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Category: Thermodynamics
III. The Laws of Thermodynamics
Updated: 2014jan13
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Outline
A.
B.
C.
D.
First Law of Thermo
Second Law of Thermo (Entropy)
Statistical Mechanics
References
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A. First Law of Thermo
Stored Heat
WORK OUT
HEAT IN
Heat into a system
is either stored as
internal energy, or
performs work on
the universe
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1. Work
W=Fx
(a)
Recall Definition
(b)
For a gas expanding against a piston the
work done is: W=Fx=(PA)x=P(Ax)
Work done by gas on environment is
pressure time change in volume:
W=+PV
(c)
The above formula is only really valid for
isobaric process (P=0 ). Generally it’s the
area under the curve of a PV diagram.
(example to right is “isothermal process”
T=0 )
1. Work continued
(a) Isobaric Process: P=0
• W is the work done BY the system
• Exact Form: W=+PV
(b) Isothermal Process (T=0 ):
dV
W   PdV  nRT 
V
 V2 
W   nRT Ln 
 V1 
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2. Internal Energy
• Adding heat “Q” to a gas (at
constant volume “isovolumetric”)
increases its internal energy “U”
• For monatomic gas (noble
gasses, He, Ne) all the internal
energy is in the kinetic energy of
molecules
• For diatomic gas (e.g. O2 or H2),
there is more energy stored in
rotation and vibration of molecule
Q  U
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2
mv  32 kT
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U  N 32 kT  32 nRT
U  52 nRT
Note: N=number of molecules, n=number of moles,
k=Boltzmann’s constant, R=gas constant, T in Kelvin, Q & U in Joules
2b Specific Heat and gamma
(a) Constant Volume Molar Specific Heat Cv
• Monatomic Ideal Gas
U  32 nRT
• More generally:
U  nC vT
• for monatomic Cv  32 R
for diatomic
Cv  52 R
(b) Constant Pressure Molar
Specific Heat Cp=Cv+R =Cv
=Cp/Cv, 5/3 for monatomic (7/5 diatomic)
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2.c Adiabatic Process
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Definition: Q=0 (no heat in)
•
A gas that expands
adiabatically will cool (by the
first law): Q=0=U+W
U   W
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2
nRT   PV
 32 TT 
V
V
Const  PV 
Const  TV  1
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3. First Law of Thermodynamics
(a) The First Law
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•
•
Q=U+W
Q is heat INTO system
U is change of internal energy of system
W is the work done by system on
environment
So its based on the law of
CONSERVATION of ENERGY
BEWARE: Some authors define work as done ON the system by environment
and heat flowing OUT of system, so equation will have minus signs.
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3.b Enthalpy
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Definition: H=U+PV
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Change in Enthalpy:
•
•
•
H=U+PV+ VP
Now insert the 2nd law to get
H=Q+ VP
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Hence Enthalpy will be conserved in a process
which is both isobaric (P=0) and adiabatic (Q=0)
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Or, for chemistry in an open beaker: H=Q
3.c Gibbs’ Free Energy
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Definition: G=U+PV-TS = H-TS
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Change in Gibbs: G=VP - ST
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Hence Gibbs’ Energy will be conserved in a
process which is both isobaric (P=0) and
isothermal (T=0) [i.e. Chemistry in an open
beaker in an ice bath]
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Spontaneous reactions occur when: G < 0
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B. Second Law of Thermo
Why does heat flow from hot to cold (instead of the
other way around?)
1. Definition of Entropy:
2. TS Diagrams
3. Entropy of Ideal Gas
S=Q/T
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1. Entropy
(a) Definition
• Rudolf Clausius (1865)
• Extrinsic Quantity:
Q
S 
• Units: Joule/Kelvin
T
For example, it takes 334 Joules to melt 1
gram of ice. The entropy of the liquid water
is bigger than that of ice:
Q 334 J
J
S 


1
.
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K

T
273 K
1.b Entropy and First Law (details)
From Definition: Q=TS
•
–
•
Isentropic Process (S=0) is equivalent to saying process is
adiabatic (if reversible)
First Law Restated in terms of Entropy:
TS = U + PV
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1.c Entropy and Gas Volume
TS = U + PV
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First Law:
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For ideal gas, isothermal process (T=0, or U=0)
reduces to: TS=PV
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From this we can deduce change in entropy in an
isothermal gas expansion is equal to isothermal
work done by gas
 V2 
W
S 
 nR Ln 
T
 V1 
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1.d Entropy of Ideal Gas
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First Law of Thermo: U=TS-PV
• For monatomic ideal gas: U=(3/2)nRT
• For ideal gas: P=(nRT)/V
• Substitute into first law and solve for entropy:
T
V
S  NR
 NR
T
V
3
2
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Integrate to get entropy of monatomic ideal gas
 
S ( n, T , V )  nR Ln V T
3
2
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2. Reversibility
(a) Second Law of Thermodynamics
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Change in (total) entropy of a closed system tends to
increase in time: S0
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Explains why heat flows from hot to cold, but not the other
way around
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Example: Heat Q removed from hot water and given to cold
water (TH>TC)
1
Q Q
1 
S  

 Q    0
TH TC
 TC TH 
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2b. Clausius Inequality
Q
S 
T
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Any process obeys the inequality:
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For a system to be reversible, all
the processes must obey the equality
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Example: Free expansion of gas in vacuum. No
heat was added (Q=0), but the entropy has
increased due to increased volume:
Q
S 
T
 V2  Q
S  nR Ln  
0
 V1  T
2c. Irreversibility
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Total entropy of universe increases (this tells us the
direction of time!)
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A open system may decrease in entropy (e.g. freeze
water), but the heat exhausted from it increases the
entropy of the environment such that the total
entropy increases
S universe  S system  S environment  0
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3. Efficiency
(a) Carnot Energy Diagram
• Sadi Carnot (1824)
• QH=Heat from fuel
• W=useful work by engine
• QC=waste heat exhausted
• Conserve Energy
QH  QC  W
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Efficiency
QC
W

 1
1
QH
QH
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3b. Engines MUST waste heat
 QH QC
S 

0
TH
TC
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From second Law,
Entropy must increase
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Must waste heat to
environment. The colder the
environment the less heat
must be wasted.
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This puts a limit on
efficiency (Carnot 1824)
TC
QC  QH
TH
QC
TC
W

 1
 1
QH
QH
TH
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3c. Carnot Cycle (1824)
The Carnot Cycle is the best efficient engine
that can be made (reversible processes)
TC
W

 1
QH
TH
qh  Th S B  S A 
qc  Tc S D  S C 
C. Statistical Mechanics
1. The 3rd Law of Thermodynamics
2. Probability and Equilibrium
3. Boltzmann’s Definition of Entropy
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1. The
3rd
law of thermodynamics
Aka “Nernst’s Theorem” (1906)
a) Impossible to reach absolute zero
b) As T0 the heat capacity
approaches zero
c) As T0 all thermodynamic
processes stop
d) At absolute zero the entropy of a
system would be minimum (S=0)
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2. Probability and Equilibrium
(a) Equilibrium is the most probable situation
(b) Consider 4 particles in a box.
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–
Each has a 50/50 chance of being on the Left half (“L”) of the box.
There are 24=16 possible microstates
The most probable situation is that 2 will be on either side.
Macrostate
Microstates
Frequency
Probability
0
LLLL
1
6.25%
1
LLLR, LLRL, LRLL,
RLLL
4
25%
2
LLRR, LRLR, RLLR,
RRLL, RLRL, LRRL
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37.5%
3
RRRL, RRLR<
RLRR, LRRR
4
25%
4
RRRR
1
6.25%
2c. Large N systems
For example, consider N=40 atoms
• Total number of microstates is
2N= 1,099,511,627,776
• Only 1 possible for all on same side:
Probability 9x10-11%
For N=1024 atoms deviations from the average
situation are only in the 12th decimal place!
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3. Boltzmann Entropy
Entropy is a measure of the
randomness (disorder) of a system
Details: S=k Ln()
• k=Boltzmann’s Constant
• =Number of Microstates (for that
macrostate), i.e. a measure of disorder.
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Maximum order: =1, so S=0
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Maximum disorder is the average
 N 2
 2 !
“equilibrium” situation of half on each side
of box
S  kN Ln( 2)
N!
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References
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http://www.chem1.com/acad/webtext/thermeq/TE3.html
http://stp.clarku.edu/simulations/
http://intro.chem.okstate.edu/1314F00/Laboratory/GLP.htm
Good One http://webphysics.davidson.edu/physlet_resources/bu_semester1/c27_process_expansion_sim.html
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Notes
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Discuss Helmholtz contraction of a star
What about Refrigerators?
Example: change in entropy by melting ice
Demos: Stirling Engine