
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The First Law

• Concepts

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Definitions

• A thermodynamic system is that part of the
  physical universe the properties of which are
  under investigation.
• The system is confined to a definite place in
  space by the boundary which separates it from the
  rest of the universe, surroundings.
• Open Closed Isolated

Surroundings
System
Matter
Energy
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Work

• Work is any quantity that flows across the
  boundary of a system during a change in its state
  and is completely convertible into the lifting of
  a weight in the surroundings.
• Work appears only at the boundary of a system.
• Work appears only during a change in state.
• Work is manifested by an effect in the
  surroundings.
• Quantity of work is equal to mgh.
• Work is positive if the weight is lifted, in
  which case we say that work has been produced in
  the surroundings or has flowed to the
  surroundings it is negative if the weight is
  lowered, in which case we say that work has been
  destroyed in the surroundings or has flowed from
  the surroundings.

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Heat

• Heat is a quantity that flows across the boundary
  of a system during a change in its state in
  virtue of a difference in temperature between the
  system and its surroundings and flows from a
  point of higher to a point of lower temperature.
• Heat appears only at the boundary of a system.
• Heat appears only during a change in state.
• Heat is manifested by an effect in the
  surroundings.
• The quantity of heat is equal to the number of
  grams of water in the surroundings which are
  increased by one degree in temperature starting
  at a specified temperature under a specified
  pressure.
• Heat is positive if a mass of water in the
  surroundings is cooled, heat has flowed from the
  surroundings it is negative if a mass of water
  in the surroundings is warmed, heat has flowed to
  the surroundings.

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Work and Heat (Joules Experiment)
surroundings
system
100 mg H2O T 90 oC
10 mg H2O T 25 oC
100 mg H2O T 89 oC
10 mg H2O T 35 oC
100 units of heat has flown from the surroundings.
system
h
10 mg H2O T 25 oC
10 mg H2O T 35 oC
There is no heat flow but work flow
Heat is equivalent to mechanical energy
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Expansion Work

• The piston is weightless and frictionless.
• M is small enough so when the S stops are pulled
  out, the piston rises.
• T is kept constant by a thermostat.
• The air above the piston evacuated so no air
  pressure is pushing down the piston

M
S
S
S
S
M
S
S
S
S
T, p1, V1
T, p2, V2
Work is produced since a mass M in the
surroundings has been lifted a vertical distance
h against the force of gravity Mg. W Mgh The
pressure that opposes the motion of the piston
pex Mg/A W pex Ah Ah DV W pex (V2-V1) In
expansion DV is , W is ,weight rises. In
compression DV is -, W is -, weight falls. It is
assumed that pex is constant throughout the
change in state.
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S
S
S
S
S
S
M
M
S
S
S
S
M
S
S
S
S
S
T, p, V
T, p2, V2
T, p1, V1
The same expansion can be carried out in two
steps. W Wfirst Wsecond pex(V-V1)
pex(V2-V) The figures show that for
the same change in state the two-stage expansion
produces more work than the single-stage
expansion. Therefore ?W pex ?V
p1, V1
p1, V1
Pex
p2, V2
p2, V2
pex
Pex
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Work of Compression

• To compress the gas, larger weights on the piston
  than those that were lifted in the expansion are
  needed. Thus, more work is destroyed in the
  compression of a gas than is produced in the
  corresponding expansion.
• We must choose a mass large enough to produce and
  opposing pressure Pex which is at least as great
  as the final pressure p1.
• W pex (V1-V2)

M
S
S
S
S
M
S
S
S
S
T, p2, V2
T, p1, V1
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p1, V1
p1, V1
Pex
pex
Pex
p2, V2
p2, V2
Work in one-stage compression Work in
two-stage compression
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Max. and Min. Amount of Work

• If the expansion were done in many stages using a
  large mass in the beginning and making it smaller
  as the expansion proceeded, the work produced can
  be maximized.
• W pex ?V
• For the integral to have a maximum value, pex
  must have a maximum value. During expansion pex lt
  pgas. So, the opposing pressure should be
  adjusted to pex p ?p.
• Wmax (p- ?p) ?V (p ?V ?p ?V)
• dp dV is an infinitesimal of higher order than
  the first so has a limit of zero. So,
• Wmax p ?V

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• In a multistage compression, least amount of work
  can be destroyed.
• For compression we set pex at each stage must be
  infinitesimally greater than the pressure of the
  gas,
• pex p ?p Wmin (p- ?p) ?V (p ?V ?p ?V)
• Wmax, min p ?V
• For ideal gases, the maximum amount of work
  produced in expansion and minimum amount of work
  destroyed in compression is equal to the shaded
  area under the isotherm.
• Wmax, min nRT/V ?V nRT ?V/V nRT In (Vf/Vi)

p1, V1
p2, V2
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Reversible and Irreversible

• Process 1 Single-stage expansion with pex p2,
  then single-stage compression with pex p1.

p1, V1
p1, V1

pex
p2, V2
pex
p2, V2
Wexp p2 (V2-V1) Wcomp p1 (V2-V1) Wcyc
(p2-p1)(V2-V1) Net work has been destroyed!
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• Process 2 Limiting multi-stage expansion with
  pex p, then limiting multi-stage compression
  with pex p.
• When a system undergoes a change in state through
  a specified sequence of intermediate states and
  then restored to its original state by traversing
  the same sequence of states in reverse order, if
  the surroundings are also brought to their
  original state, the transformation in either
  direction is reversible. If the surroundings are
  not restored to their original state, the
  transformation is irreversible. All real
  processes are irreversible!

Wexp p (V2-V1) Wcomp p (V2-V1) Wcyc 0 No
net work effect!
p1, V1
p2, V2
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Energy

• Wcyc ?W Qcyc ?Q
• If a system is subjected to any cyclic
  transformation, the work produced in the
  surroundings is equal to the heat withdrawn from
  the surroundings.
• ?W ?Q (all cycles)
• (?Q ?W) 0 (all cycles)
• ?Q ?W ?U
• ?U 0
• When small amounts of heat ?Q and work ?W appear
  at the boundary, the energy of the system suffers
  a change ?U.
• ?U q w
• Only the difference in energy in a change in
  state can be calculated but no absolute value to
  the energy of the system can be assigned in any
  particular state.

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• EXACT INEXACT DIFFERENTIALS
• meaningless as W2-W1 or Q2-Q1 does
  not exist
• the system does not have work
  or heat at any
• point but these occur during
  a change in state.
• inexact
  differentials
• 
  exact differentials
• --
  independent of the path --
• ?U Q - W

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• For a system of Fixed Mass,
• system can be
  defined by TV.
• increase in U increase
  in U
• resulting from T
  resulting from V
• alone
  alone
• (assuming only expansion work)

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• If volume is constant, then dV0 dQV
  dU
• 
  ?U QV
• 
  
• (heat capacity
  of the system)
• heat withdrawn temperature
  increase
• from the surroundings of the sysytem
  

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• dU Cv dT
• CV is always (), so if T2 gt T1 then ?U is ()
• T2 lt T1 then
  ?U is (-)

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• For a system maintained at a constant volume,
• the temperature is a direct reflection of
• the energy of the system.
• if CV is
  constant in the T
• range of
  interest, then
• ?U CV ?T

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• JOULES EXPERIMENT
  measurement
• thermometer
  stirrer
• 
  Stopcock is opened, no change in T
• 
  after the expansion of A.
• 
  Free expansion dW 0
• dU dQ no change
  of T dQ 0 dU 0
• 
  
  
• 
  
  U of gas is
• 
  as dV ? 0 then
  independent of
• 
  
  Volume!!
• 
  
  (Joules Law)
• Only true for ideal gases. Real gases (?U / ?V)T
  is a small () quantity.
• The reason for this observation is the large
  heat capacity of water and small heat capacity of
  gas reduced the magnitude of the effect below the
  limits of observation.

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• Changes in State At Constant Pressure
• The piston floats
  freely, the piston will move ppex
• By adjusting the
  mass M, the p on the system may
• be adjusted to any
  constant value.
• At constant pressure dU dQp pdV
• U2 U1 Qp p( V2 V1)
• (U2 pV2) (U1 pV1) Qp p
  p1p2
• (U2 p2V2) (U1 p2V1) Qp
• H UpV (Enthalpy)
• ?H Qp

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• In constant pressure process the heat withdrawn
  from the surroundings is equal to the increase in
  enthalpy of the system.
• ?U p?V ?H Qp
• At constant p, dp0
• At constant p dH CpdT
• If Cp is constant for the T range of interest
  then ?H Cp ?T

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• The realationship between Cp and Cv
• dQ dW dU CvdT
• At constant pressure
• 
  Energy requires to pull the molecules
• 
  further apart against the
  attractive
• 
  intermolecular forces
• Work produced per unit increase in T
• In the contant pressure process.

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• Cp is always gt Cv Why?
• When a gas expands, the average distance between
  molecules
• Energy is neede to pull the molecules to this
  greater separation against the attractive forces
• If V is constant, W0, distance remains constant
• All the heat withdrawn goes into the chaotic
  motion and results in T
• Heat capacity (CV) is small
• If P is constant, system expands and produces
  work.
• E of the
  chaotic motion only this result in T
• Heat withdrawn E neccessary for
  separating the molecules
• Work in
  surroundings
• SO To produce a temperature incement of 1o, more
  heat must be withdrawn in the constant p process
  than is withdrawn in the constant V thus Cpgt Cv

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• If ideal gas, then
• as
  (Joules Law)
• 
  only for ideal gas
• For solids liquids the Cp- Cv is very small so
  can be taken Cp Cv

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• Measurement of Joule-Thomson
  Experiment
• dH dU pdV VdP
• If dT0
• 1
  2
• For solids liquids 1ltlt2
• For ideal gas
• For real gases is very
  small but measurable

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• Adiabatic Changes in State
• If dQ 0 adiabatic
• dU - dW
• ?U - W W ?U Work is produced _at_
  the expense of a
• 
  decrease in E of the system.
• dU - PexdV
• For ideal gas CvdT - PexdV
• dT and dV have opposite
  signs.
• For a fixed change in volume, reversible
  adiabatic expansion produces the greater drop in
  T.
• For a reversible adiabatic change in state
  of the ideal gas
• Pex p and CvdT - pdV
• CvdT - nRT(dV / V)
  
• Cv(dT / T) - R(dV /
  V)

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• Heat of Reaction
• Heat flows to the surroundings - exothermic ?H lt
  0
• Heat flows from the surroundings endothermic ?H
  gt 0
• Heat of a rxn is the heat withdrawn from the
  surroundings in the transformation of reactants
  at T and p to products at the same T and p.
• Eq. Fe2O3(s) 3 H2(g) ? 2Fe(s) 3 H2O
• T, p Adiabatic
  T, p
• 1) dQ
  0
• When p is constant, ?H0 Qp but when adiabatic
  ?H1 0 (Qp)1 0
• 2Fe(s) 3H2O(l) ? 2Fe(s) 3H2O
• T, p
  T, p

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• Here ?H2 Qp
• Here ?H ?H1 ?H2 0 Qp
• ?H Qp ( Increase in enthalpy of the system
  resulting from the
• chemical rxn.)
• H U pV
• ?H Hfinal Hinitial
  Hfinal 2H (Fe(s)) 3H(H2O(l))
• 
  Hinitial H (Fe2O3(s)) 3H (H2(g))
• ?H 2H (Fe(s)) 3H(H2O(l)) H (Fe2O3(s))
  3H (H2(g))
• Its impossible to determine absolute values of
  enthalpy.
• H H(T, p) If we choose p1 atm (standard p)
• Ho H (T, 1 atm) Standard molar enthalpy is at
  only T.
• Most stable state elements have Ho 0 (we cant
  measure, we assign)

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• Formation Reaction
• ½ H2(g) ½ Br2(l) ? HBr(g)
• ?Hfo HoHBr(g) (½ HoH2(g) ½ HoBr2(l) )
  HoHBr(g)
• ?Hfo is the molar enthalpy of the compound
  relative to the element which compose it. So if
  the ?Hfo of all the compounds in a chemical rxn.
  are known, the heat of rxn. can be calculated.
• ?Hfo sometimes can be measured diractly in a
  calorimeter.
• e.g. C(grap) O2(g) ? CO2(g)
• But C(grap) 2H2(g) ? CH4(g) cannot be measured.
• C and H2 does not react readily.
• CH4 is not the only product.

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• CH4(g) 2O2(g) ? CO2(g) 2H2O (l)
• can be measured easily.
• ?Hoconv HoCO2(g) HoH2O(l) HoCH4(g)
  
• measured known known
  calculate
• So carrying out combustion rxn. helps to
  calculate ?Hfo all organic compounds which
  contain C, H, and O only.
• A calorimeter rxn must take place quickly.
  Complete within a few min.
• As few side rxn. as possible, preferably none at
  all.
• The final product mixture must be carefully
  analyzed and the thermal effect of the side rxns
  must be substracted from the measured value.

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• Heat of Solution Dilution
• Enthalpy change ass. with the addition of a
  specified amount of solute
• to a specified amount of solvent at constant T
  P.
• HCl(g) 10H2O ? HCl.10H2O ?H1 -16.608
  kcal
• HCl(g) 25H2O ? HCl.25H2O ?H2 -17.272
  kcal
• 8 H2O ? HCl. 8H2O ?H8
  -17.96 kcal
• initially dilute solution limiting value of ?H8
  .
• ?H ?H2 - ?H1 -0.664 kcal ? heat of dilution.

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• Dependence of Heat of Rxn. on Temperature
• If the ?Ho value is known for a rxn at a
  particular T, then the ?Ho at any
• other T can be calculated, if the heat cap. of
  all substances taking part
• in the rxn are known.
• Cpo
• If the T range is short, Cp of all subs. Involved
  can be considered constant. If very large then
  Cp a bT cT2 dT3 ...
• constant for secified material