FULL_Thermodynamics_lecture FULL_Thermodynamics_lecture

Classical Thermodynamics
y
CHM102
1

Quantum Chemistry: Dr. Saurav Pal
Molecular properties
Spectroscopy: Dr. Pankaj Mandal
Kinetics
Macroscopic properties
Thermodynamics
p p p
2

Thermodynamics: “Heat” + “Study of motion” Heat Transfer
Thermodynamics: Heat + Study of motion Heat Transfer
 Themodynamics: Heat, work, energy
Wide Applications:
1. Energy change associated to all chemical and physical processes.
2. Mutual transformation of different kinds of energy.
3 To predict the direction and extent of chemical reaction
3. To predict the direction and extent of chemical reaction.
Birth:
3
Industrial revolution (late 18th, early 19th century)

Thermodynamics
Basically is based on four laws.
Th l t f l t d th th li ti
These laws are not formulated, rather these are generalization
deduced from our long experience in nature.
Classical thermodynamics… is the only
physical theory of universal content which
I am convinced that, within the
I am convinced that, within the
applicability of its basic concepts, will
b h ”
never be overthrown.”
Albert Einstein
5
Albert Einstein

Classical Thermodynamics:
Describes macroscopic properties of the systems.
Entirely Empirical
 Based on 4 Laws:
0th L D fi T t (T)
0th Law ⇒ Defines Temperature (T)
1st Law ⇒ Law of conservation of energy It tells that system may
1 Law ⇒ Law of conservation of energy. It tells that system may
exchange energy with its surroundings strictly by heat flow or
work. Defines Internal Energy (U)
gy ( )
2nd Law ⇒ Defines Entropy (S)
3rd Law ⇒ Gives Numerical Value to Entropy

Systems
• A system is part of the universe chosen for observation,
separately from the rest of the universe
separately from the rest of the universe.
• The system plus surroundings comprise a universe.
The boundary between a system and its surroundings is the
• The boundary between a system and its surroundings is the
system wall.
If heat cannot pass through the system wall it is termed an
• If heat cannot pass through the system wall, it is termed an
adiabatic wall, and the system is said to be thermally isolated
or thermally insulated.
y
• If heat can pass through the wall, it is termed a diathermal
wall.
• Two systems connected by a diathermal wall are said to be in
thermal contact.
7

Isolated Closed and Open Systems
Isolated, Closed and Open Systems
• An isolated system cannot exchange mass or energy with its
di
surroundings.
• The wall of an isolated system must be adiabatic.
y
• A closed system can exchange energy, but not mass, with
y g gy, ,
its surroundings.
• The energy exchange may be mechanical (associated with a
The energy exchange may be mechanical (associated with a
volume change) or thermal (associated with heat transfer
through a diathermal wall).
• An open system can exchange both mass and energy with
8
p y g gy
its surroundings.

Closed System
10
Isolated System Isolated System

Isolated, Closed and Open Systems
Isolated
S t
Closed
S t
Open
System
Neither energy
nor mass can be
System
Energy, but not
b
System
Both energy and
11
nor mass can be
exchanged.
mass can be
exchanged.
mass can be
exchanged.

Properties of a System
Thermodynamic variables which are experimentally
measurable. Such variables are macroscopic properties
such as P V T m composition viscosity etc
such as P, V, T, m, composition, viscosity etc.
Thermodynamic variables
y
Intensive variables extensive variables
T, P, density, specific gravity,
surface tension,
specific heat capacity, dielectric
Volume, energy, enthalpy,
Entropy, free energy, mass
specific heat capacity, dielectric
constant.
Does not depend upon mass of
Depends on the mass
of the system.
12
p p
the system

States of a system
How do we define a State of a System?
How do we define a State of a System?
Macroscopic state of a system can be specified by
p y p y
thermodynamic variables which are experimentally
measurable
• Composition – mass of each chemical species that is
present in the system
present in the system.
• pressure (p or P), volume (V), Temperature (T), density,
p (p ), ( ), p ( ), y,
etc.
fi ld t th if ti / l t i l fi ld t th
• field strength, if magnetic/electrical field act on the
system
13
• gravitational field

Equilibrium States
1. Composition remains fixed and definite.
Co pos t o e a s ed a d de te
2. Temp. at all parts of the system is the same.
3. No unbalanced forces between different parts
of the system or between system and surroundings
of the system or between system and surroundings.
System at equilibrium must have definite P, T, and composition
14

Processes
A process refers to the change of a system from one equilibrium
state to another
state to another.
Isothermal (T)
Adiabatic (no heat exchange between system and surroundings).
Isobaric (p)
Isobaric (p),
Isochoric (V)
( )
Cyclic
15

Reversible Process:
• Change must occur in successive stages of infinitesimal quantities
• Infinite duration
• Changes of the thermodynamic quantities in the different stages will be the
• Changes of the thermodynamic quantities in the different stages will be the
same as in the forward direction but opposite in sign wrt forward
direction
Irreversible Process:
• Real / Spontaneous
• Occurs suddenly or spontaneously without the restriction of
• Occurs suddenly or spontaneously without the restriction of
occurring in successive stages of infinitesimal quantities.
• Not remain in the virtual equilibrium during the transition.
ot e a t e tua equ b u du g t e t a s t o
• The work (w) in the forward and backward processes would
be unequal.
16

State function: Those thermodynamic properties depends on the state of the
system, not on the path through which it has been brought in
that state. Potential energy, Internal energy, entropy,
enthalpy.
Path function: Those thermodynamic properties depends on the path
Path function: Those thermodynamic properties depends on the path
through which it has been brought in that state.
(Heat, work).
17

Mathematical formulation of State function (TUTORIAL 1):
1. If any thermodynamic property or function, z = f(x,y) depends
on the initial and final values of thermodynamic variables, then the
change of z i e dz is a perfect differential
change of z, i.e., dz is a perfect differential,
2 In that case it will follow the following mathematical relationship
2. In that case, it will follow the following mathematical relationship,
3. If Z = f(x,y) depends on the initial and final values of
thermodynamic variables, then also ∮dz= 0
18

Concept of Heat and work
Joule's experiment W = JQ
Joule s experiment W = JQ
W = work expended in the
W = work expended in the
production of heat or
obtained from heat
Heat  Work
Both heat and work represents energy in transit
Both heat and work represents energy in transit.
Work involved in a process , heat change involved during
d d h h f f i
a process depend on the path of transformation.
 Work and heat are the two methods by which energy is exchanged
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 Work and heat are the two methods by which energy is exchanged
between system and surroundings.

Sign Convention of heat and work:
Work:
Work:
The work done by the system is defined to be nagative (-).
 The work done on the system the external work of
 The work done on the system – the external work of
mechanics – is positive (+).
Heat:
The heat absorbed by the system is defined to be positive (+)
The heat absorbed by the system is defined to be positive (+).
 The given out by the system is negative (-).
20

21
U = Uf – Ui Measurable

(TUTORIAL-1)
Concept of Internal energy:
Concept of Internal energy:
Internal Energy, U. is the total energy within a system.
U is the internal energy of the body (due to molecular motions
and intermolecular interactions)
• Extensive property.
S f i i d d f h
• State function, independent of path.
• For cyclic process,
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dU is a perfect or exact differential

Thermal Equilibrium and Zero’th Law of Thermodynamics
23

First Law of Thermodynamics:
Energy cannot be destroyed, it can be transformed to
one form to another (Law of conservation of energy)
one form to another. (Law of conservation of energy).
K E 0 PE i
K.E.= 0, P.E. = maximum
K.E.= maximum, P.E. = 0
 Energy is conserved
24
gy

1st law of Thermodynamics
The first law for a closed system or a fixed mass may be expressed
as:
net energy transfer to (or from)
the system as heat and work = net increase (or decrease) in the
the system as heat and work net increase (or decrease) in the
total energy of the system
q+ w = U
25

How the pressure of a cylinder can be changed?
27

Equation of state for Ideal and Real gas
Ideal gas: PV = nRT
Real gas: van der Waal’s equation
Real gas: van der Waal s equation
  nRT
nb
V
a
n
P 








2
N·b
  nRT
nb
V
V
P 






 2
28

Work and heat change associated in Isothermal Reversible
TUTORIAL 2
Work and heat change associated in Isothermal Reversible
and irreversible expansion
w
w
(b) Irreversible process:
W = P (V V )
Wirr = Pext (V2 – V1)
29

Adiabatic process in an Ideal Gas
TUTORIAL 3
• Since dQ = 0 for an adiabatic process,
dU P dV d dU C dT th t dT (P/C ) dV
dU = – P dV and dU = CV dT, so that dT = – (P/CV) dV .
• For an ideal gas, PV = nRT,
g
so that P dV +V dP = nR dT = – (nRP/CV) dV.
Hence V dP + P (1 +nR/C ) dV = 0
Hence V dP + P (1 +nR/CV) dV = 0.
Thus, CV dP/P + (CV + nR) dV/V = 0.
For an ideal gas, CP – CV = nR.
so that C dP/P + C dV/V = 0 or dP/P + γ dV/V = 0
so that CV dP/P + CP dV/V = 0, or dP/P + γ dV/V = 0.
• Integration gives ln P + γ ln V = constant, so that
PVγ t t
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PVγ = constant.

• Work done in a reversible adiabatic process
For a reversible adiabatic process, PVγ = K.
• Since the process is reversible, W = -CV∆T =Cv(T1-T2) ,
For 1 mole of gas , T=PV/R
so that Wadi = = Cv[P1V1/R-P2V2/R]
= Cv/R[P1V1-P2V2]
 W = 1/(γ –1) [P2V2 – P1V1].
• For an monatomic gas γ = 5/3 so that
• For an monatomic gas, γ = 5/3, so that
W = –(3/2)] [P2V2 – P1V1].
31

Reversible Processes for an Ideal Gas
Reversible Processes for an Ideal Gas
Adiabatic Isothermal Isobaric Isochoric
process process process process
PVγ = K T constant P constant V constant
PVγ = K
γ = CP/CV
T constant P constant V constant
W = – [1/(γ –1)]
.[P2V2 – P1V1]
W = nRT ln(V2 /V1) W = P V W = 0
∆U = CV ∆T ∆U = 0 ∆U = CV ∆T ∆U = CV ∆T
PV = nRT, U = ncVT, cP – cV = R, γ = cP/cV.
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Monatomic ideal gas cV = (3/2)R, γ = 5/3.

Thermal insulation
F
F
O
O
C
O
C
O
5
0
4
0
3
0
2
0
1
0
0
1
0
120
100
80
0
20
60
40
Valve
0
2
0
3
0
4
0
5
0
20
40
60
A
T1, Vm,1, P1
B
Stirrer

1. No change in temperature was detected,
dq = 0
2. As expansion is taking place against zero pressure,
d 0
dw = 0,
As a result dU = 0
As a result, dU = 0
U f(V T) U
T
U



 



 



 
U = f(V,T)
V
U
T
T
U
V
T
V
U


























  T
0

 









U
U
V
T
,
34
0









T
V
U
Hence ,

0




  U
(VALID f Id l l )
0




  T
V
(VALID for Ideal gas only)
0
V
U








 (VALID for REAL gas only)
V
U








= a/v2
Actually the gas in A warmed up slightly and the one which had
V T

  T
V 
 
Actually the gas in A warmed up slightly and the one which had
expanded into B was somewhat cooler and when thermal
equilibrium was finally established the gas was at a slightly
q y g g y
different temperature from that before the expansion.
 Because the system used by Joule had a very large heat
capacity compared with the heat capacity of air, the small change
of temperature that took place was not observed
35
of temperature that took place was not observed.

Concept of Enthalpy and heat Capacities
 Prove that, H = qP
, qP
 Prove that, CP - CV = R (for one mole of Ideal gas)
36

Joule-Thompson Effect
A gas passes through a POROUS PLUG from
i h it i t hi h t i
a region where it is at high pressure to a region
where it is at lower pressure. The gas expands,
and the temperature of the gas can be lowered.
This is an important tool in low temperature
This is an important tool in low temperature
physics.
37

Background
The objective of this experiment is to quantitatively
measure the non-ideality of gases using the Joule-
Th ffi i t d l ti it t th ffi i t f
Thomson coefficient and relating it to the coefficients of
equations for non-ideality.
For an ideal gas, the internal energy is only a function
of the absolute temperature so in an isothermal
p
process ∆U = 0. The same is true for the enthalpy for
such a process: ∆H = 0.
Thus:
0







 H
H
0




 




  T
T P
V
These are non-zero for a real gas.

Suppose the volume V1 under a constant pressure P1 is
Suppose the volume V1 under a constant pressure P1 is
allowed to pass through porous plug from region of the left to
right where the constant low pressure is P2. The net volume
2
change in the right side of the chamber is V2.
Th k d th l ft id i P V
The work done on the left side is: w1 = P1V1.
The work done on the right side is:W2 = -P2V2.
The total work done is then:
The total work done is then:
w = P1V1 - P2V2 (1)
Since the process is adiabatic, the total change in U,
U = W
U U U P V P V (2) H = H
U = U2 − U1 = P1V1 - P2V2 (2)
U2 + P2V2 = U1 + P1V1 (3)
H1 = H2 ,

H



 
JT
T
P
H
C
H
P
P
T













 
















 1
T
P
P
H P
C
T
H
P 
 









 
Later we can show










 







 
V
V
T
T
JT
1
 






 



  T
C
P P
P
JT
H


P
T
 In an adiabatic throttle process, the gas pressure is reduced (P2<P1),
and thus
T
 If the temperature of the gas is reduced,T2<T1 , which
produces a cooling effect;
0
)
( 



 H
T
J
P
T

 If ,the temperature of the gas is raised, T2>T1,
0
)
( 



 H
T
J
T

which produces a heating effect;
)
(

 H
T
J
P


45
 If , the temperature of the gas has no change, i.e.,
T2=T1
0
)
( 



 H
T
J
P
T


S d l f Th d i
Second law of Thermodynamics
47

Second Law of Thermodynamics
Hot Reservoir, TH
E i
QH
W
Engine
QC
Q W
Is it possible?
Cold Reservoir, TC
Is it possible?
It is impossible for a system to undergo a cyclic process whose sole effects
are the flow of heat into the system from a heat reservoir and the performance
are the flow of heat into the system from a heat reservoir and the performance
of an equivalent amount of work by the system on the surroundings.
-------Kelvin-plank statement
48

Kelvin statement (1851)
No process can completely convert heat into work; i.e. it
is impossible to build a “perfect” heat engine.
Efficiency = work done/heat absorbed
Hot Reservoir, TH
Efficiency work done/heat absorbed
= W/QH
QH
W
H
= QH- QC/QH
Engine
QC
Later we will show,
W Q * T/T
Cold Reservoir, TC
W = QH*T/TH
49

Heat Engine
A h t i i li th t b b h t f
A heat engine is a cyclic process that absorbs heat from
a heat bath and converts it into work. We shall see that
i th li th i l di i t
in the cyclic process the engine also dissipates some
heat to a heat bath at a lower temperature.
Hot Reservoir, TH Hot Reservoir, TH
, H
QH
W
, H
QH
W
Engine
QC
W
Engine
QC = 0
W
Cold Reservoir, TC
QC
Cold Reservoir, TC
QC
50
Real engine. QH = QC + W Impossible engine. QH = W

Q1 = Q2 + W
• It is impossible for a system to undergo a cyclic process whose sole
effects are the flow of heat into the system from a cold reservoir and the
flow of an equal amount of heat out of the system into a hot reservoir.
• -------Clausius statement
Refrigeration engine takes away heat from the colder reservoir
to hot reservoir with the help of external electrical work
51
to hot reservoir with the help of external electrical work

Carnot engine or Carnot cycle
1 Th i t t i l t l t l d
1. The engine must operate in complete cycles to exclude any
work involved in its own change.
2 T bt i i k i l f ti t
2. To obtain maximum work in a cycle of operations, every step
should be carried out in a reversible fashion.
52

Carnot Cycle
Pressure
•
a
Q1
•
b
T
Step -I
Step -IV • T1
Q=0
Q=0
Step IV
•
d
Step -II
Step -III
d
Q2
T2
•
c
Volume

Step 1 a-b (Isothermal Reversible Expansion):
The gas enclosed in a cyclinder fitted with frictionless
Piston. To start with cylinder containing the gas is kept in a
large thermostat at higher temp. T1 (source), and suppose the
volume of the gas be V1.
V1 V2
As temperature remains constant hence the change in
As, temperature remains constant, hence the change in
Internal energy (U = 0).
Hence, U = q + W
, q
Heat absorbed by the system = (-) of work done by the system
The heat absorbed by the gas, Q1 = RTln(V2/V1)
54
e eat abso bed by t e gas, Q1 ( 2/ 1)
= work done by the gas
= w1

Step 2 (Adiabatic Reversible Expansion):
The cyclinder is taken out from the thermostat and kept in a
thermally insulated enclosure. The gas is allowed to expand
further from volume V2 to V3 adiabatically and reverasibly until
the temperature falls down to that of the sink T2.
The heat absorbed by the gas = nil
The work done by the gas, w2 = Cv (T2 – T1) (T1 > T2)
55

Step 3 (Isothermal Reversible Compression):
The cylinder is then placed in a thermostat at lower temp. T2
(sink), and the gas is compressed isothermally and reversibly
( ) g p y y
from V3 to V4.
V3 V4
V3 V4
Th k d th ( )RT l (V /V ) ( V V )
The work done on the gas, w3= (-)RT2ln(V4/V3) (as V4 < V3)
The heat given out by the gas, Q2 = - w3
= RT2ln(V4/V3)
2 ( 4 3)
56

Step 4 (Adiabatic Reversible Compression):
The cyclinder is taken out from the thermostat and kept in a
thermally insulated enclosure. The gas is allowed to compress
reversibly to its volume V1 and its original temperature T1 is
attained.
The heat absorbed by the gas = nil
The work done on the gas, w4 = Cv (T1 – T2)
57

Carnot Engine
1. Calculate total work involved in all these steps:
2 Net heat absorbed :
2. Net heat absorbed :
3. Efficiency of the engine:
y g
TH = Temperature of the source
H p
TL = Temperature of the sink
Efficiency of any heat engine does not depend
on the working substance rather depends on the temperature
58
on the working substance, rather depends on the temperature
of the source and sink

Second law of Thermodynamics
C f
1. Concept of time-arrow
2 C f
2. Concept of entropy.
59

64
The total energy is dispersed into random thermal motion of the
particles in the system.

• The thermodynamic property of a system that is related to its
degree of randomness or disorder is called entropy (S)
degree of randomness or disorder is called entropy (S).
E t i f th t t t hi h i
• Entropy is a measure of the extent to which energy is
dispersed.
• The entropy S and the entropy change ∆S=S2-S1 are state
functions
functions.
Th t S h i l th P
• The entropy S has a unique value, once the pressure P,
temperature T and the composition n of the system are
specified, S = S(P,T,n).
specified, S S(P,T,n).
• The entropy is an extensive property i e increases with the
The entropy is an extensive property, i.e. increases with the
amount of matter in the system. 66

Entropy Criteria in different Processes
67

Change in the extent to which energy is dispersed depends on
how much energy is transferred as heat.
In case of closed system:
68

Combination of first and 2nd law:
dS =
dS
70

Entropy change due to Phase Transition
At the transition temperature, any transfer of energy as heat
between system and its surroundings is reversible because the
two phases in the system are in equilibrium
71

Calculate entropy change in several processes
for ideal gas. (Tutorial-----)
for ideal gas. (Tutorial )
Isothermal Changes:
sot e a C a ges
(a) Reversible
(b) Irreversible process
Adi b ti Ch
Adiabatic Changes:
(a) Reversible
73

Clausius Inequality: ds ≥ dq/T
Assume reversible and irreversible paths between two states.
R ibl th d k th i ibl th
Reversible path produces more work than irreversible path,
That is /dw / ≥ /dw/ Because dw and dw are negative when
That is, /dwrev/ ≥ /dw/. Because dw and dwrev are negative when
energy leaves the system as work, this expression is same as
- dwrev ≥ dw, and hence, dw –dwrev ≥ 0
rev rev
dU is the same for both the paths.
dU = dq + dw = dqrev + dwrev
dqrev–dq = dw - dwrev ≥ 0
dqrev/T ≥ dq/T
74
dS ≥ dq/T

If N be the total number of distinguishable molecules in a
system, and N1, N2, N3……. etc. be number of molecules
distributed in different energy levels, then the number of
microstates corresponding to the given distribution is called
microstates corresponding to the given distribution is called
Thermodyamic Probability,
S= klnW
S klnW
1

FULL_Thermodynamics_lecture FULL_Thermodynamics_lecture

Let us a take a sample containing 16 distinguishable molecules,
sharing a total energy of 16 E and quantum states differing
by unit E
E
by unit E.
4
2
3
0
1
W=1 W=8.9  10⁵ W=1.44  10⁷
0
Number of microstates increases with increase in the randomness
in the distribution of the molecules. Hence, entropy increases.
S= klnW

W =1
S= klnW = 0
W >1
S= klnW  0

Energy levels of Paramagnetic Substance in Absence and
Presence of Magnetic Field:
g

Corollary of the Third LAW of Thermodynamics
Absolute zero ????

Concept of Helmholtz work Function and Gibbs Free energy

Why it is called free energy or available energy?
G = H – TS
If a system undergoes reversible change
dG = dH – d(TS)
= d(U + PV) – d(TS)
= dU + d(PV) – TdS – SdT
= dU + VdP + PdV – TdS – SdT
dU VdP PdV TdS SdT
At constant external P & Isothermal condition
dGP,T = dU + PextdV – TdS
= dU + P dV – dq
= dU + PextdV – dq
= -(dW) + PextdV
= -(dWP-V + dWnonP-V ) + PextdV
(PdV + dW ) + P dV
= -(PdV + dWnonP-V ) + PextdV
= - dWnonP-V
-dGP T = dWnonP-V
P,T nonP-V
It is the amount of work required for any external
use exclusive of the expansion work. This work may be electrical
work for pushing electron through a circuit or work required for
transmitting nerve impulse.

Isothermal processes can occur in any kind of
system, including highly structured machines,
and even living cells.
and even living cells.
Surroundings act as heat sink in most of the
Surroundings act as heat sink in most of the
cases.

Most reactions of an acid and base mixed together to form a
Example of constant T and Pressure Reaction
Most reactions of an acid and base mixed together to form a
salt are exothermic. If a strongly acidic solution is rapidly
poured into a strongly basic solution in a beaker (glass cup)
p g y (g p)
that is wrapped with thermal insulation, the heat generated
cannot escape, and the resulting system of solution plus salt
( i h di l d i i d) h Thi i
(either dissovled or precipitaed) heats up. This is not an
isothermal process. But if the thermal insulation is removed
from the beaker of basic solution and the beaker is set into a
from the beaker of basic solution, and the beaker is set into a
large tub of water, and the temperature of the basic solution is
allowed to equilibrate with the water bath (come to the same
temperature), and the acid solution is added slowly, then the
heat of reaction will have time to move through the beaker
l ll i t th l b th f t d th t t f
glass wall into the large bath of water, and the temperature of
the solution in the beaker will remain at the temperature of the
bath as the acid is slowly poured into the beaker and the acid
bath as the acid is slowly poured into the beaker and the acid
and base react. This is an isothermal process.

Problem: How much energy is available for sustaining
l d i i f h b i f 1
muscular and nervous activity from the combustion of 1
mol of glucose molecules under standard conditions at
37 °C (blood temperature)? The standard entropy of
reaction is +259.1 JK-1 mol-1. ΔrH = -2808 kJ mol-1
Method: The available energy from the reaction is
gy
equal to the change in standard Gibbs energy for the
reaction (ΔrG). To calculate this quantity, it is legitimate
reaction (ΔrG ). To calculate this quantity, it is legitimate
to ignore the temperature dependence of the reaction
enthalpy to obtain Δ H and to substitute the data into
enthalpy, to obtain ΔrH, and to substitute the data into







 S
T
H
G r
r
r

Answer: Because the standard reaction enthalpy is
-2808 kJ mol-1, it follows that the standard reaction
Gibbs energy is
gy
1
1
1
1
r )
mol
JK
1
.
259
(
)
K
310
(
kJmol
2808
G 








Therefore, W dd = -2888 kJ.mol-1 for the combustion
1
kJmol
2888 


Therefore, Wadd,max 2888 kJ.mol for the combustion
of 1 mol glucose molecules, and the reaction can use
2888 kJ for external work
2888 kJ for external work.
To place this results in perspective, consider that a
person of mass 70 kg needs to do 2 1 kJ of work to climb
person of mass 70 kg needs to do 2.1 kJ of work to climb
vertically through 3 m; then at least how much amount
f l i d d t l t th t k?
of glucose is needed to complete the task?

Gibbs Free Energy
gy
1. If G is negative, the
g ,
forward reaction is
spontaneous.
p
2. If G is 0, the system is
at equilibrium.
at equilibrium.
3. If G is positive, the
reaction is spontaneous
reaction is spontaneous
in the reverse direction.

Free Energy and Equilibrium
Free Energy and Equilibrium
Remember from above:
If G is 0, the system is at equilibrium.
So G must be related to the equilibrium constant,
q ,
K. The standard free energy, G°, is directly
linked to Keq by:

Masters Equations of Chemical Thermodynamics
A closed system (constant
composition, and only p‐V work)
28

The "cyclic relation" (sometimes called Euler's chain relation) is a calculus identity which is very
useful in thermodynamics. This relation can appear in several different forms all of which are
equivalent. The form that we will find most useful is,
30

Maxwell’s Relation:
Imp
I
Imp
31

Maxwell Relation for G
The Gibbs function (or free energy) is defined as
G = U – TS + PV
G U TS PV
dG = dU – TdS – SdT + PdV + Vdp .
dU = TdS – PdV,
so that dG = – SdT + VdP ; i.e. G = G(T,P).
so that S (G/T) and V (G/P)
so that S = – (G/T)P and V = (G/P)T...
2G/TP = 2G/PT,
 G/TP  G/PT,
so that (S/P)T = –(V/T)P .
Note that Maxwell’s relation equates (S/P)T , a theoretical
q antit to (V/T) α V both of hi h ma be meas red
32
quantity, to (V/T)P = α V , both of which may be measured.

Now you should be able to show that,
36

closed system, only p‐V work
Does not apply
pp y
• When composition is changing due to exchange of matter
with surroundings (open system)
• Irreversible chemical reaction
38
• Irreversible inter‐phase transport of matter

Physical Significance of Chemical potential
 Chemical potential as measure of a general tendency of substances to transform,
and as central concept of chemical dynamics
and as central concept of chemical dynamics.
It reflects the potential of a substance to undergo physical
Change or chemical change.
Change or chemical change.
In thermodynamics, chemical potential, also known as
partial molar free energy, is a form of potential energy
that can be absorbed or released during a chemical reaction.
41

Phase Transition
 Map showing conditions of T
and p at which various phases
are thermodynamically stable.
At any point on the phase
boundaries, the phases are in
, p
dynamic equilibrium.
Water and ice are in equilibrium at 00C and 1 atm.
Water and vapour are in equilibrium at 100 at 1 atm.
42
G = 0 at phase transition temp at 1 atm.

Effect of pressure and temperature on Phase Equilibrium
44

At point a,
µα = µβ (as system exists in equilibrium at point a)
Or
Or
Now imagine you move to point b, where new equilibrium will be established.
(We can write with respect to partial molar quantity. G bar means per mole)
g y p , q
At point b,
µα + dµα = µβ + dµβ
As µα = µβ , hence,
dµα = dµβ or
Clausius Clapeyron equation
This equation dictates how the pressure variation leads to temperature
45
q p p
variation or vice versa in order to achieve new equilibrium condition.

Phase Transition
 Map showing conditions of T
are thermodynamically stable.
At any point on the phase
boundaries, the phases are in
, p
dynamic equilibrium.
Water and ice are in equilibrium at 00C and 1 atm.
Water and vapour are in equilibrium at 100 at 1 atm.
46
G = 0 at phase transition temp at 1 atm.

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