Chapter 5: Entropyscienide2.uwaterloo.ca/~nooijen/website_new_20_10... · Winter 2013 Chem 254:...
Transcript of Chapter 5: Entropyscienide2.uwaterloo.ca/~nooijen/website_new_20_10... · Winter 2013 Chem 254:...
Winter 2013 Chem 254: Introductory Thermodynamics
46
Chapter 5: Entropy ........................................................................................................................ 46
Calculating ΔS for processes in ideal gas .................................................................................. 48
ΔS as a function of changes in P,T,V in pure substances (not mixtures or reactions) .............. 51
The Second law of thermodynamics......................................................................................... 54
Absolute Entropy (Pure Substances) ........................................................................................ 56
Microscopic (molecular) view of Entropy ................................................................................. 56
Heat Engine ............................................................................................................................... 58
Carnot Cycle .............................................................................................................................. 59
Midterm Review............................................................................................................................ 61
Chapter 5: Entropy
Entropy: the second and third law thermodynamics
First law: Energy can be converted from one form to another but the total energy for an
isolated system is constant (heat-work-chemical energies)
0U (is constant if all processes are included)
Can we obtain a law that describes what can or cannot happen?
Mixture of 2N ,
2H (raise T , catalyst) forms 3NH in equilibrium, once it reaches
equilibrium, it does not go back to reactants, again we observe an “arrow of time”.
For isolated system
- Entropy ( S ) 0S for any possible, irreversible process
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Chapter 5: Entropy 47
0S for a reversible process
- Other quantities (Ch 6)
G Gibbs free energy, used when ,T P constant
A Helmholtz free energy, used when ,T V constant
Entropy is an abstract quantity for macroscopic systems, thermodynamics
History of discovery
- discovered by engineer Carnot
- improve efficiency of steam engines (1830s)
- it is still taught like this (see 5.2 in book)
Physical insight into entropy microscopic theory
- Quantum Mechanics
- Statistical Mechanics (molecular point of view, discussed at Ch. 5 end)
Consider a reversible process for an ideal gas rev revdu q w ,rev revq w inexact differentials (depends on path)
rev
V extC dT q P dV
rev
V
nRTq C dT dV
V rev
ext gas
nRTP P
V
Differential for revq
revq : NOT a exact differential
V
T V
nRTC
V T V
0nR
V
0nR
V therefore revq is not an exact differential
rev
VCq nRdT dV
T T V
is this an exact differential?
V
VT
C nR
V T T V
0 0 therefore is an exact differential
revq
dST
entropy, an exact diff, independent of path
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Chapter 5: Entropy 48
revf
f ii
qS S S
T
entropy is a state function
1
T : is called an integrating factor changes inexact diff to exact diff
Eg. From math
dF ydx xdy
yx
y xy x
1 1 dF is an inexact diff
2 2
1dF ydG dx dy
xx x
2
1
yx
y
y x xx
2 2
1 1
x x dG is an exact diff
2
1
x: integrating factor
Calculating ΔS for processes in ideal gas
revq
dST
(exact differential)
S is a state function
, ,S T P V of single component systems
final initialS S S
Like ,H U , S independent of process or path
How to calculate S
revf
i
qS
T
What does this mean?
Choose a reversible path: at all stages, the system retains equilibrium (stays on
the ideal gas curve)
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Chapter 5: Entropy 49
Even for irreversible process calculate S for a reversible path with same initial and
final states
Ideal gas PV nRT
a)1 1 2 2TV T V
( ) ( ) ( )C A BS S S
A. Constant T (reversible isothermal) B. Constant V
0U q w 0w U q
( )f
exti
nRTq w P dV dV
V
Vq dU C dT
nRT
q dVV
VCdS dT
T
q nR
ds dVT V
ln
f fVV
ii
TCS dT C
T T
lnf f
ii
VnRS dV nR
V V
,V mnC nRdS dT dV
T V
1 1 2 2TV T V , ln lnf f
V m
i i
T VS nC nR
T V
b) 1 1 2 2PV PV
( ) ( ) ( )S F S E S D
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D. Constant P E. Constant V (same as before)
P Pq H C dT ln
f fVV
ii
TCS dT C
T T
PCdS dT
T Constant Volume ,
nRT
P constant
lnf
P
i
TS C
T
f f
i i
T P
T P
f i
f i
T T
V V since
nRTP
V constant ln
f
V
i
PS C
P
lnf
P
i
VS C
V
, ,P m V mC CdS n dV n dP
V P
, ,ln lnf f
P m V m
i i
V PS nC nC
V P
c) 1 1 2 2PT PT
f i
i f
P V
P V constant T
,P mC RdS n dT n dP
T P
, ln lnf f
P m
i i
T PS nC nR
T P
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Chapter 5: Entropy 51
d) adiabatic reversible process
0revq 0revq
T
0S
1 1 2 2TV T V adiabatic reversible process 0S
, ln lnf f
v m
i i
T VnC nR
T V
,
ln lnf fv m
i i
T VC
R T V
(see section 2.11 in book)
ΔS as a function of changes in P,T,V in pure substances (not mixtures or reactions)
Dependence of S on ,T P
P T
S SdS dT dP
T P
exact differential
P
P
C VdT dP
T T
Why is this true?
1) At constant P rev
PCqdS dT dT
T T
2) T P
S V
P T
Maxwell relation, see later
Liquids and solids:
P
VV
T
f f
i i
T PP
T P
CS dT V dP
T
f
i
TP
T
CdT V P
T
Gases:
P
V
T
a little effort
nRT
VP
P
V nR
T P
(ideal gas)
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Chapter 5: Entropy 52
,P mC RdS n dT n dP
T P
, ln lnf f
P m
i i
T PS nC nR
T P
Dependence of S on ,T V
V T
S SdS dT dV
T V
V
V
C PdT dV
T T
1) At constant V , rev
VCqdS dT
T T
2) Maxwell Relation T V
S P
V T
V T P
P P V
T V T
Liquids and solid:
f f
i i
T VV
T V
CS dT dV
T
f
i
TV
T
CdT V
T
Gases:
Calculate V
P
T
f
i
V
VV
PdV
T
Ideal gas V
nRT P nRP
V T V
f f
i i
V V
V VV
P dVdV nR
T V
,f f
i i
T Vv m
T V
C nRS n dT dV
T V
, ln lnf f
v m
i i
T VS nC nR
T V
(as we saw before)
S for phase change (single substance)
s l s g l g
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Chapter 5: Entropy 53
Reversible process: example l g
Boiling point of water, 100oC at 1 atm
373
orevvap
vap
Hq
T T
Vaporization 373 0o
vapH
o
vap T
moles
vap
HS n
T
s g s l
o
sub T
moles
sub
HS n
T
o
fus T
moles
fus
HS n
T
Example: Raise T of water at 1 atm from 25oC to 125 oC
298 K 373 K heating liquid
At 373 K liquid gas
373 K 398 K heating gas
373 398
, ,
298 373
liquid gas
vapP m P m
vap
HC CS n dT n n dT
T T T
Other example
liquidS freezing + cooling
systembath
bath
bath bath
qqS
T T
system systemq H (constant P )
OR system systemq U (constant V )
surrounding system
surrounding
surrounding surrounding
q qS
T T
(constant surroundingT )
0system surroundingS S for any possible process. (see later)
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Chapter 5: Entropy 54
The Second law of thermodynamics
Postulate by Rudolph Clausius (~1865)
- A Postulate is something you cannot derive but take to be true, because large
number of experiments support it.
0dq
T for any possible physical process
(If 0dq
T for any reproducible process we would have to rewrite
thermodynamics and all of its rules.)
Anything derived from postulate should be true also
Consequences of Postulate:
a) 2
1 2 1
dqS
T for any process
2 2
1 1
revdq dq
T T
2 1
1 20
revdq dq dq
T T T
2 2
1 10
revdq dq
T T
b) For isolated system 0S for a possible physical process
2
1
dqS
T For isolated system 0q
So 0S
c) We can always create a system and environment such that total
system is isolated
0total system surroundingS S S
If environment is kept at constant T
sysenv
env
env env
qqS
T T
We can calculate S for ANY process (allowed and unallowed)
If we calculate totalS for isolated system then:
0S irreversible, possible spontaneous (ice forming at -5oC)
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Chapter 5: Entropy 55
0S reversible, possible (ice equilibrium at 0 oC)
0S impossible (ice forming at 5 oC)
Example:
0q
T T show it?
0total
q qdS
T T
q qT T
T T
Heat flows from hot to cold (another formulation of the second law)
ln 0f
total
i
VS nR
V (at constant T )
0U w q
final initalV V
0
dq
T
2 1
1 20
irrev revdq dq
T T
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Chapter 5: Entropy 56
Absolute Entropy (Pure Substances)
Molar entropy, all 3 phases, all H and ,P mC can be measured
, , ,
0( ) (0 K)
fus vap f
fus vap
S l gT T Tfus vapP m P m P m
m mT T
tus vap
H HC C CS T S dT dT dT
T T T T T
( ) (0)m m mS S T S
Nernst and Richards (~1910)
(0 K) 0mS for perfect crystalline solids.
Today, we can calculate absolute entropy from first principles, quantum and statistical
mechanics for ideal gas. Confirms (0 K) 0mS
Non crystalline substances
Perfect arrangement, but in practice we can get random arrangements
In this case (0 K)mS 5.76 J/mol ( ln 2R )
Microscopic (molecular) view of Entropy
Diffusion of gas:
Laws of microscope physics are time reversible? Why do processes occur in particular
direction?
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Chapter 5: Entropy 57
Difficult (impossible..) to get the initial conditions to reverse such reactions
Calculate for diffusion of gas
Pr 1 1 1 1
......1 1 1Pr 1
2 2 2 2
A
full
N
half
Pr 1
ln ln ln 2 ln 2Pr 2
aN
full
B B B A
half
K K K N R
Thus Pr
ln ln 2Pr
full
B
half
K R (for 1 mol)
S for 1 2V V ln ln 2
f
i
VnR R
V
Similar idea for the crystal, only 2 orientations
Pr 1
Pr 1
2
a
ordered
N
disordered
ln 2 ln 2AN
B B AS K K N
Fully ordered state has zero entropy, ln 1 0AN
BS K
Quantum and statistical mechanic picture of entropy:
Quantum: jE quantum energy levels
Statistical: Probability to find molecule at particular state (Boltzmann
distribution)
/
/( )
j B
j B
E K T
j E K T
j
eP T
e
( ) lnB j j
j
S T K P P
This is our best understanding of entropy
Winter 2013 Chem 254: Introductory Thermodynamics
Chapter 5: Entropy 58
Heat Engine
History: Carnot investigated efficiency of heat (steam) engines ~1830
Carnot: run each cycle in reversible fashion to get max efficiency, recall that grain of
sand compression is more efficient than irreversible process.
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Chapter 5: Entropy 59
Carnot Cycle
All reversible processes:
A: 0U 0hot hotw q
B: 0q B B v cold hotw U C T T
C: 0U 0cold coldw q
D: 0q D D v hot coldw U C T T
total A Bw w w C Dw w
hot coldq q
0 0 0 0hot cold
total
hot cold
q qS
T T
hot cold
hot cold
q q
T T cold
cold hot
hot
Tq q
T
cold
total hot hot
hot
Tw q q
T
1 cold
total hot
hot
Tw q
T
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Efficiency = useful quantity
1costly quantity
total cold
hot hot
w T
q T
(best efficiency you can ever get)
Abstraction of Heat Engine
Carnot: heat cannot be converted to work perfectly
1 cold
hot
T
T
Without wasted heat 1 , then 0coldT K
Refrigerators (or heat pumps)
Can’t convert all work into heat 0S
coldq
w
How do you do it? Run Carnot in reverse
Expand at cold T (you gain very little)
Compress at hot T (takes a lot of work)
cold cold
hot cold
T q
T T w
There is a lot of waste heat, we can use that to heat our homes (Heat pump). We extract
heat from a cold environment (lake) and release the heat in our living room.
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Midterm Review 61
Midterm Review
4 questions, 50 points each, multiple parts
1) Question to derive something
2) Calculate r H ,
rU for reaction at particular T , using heats of reaction. Constant
pC value
3) Ideal gas cycle. Calculate , ,P V T , , , , ,q w U H S
4) Calculate S for spontaneous process
systemS : reversible
system
environment
bath
HS
T
More details on each
1) Derivatives:
- Manipulate derivatives
1
z
z
x
yy
x
y xz
x x y
y z z
- Use an exact differential
, , , ,dF A P V T dT B P V T dV
T V
A B
V T
Winter 2013 Chem 254: Introductory Thermodynamics
Midterm Review 62
Why? Exact differential CAN BE written as
V T
B FdF dT dV
T V
- Mixed derivatives are equal. Any variation is possible
,dT dV
,dT dP
,dV dP (in principle)
Read paragraph 6.2 for application
2) Calculating r H ,
rU
H U PV
U H PV
U H RT n
gasn n (only gases count)
Calculating r H :
298298
To
r i f i p
i i
H T H i C i dT
i in for products
in for reactants
- use general formula Calculate
r H
r H is given, calculate missing piece
- Or draw diagram for complicated reactions and process (I will ask for it if needed)
- Or balance reactions to put proper data together
[examples in book]
3) Ideal gas cycle
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Midterm Review 63
Adiabatic: 0q
Isothermal: constant T
Reversible: follows ideal gas curve
Calculate , ,P V T
Important : reversible adiabatic
,ln ln
f fV m
i i
T VC
R T V
Or ,ln ln
fP m i
i f
TC P
R T P
,V mU nC T
,P mH nC T
extw P dV
q U w (sometimes known to be 0)
,ln
f
i
T fV m
Ti
VCS n dT nR
T V
,ln
f
i
T fP m
Ti
PCn dT nR
T P
0S for adiabatic process
4) Calculate S for spontaneous process
systemS : use reversible path to calculate S always
environmentS : Often keep at constant T
system
environment
bath
HS
T
- Calculate systemH for the actual process