1. CY 11001 CHEMISTRY
Professor: Dr. Amita (Pathak) Mahanty
Office: Room C-301 (2nd Floor, Chemistry Department)
e-mail: ami@chem.iitkgp.ernet.in
phone: 83312 (office)
Ground Rules for the Class
• Mobile Phones in switched off mode
• Attendance is a must and will be checked regularly
• Vast syllabus and fast paced, advised to take class notes
4. CY 11001 CHEMISTRY
Distribution of Marks:
Mid Semester Exam (Only Physical): 30%
End Semester Exam (Inorganic & Organic): 50%
Teachers Assessment (TA): 20%
Break up: [Physical (6%), Inorganic (7%), Organic (7%)]
TA marks to be based on a CLASS-TEST: 30th January 2013
Text Book:
Atkin’s Physical Chemistry 8th Ed
Reference Books:
Physical Chemistry by Silbey, Alberty & Bawendy; Wiley.
Physical Chemistry by I. N. Levine; McGrew.
Physical Chemistry David W. Ball; Thompson.
5. CY 11001 CHEMISTRY (L-T-P : 3-1-0; Credit : 4)
Thermodynamics of Chemical Processes:
• Review of Essential Concepts and Definitions
• Revision: Heat, Work & Energy; First Law of Thermodynamics
• The Second Law of Thermodynamics • Concept of Entropy
• Gibbs Free Energy • Equilibrium conditions for Closed Systems
• Maxwell Relations • The Chemical Potential
• Definition and Concept of Open Systems
• Phase and Reaction Equilibria
Electrochemical Systems:
• Electrochemical Cells and EMF
• Applications of EMF Measurements: Thermodynamic data,
Activity Coefficients, Solubility product and pH
Kinetics of Chemical Reactions:
• Reversible, Consecutive and Parallel Reactions,
• Steady State Approximation,
• Chain Reactions, Photochemical Kinetics
6. Thermodynamic
Thermodynamics provides a framework of relating the macroscopic
properties of a system to one another.
Thermodynamic: ↔ Classical, Statistical and Irreversible
Classical Thermodynamics, to the study of macroscopic properties of
system/matter at equilibrium using the laws governing transformation
of energy from one form to another.
It is concerned only with macroscopic quantities and ignores the
microscopic variables that characterize individual molecules.
THE BASIC CONCEPTS:
System, the part of the world in which we have a special interest.
Surroundings, the region outside the system where we make our
measurements.
Boundary, a hypothetical/real barrier between system and
surroundings.
7. Thermodynamic Systems
Open systems can exchange both matter
and energy with the environment.
Closed systems exchange energy but not
matter with the environment.
Isolated systems can exchange neither
energy nor matter with the environment.
8. Thermodynamic Systems
System : Open, Closed, Isolated
Surroundings
Boundary / Wall
• Rigid and Non rigid (movable)
• Permeable and Impermeable (no matter is allowed to pass
through it)
• Adiabatic and Diathermic boundary ( - permits the passage of
energy as heat )
A system surrounded by a rigid, impermeable, adiabatic wall –
Isolated system Example: Universe as a whole, a thermo flask
9. Thermodynamic Equilibrium
For Isolated system:
• Macroscopic properties do not change with time
For Non isolated system:
a) Macroscopic properties do not change with time
b) Removal of the system from contact with its surroundings causes
no change in the properties of the system
• A system in thermodynamic equilibrium is an equilibrium system
• Thermodynamic State only defined by the present values of the
state variables, not by the history
What are conditions of Thermodynamic Equilibrium?
10. Mechanical equilibrium
The condition of equality of pressure on
either side of a movable wall / boundary.
No unbalanced forces act on or within
the system – does not undergo
acceleration or no turbulence inside the
system
11. Thermal equilibrium
Thermal equilibrium between system and
the surroundings, is a condition in which
there is no change in the properties of the
system or surroundings when they are
separated by a thermally conducting wall.
In other words, Thermal Equilibrium is a
condition in which no change of state occurs
when two objects A to B are in contact
through a diathermic boundary.
12. Material Equilibrium
Concentrations of the chemical species in the various parts of the
system are constant with time
a) No net chemical reactions are occurring in the system
b) There is no net transfer of matter from one part (phase) of the
system to another or between the system and its surroundings
Phase: a homogeneous part of a system is called a phase
For thermodynamic equilibrium, all three kinds of equilibrium
must be present
No matter what is the initial state of an isolated system,
eventually it will reach the state of thermodynamic equilibrium
13. How do we define a system? State of a System
Macroscopic state of a system can be specified by the values of a small
number of macroscopic Properties/Parameters/Physical characteristics or
attributes of a system - Thermodynamic variables
Thermodynamic variables which are experimentally measurable
• Composition – mass of each chemical species that is present in each
phases,
• pressure (P), volume (V), Temperature (T), density, etc.
• field strength, if magnetic/electrical field act on the system
• gravitational field
(N.B.: only the average values are considered, all fluctuation and
perturbations are ignored).
On the other hand, Microscopic state of a system needs description of each
molecule – a very complicated picture
14. Thermodynamic State
If the value of every thermodynamic property in system A is same as in
system B – then they are in the same thermodynamic state
The state of a thermodynamic system is defined by specifying the values of
its thermodynamic properties – However, it is not necessary to specify all
the properties to define the state of a system.
In fact, the macroscopic state of a system can be specified by the values of
a small number of macroscopic variables – these called the Properties /
Parameters / Thermodynamic variables of a system.
For instance: The thermodynamic state of a single-phase system of fixed
amounts of non reacting substances can be described by any two of the
three state functions (P, V and T)
Equation of state, an equation that interrelates pressure, volume,
temperature, and amount of substance: P = f (T, V, n).
15. The Equation of State of Ideal Gases
An equation of state - an equation that relates macroscopic variables (e.g.,
P, V, and T) for a given substance in thermodynamic equilibrium.
In equilibrium (≡ No macroscopic motion), just a few macroscopic
parameters are required to describe the state of a system.
Geometrical representation of the equation of state: f (P,V,T) = 0
The ideal gas equation of state: PV = nRT
an equilibrium
P state
P – pressure [Newtons/m2]
V – volume [m3]
the equation-
n – number of moles of gas of-state surface
T – the temperature in Kelvins [K]
J
R – a universal constant R ≈ 8.31 V
mol ⋅ K
T
16. Recap
The gas laws
Boyle’s law: At constant temperature, the pressure of a
sample of gas is inversely proportional to its volume, p ∝
1/V at constant temperature.
Charles’ law: At constant pressure, the volume of a gas is
proportional to the absolute tem perature, V ∝ T at constant
pressure.
Avogadro’s principle: Equal volumes of gases at the same
pressure and temperature contain the same numbers of
molecules.
Dalton’s law: The pressure exerted by a mixture of gases is
the sum of the partial pressures of the gases.
17. Thermodynamic Systems
Homogeneous system, a system having its properties uniform
throughout – a single phase solution / pure.
Heterogeneous system, a system with more than one phase.
Intensive properties
Does not depend upon size (the quantity/amount of matter present) of
the system [e.g. Pressure, Temp, Viscosity, Density etc.]
Total Property = property of part of the system
Extensive properties
Depends upon size of the system [e.g. n, V, surface area etc ]
Extensive properties depend on the amount of substance in the system.]
Total Property = Σ property (in each part of the system)
Other variables:
Compositions, Surface area, Electrical/magnetic strength
Gravitational field can/are generally ignored
18. A very important macroscopic parameter: Temperature
Temperature is a property associated with random motion of
many particles. Introduction of the concept of temperature in
thermodynamics is based on the the 0’th law of thermodynamics:
B acts as Thermometer.
A, B, C all are at the same “temperature”.
A well-defined quantity called Temperature exists such that two
systems will be in thermal equilibrium if and only if both have the
same temperature.
20. Thermodynamic State Variables / Properties
Therefore, For describing the State of a System we require:
• A few experimentally measureable macroscopic properties: p, T, V, n, m, ...
• Knowledge if System is Homogeneous or Heterogeneous
• Knowledge if System is in Equilibrium State
• Knowledge of the number of components
21. Thermodynamic Process: Operation for Change in State
Change of State: Transformation
e.g., 3 H2 (g, 5 bar, 100 °C) → 3 H2 (g, 1 bar, 50 °C)
(initial state) (final state)
Path: Sequence of Intermediate states through which a system passes
to attain the final state
Process: Operation through which a system undergoes a change in state.
Process describes the path.
Path
Path can be:
Reversible (always in Equilibrium)
Irreversible (describes direction of time)
Cyclic(initial and final state have the same value)
Processes can be:
• Adiabatic (No heat transfer between system and surrounding)
• Isobaric (Constant pressure process)
• Isothermal (Constant temp process)
• Constant Volume process • Endothermic & Exothermic process (q<0 & q>0)
22. Essential concepts / Definitions
Processes : When one or more of the parameters of a system changes, the
state of the system also changes and the system is said to have undergone
a process.
Various types of processes:
• Isothermal (T),
• Isobaric (P),
• Isochoric (V)
• Adiabatic (q),
• Isenthalpic (H)
• Exothermic & Endothermic
• Cyclic
Each process can be
either carried out
Reversibly or
Irreversibly
23. Thermodynamic Processes can be :
Reversible process: Ideal process
• Change must occur in successive stages of infinitesimal quantities
• Infinite duration
• Virtual thermodynamic equilibrium, at each of the small stages.
• Changes of the thermodynamic quantities in the different stages will be
the same as in the forward direction but opposite in sign w.r.t. forward
direction. The system and surroundings can be put back in their original
states by exactly reversing the process.
Irreversible Process: Real / Spontaneous
• Occurs suddenly or spontaneously without the restriction of occurring in
successive stages of infinitetesimal quantities. Irreversible processes
cannot be undone by exactly reversing the change to the system.
• Do not remain in the virtual equilibrium during the transition.
• The work (w) in the forward and backward processes would be unequal.
Cyclic process: • Is one in which the initial and final states are the same.
In such process:
• There is no change in the state variables (e.g., ∫dU = Uf-Ui = 0)
• In contrast, path functions generally have non-zero values for cyclic
processes, dependent on the path (e.g., ∫δw ≠ (wf – wi) ≠ 0]
24. Spontaneous Processes
• Spontaneous processes are those
that can proceed without any
outside intervention.
• The gas in vessel B will
spontaneously effuse into vessel
A, but once the gas is in both
vessels, it will not spontaneously
go back to B
25. Reversible Processes
In a reversible process the
system changes in such a way
that the system and
surroundings can be put back
in their original states by
exactly reversing the process.
Changes are infinitesimally
small in a reversible process.
26. Irreversible Processes
• Irreversible processes cannot be undone by exactly reversing
the change to the system.
• All Spontaneous processes are irreversible.
• All Real processes are irreversible.
27. Energy, Work and Heat
Energy is the capacity to do work.
Energy classification into: kinetic potential
(by motion) (by position)
Is purely arbitrary! e.g.thermal chemical, electrical
• Heat and work are NOT “types” of ENERGY, but are processes involving
TRANSFER of energy.
• They appear and disappear at the system boundary. They are path variables.
• Heat is the transfer of energy from one body to another based on temperature
difference. They are path variables.
• Heat stimulates random motion.
• Convention: if heat flows into the system, q > 0.
• Work is the transfer of energy by some mechanism other than temperature
difference.
• They appear and disappear at the system boundary. They are path variables.
• Convention: if work is done on the system, w > 0.
• Work stimulates organized motion.
• Work “degrades” into heat.
Qualitative observations by Count Rumford (Ben Thompson)
Quantitative measurements by James Joule
28. Energy
Concept of Internal Energy
• E = K + V + U where, K and V are the macroscopic (not
molecular) kinetic and potential energies of the system
• U is the internal energy of the body (due to molecular motions
and intermolecular interactions)
• Internal Energy,U, is the total kinetic energy due to motion of
molecules (translational, rotational, vibrational) the total potential
energy associated with the vibrational and electric energy of
atoms within molecules or crystals.
• Um (molar internal energy) = U/n, where n= no. of moles of the
substance
• Um = Utr, m + Urot, m + Uvib, m + Uinternal, m + Urest, m (where,
Urest,m = rest mass energy of electrons & nuclei = mrelC2= constant)
29. Energy
Properties of Internal Energy
• U is a State function, a property that: Depends only on the
current state of the system and is independent of how or
through path that state has been reached.
• It is an exact differential and its fundamental variables are
V and T ⎛ ∂U ⎞ ⎛ ∂U ⎞
U = f (V, T); dU = ⎜
∂V
⎟ dV +⎜
∂T
⎟ dT
⎝ ⎠T ⎝ ⎠V
• For cyclic process, ∫ dU =0
• U is an Extensive property
30. The First Law of Thermodynamics: Work, Heat & Energy
T1 → T2
T1 → T2 h
Joule Experiment
Joule Equivalent of Heat: 1 cal = 4.184 J
31. Work
• Expansion work, work of expansion (or compression)
• Non-expansion / additional work: any other work
where
• pex = external pressure
• A = piston area
• dz = displacement
• dV = Adz = volume change for the gas
vf
Total work done, w = − ∫ p ex dV
vi
• Work done on the system + ve i.e., Compression
• Work done by the system - ve i.e., Expansion
32. 1st Law of Thermodynamics
Consider an example system of a piston and
cylinder with an enclosed dilute gas
characterized by P,V,T & n.
What happens to the gas if the piston is moved
inwards? Compression
•If the container is insulated the temperature
will rise, the atoms move faster and the
pressure rises.
•Is there more internal energy in the gas? Yes
∆x
External agent did work in pushing the piston
inward.
w = - F d = -(PA) (-Δx)
w = P Δ V
Work done on the gas equals the change in the
internal energy of the gas w = Δ U
33. 1st Law of Thermodynamics
• Change the situation:
• Keep the piston fixed at its
original location.
• Place the cylinder on a hot plate.
What happens to gas?
• Heat flows into the gas.
• Atoms move faster, internal energy
increases.
• Q = heat (in Joules) flows in
∆U = change in internal energy in
Joules.
Q = ∆U
34. 1st Law of Thermodynamics
F
What if we added heat and pushed the piston
in at the same time?
• Work is done on the gas,
• Heat is added to the gas and
• the internal energy of the gas
increases:
ΔU = w + Q
U INCREASES with INCREASE in TEMP. of the system, when:
1. work is done on the system by the surrounding or,
2. heat flows into the system from the surrounding or,
3. both, heat and work flows from the surr. into the system
35. The First Law: Sign convention
• Work done on the system by surr + ve i.e., Compression
• Work done by the system by surr - ve i.e., Expansion
36. Partial Differentiation: A primer
Consider a function of two variables f (x,y)
Definition: A partial derivative of one of the
variables: the rate of change of the function
w.r.t. that variable with all other variables
kept fixed.
Suppose: f = f (x,y),
⎛ df ⎞ f ( x + Δx,y ) − f ( x,y )
⎜ dx ⎟ = LimΔx →0
⎝ ⎠y Δx
⎛ df ⎞ f ( x ,y + Δy ) − f ( x,y )
⎜ ⎟ = LimΔy →0
⎝ dy ⎠x Δy
• For example: f (x, y) = x2y ⎛ ∂f ⎞ ⎛ ∂f ⎞
⎜∂ ⎟ = 2xy, ⎜ ⎟ = x 2
⎝ x ⎠y ⎝ ∂y ⎠x
37. Properties of partial derivatives
Consider a function of two variables f (x, y)
∴ Its differential form will be:
⎛ df ⎞ ⎛ df ⎞
df = ⎜ ⎟ dx + ⎜ dy ⎟ dy
⎝ dx ⎠ y ⎝ ⎠x
•The Chain rule:
Let s be an independent parameter and if change of s results in
changes in the values of x and y.
i.e., x = x (s); y = y (s); ∴ we can write, f (x, y) ≡ f (s)
df ⎛ df ⎞ dx ⎛ df ⎞ dy
= ⎜ ⎟ + ⎜ ⎟
ds ⎝ dx ⎠ y ds ⎝ dy ⎠ x ds
38. Properties of partial derivatives
• The Cyclic Rule:
If z is a function two variable, then we can write: z = z (x, y),
⎛ dy ⎞ ⎛ dz ⎞ ⎛ dx ⎞
then, ⎜ ⎜ ⎟ = −1
⎝ dz ⎟ x ⎜ dx ⎟ y
⎠ ⎝ ⎠ ⎝ dy ⎠ z
Example:
If V = f (p, T); p = f (V, T); T= f (p, V), then we can write
⎛ dp ⎞ ⎛ dT ⎞ ⎛ dV ⎞
⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = −1
⎝ dT ⎠V ⎝ dV ⎠ p ⎝ dp ⎠T
Example 1: ⎛ dp ⎞ nR ⎛ dT ⎞ p ⎛ dV ⎞ nRT
⎜ dT ⎟ = ;⎜ ⎟ = ;⎜ ⎟ =− 2
⎝ ⎠V V ⎝ dV ⎠ p nR ⎝ dp ⎠T p
Use the Equation of state for
Ideal gas to verify the cyclic ⎛ dp ⎞ ⎛ dT ⎞ ⎛ dV ⎞ ⎛ nR ⎞ ⎛ p ⎞ ⎛ nRT ⎞
∴⎜ ⎟ ⎜ ⎟ ⎜ ⎟ =⎜ ⎟⎜ ⎟ ⎜− 2 ⎟
rule ⎝ dT ⎠V ⎝ dV ⎠ p ⎝ dp ⎠T ⎝ V ⎠ ⎝ nR ⎠ ⎝ p ⎠
p = nRT/V; V= nRT/p; T= pV/nR ⎛ nRT ⎞
= −⎜ ⎟ = −1
⎝ pV ⎠
39. Properties of partial derivatives
•Two partial derivatives operating on a function:
Let f be function of two variables, i.e., f = f (x, y);
Double partial derivative of the function f can be written as:
∂ ⎛ df ⎞ ∂ 2f ∂2f ∂2f
⎜ ⎟ = =
∂x ⎝ dy ⎠ x ∂ x ∂y ∂x ∂ y ∂ y ∂x
Verification: Let the function be: f ( x , y ) = x 2 y ,
∴
∂ f
2
∂ 2f The exact differential
= 2x, = 2x,
∂y ∂x ∂y ∂x
•In general, the total differential form of f(x,y) may be written as:
df = M( x , y )dx + N ( x , y )dy
where , M = ⎛ ∂f ⎞ and ⎛ ∂f ⎞
N = ⎜
⎜ ⎟ ⎟
⎝ ∂x ⎠ y ⎝ ∂y ⎠ x
40. Properties of partial derivatives
• Does any differential form of type [M(x,y)dx + N(x,y)dy]
always represent the total differential form of a function?
Only if ⎛ ∂M ⎞ ⎛ ∂N ⎞
⎜ ⎟ = ⎜ ⎟
⎝ ∂y ⎠ x ⎝ ∂x ⎠ y
Then, df = [M(x,y)dx + N(x,y)dy] represents an exact differential.
Example:
Let us consider: 3 x 2 y 2 dx + 2 x 3 ydy where,
M( x , y ) = 3 x 2 y 2 ,
N ( x ,y ) = 2 x 3y
It can be easily verified that the above expression corresponds
to an exact differential function of the type:
3 x 2 y 2 dx + 2 x 3 ydy = d ( x 3 y 2 )
41. Properties of partial derivatives
Consequence:
Let us move from a point (xi,yi) to another point (xf,yf)
The corresponding change in f(x,y) can be written as:
f f
Δf = f ( xf , yf ) − f ( xi , yi ) = ∫ df
i
= ∫ ( Mdx
i
+ Ndy )
• The VALUE OF THE INTEGRAL OF AN EXACT DIFFERENTIAL is
INDEPENDENT OF THE PATH taken in going from initial to final
point and DEPENDS only on the values of f(x, y) at the initial and
the final points.
NOTE:
Integral of an exact differential function,df, for a closed path
(i.e., cyclic) where, i (initial state) = f (final state): ∫ df = 0
42. Test for Exactness (State Variable)
If z = f( x,y), dz will be a perfect differential
when it is found
∂z ⎞ ⎛∂z ⎞
then , dz = ⎛
⎜ ⎟ dx + ⎜ ⎟ dy
⎝ ∂ x ⎠y ⎝ ∂ y ⎠x • dz is a single-valued function
∂z ⎛ ∂z ⎞ depending entirely on the
if, ⎛ ⎞ = M ; and ⎜ ⎟ = N
⎜ ⎟ instantaneous values of x and y.
⎝ ∂x ⎠y ⎝ ∂y ⎠x
then, dz = M( x,y)dx + N( x,y)dy • dz between any two specified
points or states is independent of
the path of transition.
Then for exact differential,
⎛ ∂ M⎞ ⎛ ∂N ⎞ • dz for a complete cyclic process
⇒⎜ ⎟ = ⎜ ⎟ is equal to zero
⎝ ∂ y ⎠x ⎝ ∂ x ⎠y
⎡ ∂ ⎛∂z ⎞ ⎤ ⎡ ∂ ⎛∂z ⎞ ⎤ • δq and δw are NOT EXACT
i.e., ⎢ ⎜ ⎟ ⎥ = ⎢ ⎜ ⎟ ⎥ differential quantities therefore
⎣∂ y ⎝ ∂ x ⎠y ⎦x ⎢∂ x ⎝ ∂ y ⎠x ⎦y
⎣ ⎥ they are NOT STATE Variables.
44. The gas laws
isotherm, curves in a graph corresponding to a single
temperature.
isobar, curves in a graph corresponding to a single pressure.
isochore, curves in a graph corresponding to a single
volume.
gas constant, R (with R = NAk, where NA is Avogadro’s
constant and k is Boltzmann’s constant).
45. The gas law
perfect/ideal gas equation, pV = nRT.
perfect/ideal gas, a gas that obeys pV = nRT exactly under
all conditions.
real gas, an actual gas. e.g.,
standard ambient temperature and pressure (SATP),
298.15 K (25 °C) and 1 bar (or, 0.986 atm or 100 kPa).
standard temperature and pressure (STP)
273.15 K (0 °C) and 1 bar (100 kPa, 0.986 atm.)
Normal temperature and pressure (NTP)
273.15 K (0 °C) and 1 atm (101.325 kPa, 760 Torr)
1 atm =(101.325 kN/m2, 101.325 kPa, 14.7 psi, 29.92 in Hg, 760 torr).
46. Partial pressure , of a gas is the pressure the gas would exert if it
occupied the container alone; in general, pi = xi P
Vapour pressure, the pressure of a vapour in equilibrium with its
condensed phase
Mole fraction, the amount of i species, expressed as a fraction of moles
of the moles of ith species (ni) to the total amount of moles, n, in the
sample, xi = ni/n
47. partial pressure: of an ideal gas, the pressure a gas would
exert if it occupied the container alone; in general, pi = xi p.
Dalton’s law: The pressure exerted by a mixture of gases is
the sum of the partial pressures of the gases.
48. Thermodynamics
Thermodynamics:
→ Describes macroscopic properties of equilibrium
systems
→ Entirely Empirical
→ Built on 4 Laws and “simple” mathematics
0th Law Defines Temperature (T)
1st Law Defines Energy (U)
2nd Law Defines Entropy (S)
3rd Law Gives Numerical Value to Entropy
These laws are UNIVERSALLY VALID, they cannot be circumvented
49. Work
• Expansion work, work of expansion (or compression)
• Non-expansion / additional work: any other work
where
• pex = external pressure
• A = piston area
• dz = displacement
• dV = Adz = volume change for the gas
vf
Total work done, w = − ∫ p ex dV
vi
• Work done on the system + ve i.e., Compression
• Work done by the system - ve i.e., Expansion
50. 1st Law of Thermodynamics
F
Consider an example system of a piston and
cylinder with an enclosed dilute gas
characterized by P,V,T & n.
What if we added heat and pushed the piston
in at the same time?
• Work is done on the gas,
• Heat is added to the gas and
• the internal energy of the gas
increases:
ΔU = w + Q
Statement: Change of internal energy in a closed system of is equal to the energy
flow across its boundary as heat or work
U INCREASES with INCREASE in TEMP. of the system, when:
1. work is done on the system by the surrounding or,
2. heat flows into the system from the surrounding or,
3. both, heat and work flows from the surr. into the system
51. The First Law: Sign convention
• Work done on the system by surr + ve i.e., Compression
• Work done by the system by surr - ve i.e., Expansion
Alternate Statement: If a system is subjected to a CYCLIC
Transformation, the work produced in the surroundings is equal to
the heat withdrawn from the surroundings.
Q = -w (change in U is zero as it’s a State function)
If w is +ve (compression, work done on the system) then q will be –ve
(heat will be released by the system) and vice versa
52. Work
• Expansive work, work of expansion (or compression)
• Non-expansive / additional work : any other work
piston held down by
F= mg pins
pex= mg/A
F= pex.A
m dz
Area =A
Pi , Vi Pf , Vf
w = - mg.A.dz
w = - mg.dV
w = - pex.dV =-pex (Vf-Vi)
53. WORK
Reversible Isothermal Expansion :
w = − ∫ ( pex − dp)dV = − ∫ ( pdV − dpdV ) = − ∫ pdV
dV Vf where, p = p(V)
w = −nRT ∫ = −nRT ln p = nRT/V
V Vi
Irreversible Isothermal Expansion:
Vf
w = − pex ∫ dV = − pex (V f − Vi ) = − pex ΔV
Vi
57. Irreversible Reversible
Expansion
Irreversible
Reversible
Compression
• For a reversible process Work on expansion = work of compression
(same path)
• For an irreversible process Work on expansion ≠ work of compression
(same path)
58. Work
wrev > wirr
• Work done on expansion by the system on surroundings (sign negative)
operating between specified initial and final states and passing through
and specified path is maximum work available from a system when the
changes take place reversibly
• For a reversible process Work on expansion = work of compression
(same path)
• For an irreversible process Work on expansion ≠ work of compression
(same path)
• Work is a path dependent parameter, path function(so as Heat)
59. State and Path Functions
• Some state functions, e.g., U and H etc.
– Value of a state function is independent of Path 2 - Non-
path adiabatic
w≠0;q≠0
– Depends only on initial and final state, Path 1 -
e.g. ΔU = Ufinal - Uinitial Adiabatic
w≠0;q=0
– Overall Change
final
ΔU = ∫ dU ⇒ dU is an exact differential
initial
• Path from initial to final state has a functional
dependence
– For ∆U, work may be done, or, heat may
flow or, both
– Equations that describe how you get to final ΔUpath1 = ΔUpath2 = Uf -Ui
state are path functions
– Path functions are path specific
∂q ⇒ ∂q is
final
q=∫ an inexact differential
initial,path
60. Work is a STATE FUNCTION?
Assume a Reversible process so that pext = p where, Ar (gas)
under goes change of state through compression, from
Ar (g, p1, V1) (initial) → Ar (g, P2, V2) (final);
such that V1 > V2 and p1< p2
Compression through two paths
1st Path:
Ar(g, p1, V1) → Ar(g, p1, V2) → Ar(g, p2, V2)
First V1 → V2 when p held constant at p1;
then p1 → p2 when V held constant at V2
2nd Path:
Ar(g, p1, V1) → Ar(g, p2, V1) → Ar(g, p2, V2)
First p1 → p2 when V held constant at V1;
then V1 → V2 when p held constant at p2
Note for the closed cycle, total work = w[path (1)] - [path (2)]
61. Work is a STATE FUNCTION?
V2 V2
w( 1) = − ∫ pextdV − ∫ pextdV
V1 V2
V2
w( 1) = − ∫ p1dV = − p1(V − V ) = p1(V − V )
2 1 1 2
V1
V1 V2
w( 2 ) = − ∫ pextdV − ∫ pextdV
V1 V1
V2
w( 2 ) = − ∫ p2dV = − p2 (V − V ) = p2 (V − V )
2 1 1 2
V1
NOTE:
• Work done on the system to compress it is NOT EQUAL through
the two paths. i.e., w(1) ≠ w (2)
• ∴ Net work done on the system in the complete cycle ≠ 0
i.e., [w(1) - w(2)] ≠ 0; ∫ ∂w ≠ 0 (closed cycle)
• This means w is NOT a state function ∴ we CAN NOT write
w = f(V, p)
62. Ideal gas, closed system, reversible process
Calculate wf and wb
isothermal
10 Pa, 1 m3 , T 1 Pa, 10 m3, T
isochoric
10 Pa, 1 m3, T 1 Pa, 1 m3, T2
1 Pa, 10 m3, T
isobaric
isobaric
10 Pa, 1 m3, T 10 Pa, 10 m3, T3
1 Pa, 10 m3, T
isochoric
63. 25 L of gas is enclosed in a cylinder/piston apparatus at
2 atm of pressure and 300 K. A 100 kg of mass is
placed on the piston causing the gas to compress to 20 L
at constant pressure. This is done by allowing heat to
flow out of the gas. What is the work done on the gas?
What is the change in internal energy of the gas? How
much heat flowed out of the gas?
• Po = 202,600 Pa, Vo = 0.025 m3,
• To = 300 K, Pf = 202,600 Pa,
• Vf=0.020 m3, Tf= ?
• n = PV/RT W = -P∆V ∆ U = 3/2 nR ∆T Q = W + ∆ U
W = - P ∆ V = -202,600 Pa (0.020 – 0.025)m3
= 1013 J energy added to the gas
∆ U =3/2 nR ∆T=1.5(2.03)(8.31)(-60)= -1518 J
Q = W + ∆ U = 1013 – 1518 = -505 J heat out
64. Heat Transactions
In general change in the internal dU = dq + dw exp + dw add
energy of a closed system is given by
- For dV = 0; dwexp = 0 and if dwadd = 0, then
dU = dq (constant volume, no additional work) = dqv
For a measurable change, ΔU = qv
-heat capacity, C, the ratio of heat supplied to
the temperature rise it causes. C = dq/dT;
C = q/ ΔT
- heat capacity at constant volume, CV = (∂U/∂T)V
- molar heat capacity, the heat capacity divided
by the amount of substance, Cm = C /n. adiabatic bomb
calorimeter
- specific heat capacity, the heat capacity
divided by the mass, Cs = C /m
65. Heat Capacity for Constant Volume Processes (Cv)
insulation ΔT
Heat, Q
m added m
• Heat is added to a substance of mass m in a fixed volume
enclosure, which causes a change in internal energy, U. Thus,
Q = U2 - U1 = ∆U = m Cv ∆T
The v subscript implies constant volume
66. Heat Capacity for Constant Pressure Processes (Cp)
Δx
ΔT
Heat, Q
m added m
• Heat is added to a substance of mass m held at a fixed
pressure, which causes a change in internal energy, U,
AND some PV work.
67. Heat Transactions and Enthalpy
For dV ≠ 0; dwexp≠ 0 even if dwadd = 0, then dU ≠ dq
ΔU = qp - pΔV qp = ΔU +pΔV = U2 – U1 + p(V2 – V1)
= (U2 + pV2) – (U1 + pV1)
Define H = U + pV
•Differential scanning calorimeter
dH = dU + pdV + Vdp •Isobaric calorimeter
= dU - w (at constant p) •Adiabatic flame calorimeter
= dqp
At constant p, dH = dqp (constant pressure, no additional work)
- Heat capacity at constant pressure, Cp = (∂H/∂T)p
- Enthalpy change and temperature change, dH = nCp,m dT;
ΔH = nCp,m ΔT (if Cp independent of T)
Empirical expression,
Cp,m = a + bT + c/T2
68. Heat Transactions
- for ideal/perfect gas; relation between ∆U and ∆H,
∆H = ∆U + ∆ngRT.
- relation between heat capacities of a perfect gas,
Cp – CV = nR
69. For any process
U = f (V , T ) H = f ( P, T )
⎛ ∂U ⎞
dU = ⎜
⎛ ∂U ⎞
⎟ dV + ⎜
⎛ ∂H ⎞ ⎛ ∂H ⎞
⎟ dT dH = ⎜ ⎟ dP + ⎜ ⎟ dT
⎝ ∂V ⎠T ⎝ ∂T ⎠V ⎝ ∂P ⎠ T ⎝ ∂T ⎠ P
= π T dV + CV dT ⎛ ∂H ⎞
=⎜ ⎟ dP + C P dT
⎛ ∂U ⎞ ⎝ ∂P ⎠T
πT = ⎜ ⎟ Internal pressure
⎝ ∂V ⎠ T
dU = CV dT (for ideal gas) dH = Cp dT (for ideal gas)
For ideal gas ⎛ ∂U ⎞ For ideal gas
πT = ⎜ ⎟ =0 ⎛ ∂H ⎞
⎝ ∂V ⎠T ⎜ ⎟ =0
⎝ ∂P ⎠T
Joule Experiment
70. Adiabatic Reversible Process for ideal gas,
Cv independent of Temp
1
dU = dw + dq = dw
CV nR
⎛ T2 ⎞ ⎛ V1 ⎞ T2 ⎛ V1 ⎞ C
⎜ ⎟ =⎜ ⎟
c
⎜T ⎟ ⎜V ⎟ = ⎜ ⎟ where c = V ,m
Cv dT = − pdV ⎝ 1⎠ ⎝ 2⎠ T1 ⎜ V2 ⎟
⎝ ⎠ R
CV nR C p −CV
T2 dT V2 dV ⎛ T2 ⎞ ⎛V ⎞ ⎛V ⎞
Cv ∫ = - nR ∫ ⎜ ⎟
⎜T ⎟ =⎜ 1 ⎟
⎜V ⎟ =⎜ 1 ⎟
⎜V ⎟
T1 T V1 V ⎝ 1⎠ ⎝ 2⎠ ⎝ 2⎠
γ −1
T2 V1 T2 ⎛ V1 ⎞ C p ,m
Cv ln = nR ln =⎜ ⎟ where γ =
T1 V2 T1 ⎜ V2 ⎟
⎝ ⎠ CV ,m
TV γ −1 = constant
PV γ = constant
VT c = constant
T γ P 1−γ = constant
71. Conclusion of Ist Law of Thermodynamics
• The internal energy of an isolated system is a constant
For isolated system : dU = 0; (as q = 0, w = 0» No perpetual
motion machines! )
• The energy of the universe is constant. ∆Usystem = −∆Usurroundings
• When a system undergoes a process that leads to a change of
state then the sum of the heat q transferred and the work w
performed, is equal to the change in the internal energy dU of
the system.
dU = ∂q + ∂w (Closed system, through a number of steps)
• A system contains ONLY internal energy. A system does not
contain heat or work.
72. Conclusion of Ist Law of Thermodynamics
• Cyclic Process ∫ dU =0 (Exact differential)
• In a cyclic process the work done by a system on its
surroundings is equal to the heat withdrawn from the
surroundings .
- δ w = δq
• For an Adiabatic Process δ wad = dU ; (as δq = 0)
• For a Non-Adiabatic Process
δq = dU − δw ≡ δq = (δwad – δw); (as, δwad = dU)
73. Calculation of thermodynamic functions
for various processes
reversible phase const p heating, no const V heating, no
change at const. T, p phase change phase change
w − ∫ p dV = − pΔV − ∫ p dV = − pΔV 0
q latent heat q p = ΔH qV = ΔU
ΔU q + w q + w = ∫ CV (T )dT
ΔH q = ∫ C p (T )dT ΔU + VΔp
74. Calculation of thermodynamic function for
various processes for Ideal Gas
Reversible, no phase change rev, isothermal rev, adiabatic
rev rev, const irrev, const
ext. p final p
w − nRT ln
V2 = ∫ CV (T )dT
− ∫ PdV − pex ΔV − p f ΔV V1
q ΔU-w nRT ln
V2
0
V1
ΔU
= ∫ CV (T )dT 0 = ∫ CV (T )dT
ΔH
= ∫ C p (T )dT 0 = ∫ C p (T )dT
75.
76.
77.
78.
79. SUMMRAY OF ADIABATIC WORK (with solved example)
• Reversible Adiabatic Expansion (or compression) of an Ideal Gas
1 mole gas (V1,T1) = 1 mole gas (V2,T2)
adiabatic ⇒ đq = 0 Reversible ⇒ đw = -pd V
Ideal gas ⇒ dU = Cv d T
∴ From 1st Law dU = -pd V ⇒ CvdT = -pdV along path
dT dV
CV dT = − pdV ⇒ CV = −R
p =RT T V
V
Cp
R CV −1
T2 dT V dV ⎛T2 ⎞ ⎛V1 ⎞ C p −CV =R for i.g. ⎛T2 ⎞ ⎛V1 ⎞CV
CV ∫ = −R ∫
2
⇒ ⎜ ⎟=⎜ ⎟ ⎯⎯⎯⎯⎯⎯→ ⎜ ⎟=⎜ ⎟
T1 T V V
1
⎝T1 ⎠ ⎝V2 ⎠ ⎝T1 ⎠ ⎝V2 ⎠
γ −1
Cp ⎛T2 ⎞ ⎛V1 ⎞
Define γ ≡ ⇒ ⎜ ⎟ =
⎜ ⎟
CV ⎝T1 ⎠ ⎝V2 ⎠
3 ⎫
CV = R ⎪
2 ⎪ 5
For monatomic ideal gas: ⎬ γ = ( > 1 generally)
5 ⎪ 3
Cp = R
2 ⎪⎭
In an adiabatic expansion (V2 > V1), the gas cools (T2 > T1).
And in an adiabatic compression (V2 < V1), the gas heats up.
γ
pV ⎛ p2 ⎞ ⎛V1 ⎞
For an ideal gas (one mole) T = ⇒ ⎜ ⎟=⎜ ⎟ ⇒ pV1 γ = pV2γ
R ⎝ p1 ⎠ ⎝V2 ⎠
1 2
pV γ is constant along a reversible adiabat
For an isothermal process T = constant ⇒ pV = constant
80. p
p1 V2adiabat < V2isotherm
because the gas
cools during reversible
p2 adiabatic expansion
V1 V2ad V2iso
• Irreversible Adiabatic Expansion of an ideal gas against a constant
external pressure
1 mol gas (p1,T1) = 1 mol gas (p2,T2) (pext=p2)
adiabatic ⇒ đq = 0
Constant pext = p2 ⇒ đw = -p2dV
Ideal gas ⇒ dU = CvdT
1st Law ⇒ dU = -p2dV
∴ Cv d T = - p2 dV
Integrating: Cv ( T2 - T 1 ) = - p2 ( V2 - V1 )
⎛ p2 ⎞
Using pV = RT T2 (CV + R ) =T1 ⎜ CV + R⎟
⎝ p1 ⎠
Note p2 < p1 ⇒ T2 < T1 Again, expansion cools
Note also (-wrev) > (-wirrev) Less work is recovered through
an irreversible process
81. Some Thermodynamic Cycles
đq
• Reversible Ideal Gas processes: Find ∆U , ∆H , q , w , ∫T
(I
) (II)
p p
(T )
1 (T )
1 D (const. p) (T )
3
p1 p1
A (isotherm)
B (T ) A E (const. V)
p2 1 p2
p3 (rev. C (const. V) (isotherm) (T )
1
adiabat) (T )
2
V1 V2
V1 V2
const. T
[A] 1 mol gas (p1,V1,T1) = 1 mol gas (p2,V2,T1)
Ideal gas isotherm: ∆UA = 0 ∆HA = 0 as, ΔH = Cp ΔT = 0
V V đ qA V2
wA = −RT1 ln 2 qA = RT1 ln 2 ∫T = R ln
V1 V1 V1
rev.adiabat
[B] 1 mol gas (p1,V1,T1) = 1 mol gas (p3,V 2,T2)
Adiabat: qB = 0
∆UB = CV (T2 −T1 )
Ideal gas:
∆HB = C p (T2 −T1 )
1st Law: w B = CV (T2 −T1 )
đqB
∫T =0
82. reversible
[C] 1 mol gas (p3,V2,T2) = 1 mol gas (p2,V2,T1)
const. V
∆UC = CV (T1 −T2 )
Constant V: wC = 0 Ideal gas:
∆HC = C p (T1 −T2 )
1st Law: qC = CV (T1 −T2 )
đqC ⎛T ⎞
∫T = CV ln ⎜ 1 ⎟
⎝T2 ⎠
[A] vs. [B] + [C]
∆UA = 0 ∆UB + ∆UC = 0 = ∆UA
∆HA = 0 ∆HB + ∆HC = 0 = ∆HA
V
qA = RT1 ln 2 qB + qC = CV (T1 −T2 ) ≠ qA
V1
V
w A = −RT1 ln 2 wB + w C = CV (T2 −T1 ) ≠ w A
V1
đqA V2 đqB đqC ⎛V ⎞ đq
∫T = R ln ∫T +∫ = R ln ⎜ 2 ⎟ = ∫ A
V1 T ⎝V1 ⎠ T
⎛ đq ⎞
This result suggests that ⎜ ∫ rev ⎟ is a state function!
⎝ T ⎠
∆UD = CV (T3 −T1 )
[D] qD = C p (T 3 − 1
T )
∆H D = C p(T 3−T 1)
đqD ⎛T ⎞
wD = −R (T3 −T1 ) ∫T = C p ln ⎜ 3 ⎟
⎝T1 ⎠
83. ∆UE = CV (T1 −T3 )
[E] wE = 0 qE = CV (T1 −T3 )
∆HE = C p (T1 −T3 )
đqE ⎛T ⎞
∫T = CV ln ⎜ 1 ⎟
⎝T3 ⎠
[A] vs. [D] + [E]
∆UA = 0 ∆UD + ∆UE = ∆UA
∆HA = 0 ∆HD + ∆HE = ∆HA
V
qA = RT1 ln 2 qD + qE = R (T3 −T1 ) ≠ qA
V1
V
w A = −RT1 ln 2 wD + w E = −R (T3 −T1 ) ≠ w A
V1
đqA V2 đqD đqE ⎛V ⎞ đq
∫T = R ln ∫T +∫ = R ln ⎜ 2 ⎟ = ∫ A
V1 T ⎝V1 ⎠ T
⎛ đq ⎞
Here again ⎜ ∫ rev ⎟ looks like a state function!
⎝ T ⎠
84. Conclusion of Ist Law of Thermodynamics
• The internal energy of an isolated system is a constant
For isolated system : dU = 0; (as q = 0, w = 0» No perpetual
motion machines! )
• The energy of the universe is constant. ∆Usystem = −∆Usurroundings
• When a system undergoes a process that leads to a change of
state then the sum of the heat q transferred and the work w
performed, is equal to the change in the internal energy dU of
the system.
dU = ∂q + ∂w (Closed system, through a number of steps)
• A system contains ONLY internal energy. A system does not
contain heat or work.
85. Conclusion of Ist Law of Thermodynamics
• Cyclic Process ∫ dU =0 (Exact differential)
• In a cyclic process the work done by a system on its
surroundings is equal to the heat withdrawn from the
surroundings .
- δ w = δq
• For an Adiabatic Process δ wad = dU ; (as δq = 0)
• For a Non-Adiabatic Process
δq = dU − δw ≡ δq = (δwad – δw); (as, δwad = dU)
86.
87. Calculation of thermodynamic functions
for various processes
reversible phase const p heating, no const V heating, no
change at const. T, p phase change phase change
w − ∫ p dV = − pΔV − ∫ p dV = − pΔV 0
q latent heat q p = ΔH qV = ΔU
ΔU q + w q + w = ∫ CV (T )dT
ΔH q = ∫ C p (T )dT ΔU + VΔp
88. Calculation of thermodynamic function for
various processes for Ideal Gas
Reversible, no phase change rev, isothermal rev, adiabatic
rev rev, const irrev, const
ext. p final p
w − nRT ln
V2 = ∫ CV (T )dT
− ∫ PdV − pex ΔV − p f ΔV V1
q ΔU-w nRT ln
V2
0
V1
ΔU
= ∫ CV (T )dT 0 = ∫ CV (T )dT
ΔH
= ∫ C p (T )dT 0 = ∫ C p (T )dT
89. Variation of U with system variables:
U = f (V , T ) ⎛ ∂U ⎞
dU = ⎜
⎛ ∂U ⎞
⎟ dV + ⎜ ⎟ dT
⎝ ∂V ⎠T ⎝ ∂ T ⎠V
1. Change in U w.r.t. T at constant volume
⎛ ∂U ⎞
Cv = ⎜ ⎟ ; qv = Cv ΔT = ΔU
⎝ ∂T ⎠ V
⎛ ∂U ⎞
2. Change in U w.r.t. V at constant temp: ⎜ ⎟
⎝ ∂V ⎠ T
= π T dV + C V dT For real gas
dU = CV dT (for ideal gas) As For ideal gas π T = ⎜
⎛ ∂U ⎞
⎟ =0
⎝ ∂V ⎠T
3. Change in U w.r.t. T at constant pressure
⎛ ∂U ⎞
dU = Cv dT + ⎜ ⎟ dV ⎛ ∂U ⎞ ⎛ ∂U ⎞
⎝ ∂V ⎠ T ⎜ ⎟ = Cv + αV ⎜ ⎟
⎝ ∂T ⎠ P ⎝ ∂V ⎠ T
⎛ ∂U ⎞ ⎛ ∂U ⎞ ⎛ ∂ V ⎞
⎜ ⎟ = Cv + ⎜ ⎟ ⎜ ⎟
⎝ ∂T ⎠ P ⎝ ∂V ⎠ T ⎝ ∂T ⎠ P where
90. Change in Internal Energy at Constant P
• Suppose we want to know how the internal energy, U, changes with
temperature at constant pressure dU = π dV + C dT
T V
⎛ ∂U ⎞ ⎛ ∂ (π T dV + C V dT)⎞
⎜ ⎟ =⎜ ⎟
⎝ ∂T ⎠ p ⎝ ∂T ⎠p
⎛ ∂U ⎞ ⎛ ∂V ⎞
⎜ ⎟ = π T ⎜ ⎟ + CV
⎝ ∂T ⎠ p ⎝ ∂T ⎠ p
– But the change in volume with temperature at constant
pressure is related to the Isobaric thermal expansion /
expansion coefficient of the gas, α
1 ⎛ ∂V ⎞ ⎛ ∂V ⎞
α= ⎜ ⎟ or ⎜ ⎟ = αV
V ⎝ ∂T ⎠ p ⎝ ∂T ⎠ p
⎛ ∂U ⎞
so ⎜ ⎟ = π Tα V + C V
⎝ ∂T ⎠ p
• Large α means sample responds strongly to changes in T
• For an ideal gas, πT = 0 so
⎛ ∂U ⎞
⎜ ⎟ = CV [definition ]
⎝ ∂T ⎠ p
91. Variation of V with system variables: Experimentally known that the
Volume of an isotropic material is a function of Temperature and Pressure.
isothermal compressibility
relative/fractional coefficient of Isobaric
change of Volume thermal expansion /
expansion coefficient
For an ideal gas : PV = nRT
similarly κT = 1/p
For solid and liquid α and κT are approx.
independent of T & p
92. Another quantity of interest which can be obtained from α and κT is:
This derivative tells us how fast the pressure rises when we try
to keep the volume constant while increasing the temp.
Exercise:
How much pressure would be generated in a mercury thermometer if we tried to
heat the thermometer higher than the temperature where the mercury has
reached the top of the thermometer. Given: For Hg, α = 1.82 Ø 10×4 K×1 and κT
= 3.87 Ø 10×5 atm×1
We see that each 1 C increases the pressure by
4.7 atm, about 69 lb/sq in. This is a lot of
pressure for a glass tube to withstand. It
wouldn't take very many degrees of temperature
increase to break the glass thermometer.!!!
Using a variation of Euler's
chain rule we can write
93. Variation of H with system variables: H = f ( P, T ) dH = ⎛ ∂H ⎞ dP+⎛ ∂H ⎞ dT
⎜ ⎟ ⎜ ⎟
⎝ ∂P ⎠T ⎝ ∂T ⎠ P
1. Change in H w.r.t. T at constant pressure
⎛ ∂H ⎞
C p = ⎜ ⎟ ; q p = C p ΔT = ΔH
⎝ ∂T ⎠ P
⎛ ∂H ⎞
2. Change in H w.r.t. P at constant temperature: ⎜ ⎟⎝ ∂P ⎠ T
H = f ( P, T )
⎛ ∂H ⎞
Alternately, = ⎜ ⎟ dP + CP dT
⎛ ∂H ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎝ ∂P ⎠T
⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = −1
⎝ ∂P ⎠ T ⎝ ∂T ⎠ H ⎝ ∂H ⎠ P
dH = Cp dT (for ideal gas)
⎛ ∂H ⎞ ⎛ ∂T ⎞ ⎛ ∂H ⎞
∴ ⎜ ⎟ = −⎜ ⎟ ⎜ ⎟ = − μ JT C p ∂H ⎞
⎝ ∂P ⎠ T ⎝ ∂P ⎠ H ⎝ ∂T ⎠ P For ideal gas ⎛
⎜ ⎟ = 0
⎝ ∂P ⎠T
3. Change in H w.r.t. T at constant volume: ⎛ ∂H ⎞
⎜ ⎟
⎛ ∂H ⎞ ⎝ ∂T ⎠ V
dH = C P dT + ⎜ ⎟ dP
⎝ ∂P ⎠ T
⎛ ∂H ⎞ ⎛ ∂P⎞ α
⎛ ∂H ⎞ ⎛ ∂H ⎞ ⎛ ∂ P ⎞
From earlier results ⎜ ⎟ = −μ JT Cp ; ⎜ ⎟ =
⎜ ⎟ = CP + ⎜ ⎟ ⎜ ⎟ ⎝ ∂P ⎠ T ⎝ ∂T ⎠ V κ
⎝ ∂T ⎠ V ⎝ ∂P ⎠ T ⎝ ∂T ⎠ V
⎛ ∂H ⎞ ⎡ αμ ⎤
∴⎜ ⎟ = CP ⎢1 − JT ⎥
⎝ ∂T ⎠ V ⎣ κ ⎦
94. The Joule-Thomson Expansion Expt
The Joule-Thomson experiment consists of forcing a gas at constant
backing pressure passing through a porous plug under adiabatic condition
and measuring the change in temperature.
⎛ ∂T ⎞ ≈ ΔT
The Joule-Thomson coefficient (μJT ) = ⎜ ⎟
⎝ ∂P ⎠ H Δp
The most straightforward way to measure the coefficient is to measure
the temperature of the gas entering and leaving the porous plug and
divide the change in temperature by the
change in pressure:
we need to decide what is being held constant when we take the partial
derivative. In this case, since q =0, the change in internal energy of the
gas is equal to w.
95. Joule-Thompson Experiment: Inversion Temperature
In this experiment q = 0
w = - pf (Vf – 0) - pi (0 - Vi)
ΔU = Uf – Ui = q + w = - pfVf + piVi
Uf + pfVf = Ui + piVi
Hf = Hi , ΔH = 0 Isenthalpic process
⎛ ∂H ⎞ ⎛ ∂H ⎞
H = f ( P, T ); dH = ⎜ ⎟ dP + ⎜ ⎟ dT
⎝ ∂P ⎠T ⎝ ∂T ⎠ P
⎛ ∂H ⎞ ⎛ ∂H ⎞ ⎛ ∂T ⎞
⎜
⎜ ∂p ⎟ ⎟ = −⎜ ⎟ ⎜ ⎟ = −C p × μ
⎜ ⎟
Applying Euler’s Chain Rule, we get: ⎝ ⎠T ⎝ ∂T ⎠ p ⎝ ∂p ⎠ H
Joule-Thompson Coefficient , μ (or μJT ) = ⎛ ∂T ⎞ ≈ ΔT in this J-T expt.
⎜ ⎟
⎝ ∂P ⎠ H Δp
⎛ ∂H ⎞ ⎛ ∂H ⎞
H = f ( P, T ); dH = ⎜ ⎟ dP + ⎜ ⎟ dT dH = − μC p dP + C p dT
⎝ ∂P ⎠T ⎝ ∂T ⎠ P
96. THE JOULE-THOMPSON EXPERIMENT
⎛ ∂T ⎞
μ =⎜ ⎟
⎜ ∂p ⎟
⎝ ⎠H
If μ > 0 then this indicates that
the gas cools (-ve δT) when gas
expanded (-ve δp).
If μ < 0 then this indicates that
the gas heats (+ve δT) when
expanded (-ve δp).
The "inversion temperature"
indicates where on a (T, p)
isenthalpic plot the (dT/dP)H flips
from +ve to –ve. Or,point where µ
changes sign from + to –
ve and the point is the maximum
inversion temperature
For a real gas µ is non-zero
(except at the maximum inversion
temperature)
97. 2
⎛ ∂T ⎞
2
μ=⎜
⎜ ⎟
⎝ ∂p ⎟ H
⎠
The Linde process
98. Joule-Thompson ⎛ ∂T ⎞ ≈ ΔT
Coefficient , μ (or μJT ) = ⎜ ∂P ⎟
⎝ ⎠ H Δp
• µ can be either (+) or (-)
– Positive µ means dT is negative when
dp is negative
•Gas cools on expansion
– Negative µ means that means dT is
positive when dp is negative
•Gas warms on expansion
– Transition between positive and
negative µ is called the Joule-
Thompson inversion temperature
•Gases have both high and low
inversion temperatures
– Ideal gas µ = 0, T unchanged by Tinversion µ
change in p on expansion Gas (K) (K atm-1)
CO2 1500 1.11
• Practical applications: N2 621 0.25
– Gas liquefaction He 621 0.25
– Isotope separation
99. ⎛ ∂T ⎞ T ( ∂V ∂ T )P −V
μJT =⎜ ⎟ =
⎝ ∂ P ⎠H CP
100. ⎛ ∂T ⎞ T ( ∂V ∂ T )P −V
μJT =⎜ ⎟ =
⎝ ∂ P ⎠H CP
101. For a van der Waal gas:
Therefore, μJT depends critically on
⎛ ∂T ⎞ T ( ∂V ∂ T )P − V
μJT =⎜ ⎟ = the constants a and b, which are the
⎝ ∂ P ⎠H CP
indices of non zero interaction and
μJT =
1 ⎛ 2a ⎞
− b⎟ finite volume of the gas molecules
⎜
CP ⎝ RT ⎠
2a
If , > b then μ > 0 (cooling observed )
RT • If initial Temp (Ti) is below
2a Tinversion only then cooling is
If , < b then μ < 0 ( heating observed )
RT observed.
Inversion temperature • And the final temp T2 is
μ JT = 0 when T = Tinversion lower than T1
1 ⎡ 2a ⎤ ⎛ 2a ⎞
∴ 0= ⎢ − b⎥ or , ⎜
⎜ RT ⎟=b
⎟
Cp ⎣ RT ⎦ ⎝ inversion ⎠
2a
∴ Tinversion =
Rb
102. N2 and O2 will cool upon expansion at
room temperature, but He and Ne will
warm upon expansion at room
temperature.
103. Derive the Following expressions
For ideal gas
⎛ ∂P ⎞ α 1 ⎛ ∂H ⎞
⎜ ⎟ = ∴μ JT =− ⎜ ⎟
⎝ ∂T ⎠V κ C p ⎝ ∂P ⎠ T
⎛ ∂T ⎞ T ( ∂V ∂ T )P −V
α TV
2
μJT =⎜ ⎟ =
C p − CV = ⎝ ∂ P ⎠H CP
κ for ideal gases : μJT = 0
1 ⎛ 2a ⎞
μJT = − b⎟
C P ⎜ RT
⎝ ⎠
⎛ ∂H ⎞
Isothermal Joule-Thompson Coefficient, μT = ⎜ ⎟ μT = - Cpμ
⎝ ∂P ⎠T
Monoatomic ideal gas CV,m = 3R/2, Cp,m = 5R/2
104. The Perfect gas and πT, α, κT, μ
Property PD Value for Perfect Gas Info:
Internal ⎛ ∂U ⎞
Pressure πT = ⎜ ⎟
∂V ⎠T
πT=0 Strength/nature of
⎝ interactions between
molecules
Expansion 1 ⎛ ∂V ⎞ α =1 / T
The higher T, the less
Coefficient α= ⎜ ⎟ responsive is its volume
V ⎝ ∂T ⎠ p to a change in
temperature
1 ⎛ ∂V ⎞
Isothermal κT = −
V⎝
⎜
⎜ ∂p ⎟
⎟ κT=1 / p The higher the p, the
Compressibility ⎠T lower its compressibility
⎛ ∂T ⎞
Joule-Thomson
μ =⎜
⎜ ⎟
⎟ μ=0 Another indication of
Coefficient ⎝ ∂p ⎠ H molecular interactions.
105. Ist Law of Thermodynamics:
• The different forms of energy are inter convertible.
• When one form of energy disappears an equivalent of another kind
must appear. Conversion of energy.
• Energy can be transferred / transformed, keeping the total energy
fixed.
qsys = - qsurr ΔU = q + w (= – pexdV)
wsys = - wsurr ∆U = q – pexdV
Incompleteness of Ist Law: It Does not say anything about:
• Whether any transfer / transformation of energy would occur at all?
• If it does, then in which direction would the change occur?
• If any transfer / transformation of energy occurs in a particular
direction, then how long will it be sustained?
• If any transfer / transformation of energy occurs in a particular
direction, then how fast will it proceed?
106. What happens when ? Free Expansion of Gas
Gas Gas
Valve Gas
Vacuum Valve
Closed
Open
Gas 1 Gas 2
Gases tend to occupy the whole available
Ideal gas mixture ΔH=0 volume of the container.
Two questions:
1) Does the enthalpy of the system increase or decrease and how
much ?
2) Does the system become more or less disordered and by how
much?
The driving force of such a process is the increase of the disorder
in the system.
107. Warm
Thermal Conduction
The two liquids (with very similar molecules) form
an ideal solution (ΔH ≈ 0), in which the molecules
are randomly and uniformly mixed.
Hot dQ Cold
Suppose two bodies are brought in contact. Then
from the Ist law, we ONLY know if q heat was lost
by one, exactly q heat would be gained by the other.
The law does not specify if the Ist one will lose heat
or, the other one. To know the direction of flow of
heat, we need another information, namely the
temp of the two bodies.
108. First Law of Thermodynamics
• Conservation of Energy
• Says Nothing About Direction of Energy Transfer
Second Law of Thermodynamics
• Preferred (or Natural) Direction of Energy Transfer
• Determines Whether a Process Can Occur
• Thermodynamic Processes can be classified into Three Types :
– Natural (or Irreversible)
– Impossible
– Reversible
109. Disorder increase for the following processes:
• Solids melt to liquids
• Solids or liquids vaporize to form gases
• Solids or liquids dissolve in a solvent to form non-electrolyte
solution
• A substance is heated (increased temperature increases the
molecular motions and the disorder)
• A chemical reaction produces an increase in the number of
molecules of gases
• The 2 Law puts forward the criteria for the direction of the
nd
spontaneous chemical/ physical change of a thermodynamic system.
• The historical roots of the second law stems from the invention of
heat engine (namely, the steam engine). It is known that work can
be fully converted to heat (joule’s expt), the question is can heat
be converted to work???
110. Entropy
• The thermodynamic property of a system that is related to its
degree of randomness or disorder is called entropy (S).
• The entropy S and the entropy change ΔS=S2-S1 are state
functions
• The entropy S has a unique value, once the pressure P, the
temperature T and the composition n of the system are specified,
S = S (p,T,n)
• The entropy is an extensive property, i.e. increases with the
amount of matter in the system.
111. Carnot Engine: Introduction to Heat Engines
• One of the primary applications of
thermodynamics is to Turn heat into work
• The standard heat engine works on a
reversible cyclic process:
1. extract heat from a hot reservoir,
2. perform work,
3. dump excess heat into a cold
reservoir (often the environment).
A “reservoir” is a large body whose
temperature doesn’t change when it absorbs
or gives up heat
Carnot and then by Clausius: Statement of the Second law :
It states that it is impossible to construct a device that operates in
a cycle and produces no effect other than the transfer of heat from
a lower-temperature body to a higher-temperature body.
112. TH
TH
QH QH
R WOUT
HE WOUT
QL QL
TL
TL
−W
Thermal efficiency = ε =
QH
For a complete cycle: dU = 0
∴ Qcy = - Wcy
Qcy = QH - Qc
−W QH − Qc
∴ε cyc =
QH
=
QH
Q
= 1− c
QH
The engine will have 100% efficiency only when Qc = 0
113. Heat Engines
• By conservation of energy:
|QH| = |W| + |QL|.
• The high and low temperatures TH and TL are called
the operating temperatures of the engine.
• We will considering only engines that run in a
repeating cycle, that is, the system returns
repeatedly to its starting point, and thus can run
continuously.
• Absolute value signs are used because we are
interested only in the magnitudes.
114. Second Law of Thermodynamics.
The Carnot Cycle.
• The Carnot cycle is an
example of a reversible
process, which is a process
that can be done in
reverse.
• A reversible process
requires that any changes
are made infinitely slowly.
• Real processes are not
reversible due to friction ,
turbulence in the gas, etc.
116. • Four steps
• Isothermal expansion: a to b
• Adiabatic expansion: b to c
• Isothermal compression: c to d
• Adiabatic compression: d to a
Define the efficiency ε
work done by system " work"
ε ≡ =
heat extracted from reservoir " cost"
Valid for all heat engines.
W QC
efficiency = ε = = 1− (∆U = 0 for closed cycle)
QH QH
where, W = │QH│ - │Qc│; following conservation of energy
117. The Carnot Cycle
• Step 1: a to b.
– The gas is in contact with a
heat bath at temperature TH
and weight is removed from
the piston.
– The gas expands, while
maintaining a constant
Vb
temperature, isothermal w1 = − nRTH ln
Va
– the change in the internal
energy is thus equal to zero dU = 0
w1 = − q1
118. The Carnot Cycle
• Step 2: b to c.
– The gas is isolated from
the environment and some
more weight is removed
from the piston.
– The gas expands and during
the adiabatic expansion, the
temperature of the gas will
decrease.
γ −1 γ −1
TH Vb = TLVc
– For adiabatic expansion pVγ
is constant, and we can w2 = nCV (TL − TH )
thus related state b to
state c:
q2 = 0
119. The Carnot Cycle
• Step 3: c to d.
– The gas is in contact with
a heat bath at
temperature TC and weight
is added to the piston.
– The gas is compressed,
while maintaining a constant Vd
temperature, isothermal w3 = − nRTL ln
Vc
– the change in the internal
energy is thus equal to dU = 0
zero
w3 = − q3
120. The Carnot Cycle
• Step 4: d to a.
– The gas is isolated from the
environment and some more
weight is added to the piston.
– The gas is compressed and
during the adiabatic
compression, the temperature
of the gas will increase.
γ −1 γ −1
TH Va = TLVd
w4 = nCV (TH − TC )
q4 = 0
121. Vb
• Four steps w1 = − nRTH ln
• Isothermal expansion: a to b Va
dU = 0
w1 = − q1
γ −1 γ −1
• Adiabatic expansion: b to c TH Vb = TLVc
w2 = nCV (TL − TH )
q2 = 0
Vd
• Isothermal compression: c to d w3 = −nRTL ln
Vc
dU = 0
w3 = −q3
• Adiabatic compression: d to a γ −1 γ −1
TH Va = TLVd
w4 = nCV (TH − TC )
q4 = 0
122. wcycle = w1 + w2 + w3 + w4
Vb V
wcycle = − nRTH ln + nC v (TL − TH ) − nRTL ln d + nC v (TH − TL )
Va Vc
Vb V
wcycle = − nRTH ln − nRTL ln d
Va Vc
∫ dU = 0
Vb Vd TH Vb γ −1 = TLVc γ −1
qcycle = nRTH ln + nRTH ln
Va Vc TLVd γ −1 = TH Va γ −1
γ −1 γ −1
TH Vc Vd Vc Vd
wcycle
V V
= nRTH ln a − nRTL ln d = γ −1 = γ −1 ∴ =
TL Vb Va Vb Va
Vb Vc
Va
wcycle = nR(TH − TL ) ln
Vb
123. W QC
efficiency = ε = = 1−
QH QH
But, QH = q1 = -w1 And, Qc = q3 = -w3
Vb Vd TH Vb γ −1 = TLVc γ −1
w1 = −nRTH ln w3 = −nRTL ln
Va Vc T V γ −1 = T V γ −1
L d H a
dU = 0 dU = 0 TH Vc
γ −1
Vd
γ −1
Vc Vd
= = ∴ =
w1 = −q1 w3 = −q3 TL Vb
γ −1
Va
γ −1
Vb Va
ε = 1 – │Qc│/ │QH│ = 1 – TLln (Vd/Vc)/ THln (Vb/Va)
1 – │Qc│/ │QH│ = 1 – TL/ TH
│Qc│/ │QH│ = TL/ TH │Qc│/ │ TL │ = │QH│ / TH
│Qc│/ │ TL │ – │QH│ / TH = 0 (here, Qc ≡ QL)
qi ∂q
∑ =0 ⇒∫ =0 Thus, the term δq/T is a state function
Ti T
124. Carnot’s theorem tells us that no real engine can have an efficiency more
than that of the Carnot engine.
TH − TL T
η= = 1− L
TH TH
qH − qL
η=
qH
qH − qL TH − TL
∴ =
qH TH
qL TL qL TL
1− = 1− ⇒− + =0
qH TH qH TH
Multiply by qH TL
qL qH qH qL q
− + = 0; = ⇒ = cons tan t
TL TH TH TL T
125. When the Carnot cycle is operating as a heat engine (going
around clockwise), w < 0, so that -w > 0 and q > 0.
Va
wcycle = nR(TH − TL ) ln
Vb
− wcycle wcycle
Efficiency, ε or η = =
q1 q1
Va
− nR(TH − TL ) ln
Vb TH − TL TL
η= = = 1−
Vb TH TH
nRTH ln
Va
• Efficiency of Carnot engine < 1
• No cycle can have an efficiency greater than a Carnot cycle
• All reversible engines operates between same 2 temperatures have same
efficiency
126. Carnot Cycle (Ideal gas, All the steps are reversible)
1 (A to B) 2 (B to C) 3 (C to D) 4 (D to A) Total (A to A)
W - nRThln(VB/V
A) CV(T-T )
C h - nRT ln(VD/V
C C) CV(T-T )
h C + nR(T – T) ln(V/V )
C h B A
q + nRT ln(V/V )
h B A Zero + nRT ln(V/V )
C D C Zero + nR h – T) ln(V/V )
(T C B A
DU Zero CV(T-T )
C h Zero CV(T-T )
h C Zero
+ nR ln(V A)
B/V Zero + nR ln(V C)
D/V Zero Zero
(qrev/ T)
A (pA, VA, Th)
B (pB, VB, Th)
γ −1 γ −1 γ −1 γ −1 γ −1
TV = const.; ThVB = TCVC ThVA = TCVD (pD, VD, TC) D
C (pC, VC, TC)
γ −1 γ −1
⎛ VB ⎞ ⎛V ⎞ ⎛ VB ⎞ ⎛ VC ⎞
⎜ ⎟
⎜V ⎟ =⎜ C ⎟
⎜V ⎟ ⎜ ⎟ =⎜ ⎟
⎜V ⎟ ⎜V ⎟ Th > Tc
⎝ A⎠ ⎝ D⎠ ⎝ A⎠ ⎝ D⎠ ln (VD / VC ) = − ln (VB / VA )
127. Entropy
• Our study of the Carnot cycle showed that
over the cycle q/T = 0.
• Since any reversible cycle can be
approximated by a series of Carnot cycles, we
expect that the integral of dQ/T along the
closed path of the cycle = 0.
• One consequence is that the integral of dQ/T
between a and b is independent of the path.
• The quantity dQ/T is called the entropy dS.
• The entropy difference between a and b is
thus path independent, and entropy is a state
variable.
128. Entropy
• Entropy is a state variable, but Q and W
are not state variables since they depend
on the path used to get from a to b.
• For a reversible process, the change in
the entropy of the engine is opposite to
the change in entropy of the
environment that provides the heat
required (or absorbs the heat
generated). The net change in the total
entropy = 0.
• For an irreversible process, the net
change in the total entropy will be
larger than 0.
129. Entropy :
– Is a measure of the disorderness of a system.
– In a reversible thermodynamic process, entropy of a system
is related the change in the extent to which energy is
dispersed in a disorderly manner, which in turn depends on
the amount of energy transferred as heat (as heat
stimulates motion) to the surroundings.
– Is a measure of the energy in a system or process that is
unavailable to do work.
• For a process occurring at constant temperature (an
isothermal process):
qrev = the heat that is transferred when the process
is carried out reversibly at a constant temperature.
T = temperature at which the heat is transferred (in
Kelvin).
130. Second Law of Thermodynamics
Entropy of the universe (i.e., Total Entropy):
• Does not change for reversible processes
• Increases for spontaneous processes.
• But, entropy can decrease for individual systems
Reversible (ideal):
Irreversible (real, spontaneous):
131. 2nd law of thermodynamics
dSUniv = dSsys + dSsurr > 0 Spontaneous process
dSUniv = dSsys + dSsurr = 0 Equilibrium process
dSUniv = dSsys + dSsurr < 0 Impossible process
For any process
“=“ for reversible, equilibrium
dSsys + dSsurr ≥ 0 “>” spontaneous (irreversible) (real)
all spontaneous processes are irreversible
132. Example
• Show the Carnot cycle on a T-S diagram and identify the heat
transfer at both the high and low temperatures, and the work
output from the cycle.
• 1-2, reversible isothermal heat transfer
T 1 2 QH = ∫TdS = TH(S2-S1) area 1-2-B-A
TH
• 2-3, reversible, adiabatic expansion
isentropic process, S=constant (S2=S3)
TL
4 3
• 3-4, reversible isothermal heat transfer
QL = ∫TdS = TL(S4-S3), area 3-4-A-B
A B
S1=S4 S2=S3 S • 4-1, reversible, adiabatic compression
isentropic process, S1=S4
• Net work Wnet = QH - QL, the area enclosed by 1-2-3-4, the shaded
area
133. Efficiency of an heat engine
TH
Consider a Carnot heat engine.
1.What fraction of the total heat transferred Qh
Qh
to engine is converted into work in the cycle?
Wby
QC
a. ∆T/TH b. ∆T/TC c. TC/TH
TC
2. What is efficiency of reverse engine (refrigerator)? Coefficient of
Performance
a. QC / W b. Tc/ (TH-TC) c. 1-TH/TC
3. When can the heat engine convert the absorbed heat into work
completely?.
a. At TH = -273 ˚C b. At Tc = O ˚C c. when TC =-273 ˚C
134. A Carnot heat engine receives 500 kJ of heat per cycle from a high-
temperature heat reservoir at 652oC and rejects heat to a low-
temperature heat reservoir at 30oC. Determine
(a) The thermal efficiency of this Carnot engine.
(b) The amount of heat rejected to the low-temperature heat reservoir.
a. TL
η th , rev = 1 −
TH TH = 652oC
( 30 + 273) K
= 1−
( 652 + 273) K QH
= 0.672 or 67.2% WOUT
HE
QL
QL T
b. = L TL = 30oC
QH TH
( 30 + 273) K
= = 0.328
( 652 + 273) K
Q L = 500 kJ ( 0.328)
= 164 kJ
135. Suppose a mole of a diatomic gas, such as O2, is compressed adiabatically so
the final volume is half the initial volume. The starting state is V = 1 liter,
T = 300 K. What are the final temperature and pressure?
p iV i γ = p f V fγ
Equation relating p,V for adiabatic process in α-
ideal gas γ =
7/2
= 1 .4
γ is the ratio of Cp/CV for diatomic gas in this case 5/2
γ
⎛V ⎞
Solve for pf from first equation pf = pi ⎜ i ⎟
⎜V ⎟
⎝ f ⎠
γ
Substitute for pi nRT i ⎛ V i
⎜
⎞
⎟
=
Vi ⎜ V f
⎝
⎟
⎠
= 6 . 57 x10 6 Pa
p fV f
Use ideal gas law to calculate final temperature Tf =
nR
OR use the equation relating T,V for an adiabatic = 395 K
process to get the final temperature (α = 5/2 for T i α V i = T fα V f
diatomic gas) 1
⎛V ⎞ α
T f = Ti ⎜ i ⎟
Solve for Tf ⎜V ⎟
⎝ f ⎠
= 395 K
136. Entropy change in heat engine
Th
Consider a Carnot heat engine.
1. What is the sign of the entropy change of Q
the hot reservoir during one cycle? Wby
h
a. ΔSh < 0 b. ΔSh = 0 c. ΔSh > 0 QC
Because energy is flowing out of the hot reservoir, its Tc
entropy (the number of microstates) is decreasing.
2. What is the sign of the entropy change of the cold reservoir?
a. ΔSc < 0 b. ΔSc = 0 c. ΔSc > 0
Because energy is flowing into the cold reservoir, it’s entropy (number
of microstates) is increasing.
3. Compare the magnitudes of the two changes.
a. |ΔSc| < |ΔSh| b. |ΔSc| = |ΔSh| c. |ΔSc| > |ΔSh|
