Basics_of_Thermodynamics

1. Basics of Thermodynamics
Four Laws that Drive the Universe
Peter Atkins*
Oxford University Press, Oxford, 2007
Reading
Some of the material covered here is also covered in the chapter/topic on:
Equilibrium
*It is impossible for me to write better than Atkins- his lucid (& humorous) writing style is truly impressive- paraphrasing may lead to loss of
the beauty of his statements- hence, some parts are quoted directly from his works.
MATERIALS SCIENCE
&
ENGINEERING
Anandh Subramaniam & Kantesh Balani
Materials Science and Engineering (MSE)
Indian Institute of Technology, Kanpur- 208016
Email: anandh@iitk.ac.in, URL: home.iitk.ac.in/~anandh
AN INTRODUCTORY E-BOOK
Part of
http://home.iitk.ac.in/~anandh/E-book.htm
A Learner’s Guide
Physical Chemistry
Ira N Levine
Tata McGraw Hill Education Pvt. Ltd., New York (2002).
Instructors/students may download the
appropriate files and delete the portion not
needed. This will help tailor the contents for any
specific syllabus or need. (I.e. you may copy left,
right and centre!!).
Web resource: https://www.khanacademy.org/science/physics/thermodynamics

2.  Thermodynamics deals with stability of systems. It tells us ‘what should happen?’.
‘Will it actually happen(?)’is not the domain of thermodynamics and falls under
the realm of kinetics.
 At –5C at 1 atm pressure, ice is more stable then water. Suppose we cool water to
–5C. “Will this water freeze?” (& “how long will it take for it to freeze?”) is (are)
not questions addressed by thermodynamics.
 Systems can remain in metastable state for a ‘long-time’.
 Window pane glass is metastable– but it may take geological time scales for it
to crystallize!
 At room temperature and atmospheric pressure, graphite is more stable then
diamond– but we may not lose the glitter of diamond practically forever!
Thermodynamics versus Kinetics
* The term metastable is defined in the chapter on equilibrium.

3.  One branch of knowledge that all engineers and scientists must have a grasp of (to
some extent or the other!) is thermodynamics.
 In some sense thermodynamics is perhaps the ‘most abstract subject’ and a student
can often find it very confusing if not ‘motivated’ strongly enough.
 Thermodynamics can be considered as a ‘system level’ science- i.e. it deals with
descriptions of the whole system and not with interactions (say) at the level of
individual particles.
 I.e. it deals with quantities (like T,P) averaged over a large collection of entities
(like molecules, atoms)*.
 This implies that questions like: “What is the temperature or entropy of an
atom?”; do not make sense in the context of thermodynamics (at lease in the usual way!).
 TD puts before us some fundamental laws which are universal** in nature (and
hence applicable to fields across disciplines).
Thermodynamics (TD): perhaps the most basic science
* Thermodynamics deals with spatio-temporally averaged quantities.
** They apply to the universe a whole as well! (Though the proof is lacking!).
 TD parameters are measureable macrocopic quantities, which are characterize (/associated
with) the system. These include: P, T, V, H (magnetic field). (Note: non-bolded H will be used for enthalpy)

4.  To understand the laws of thermodynamics and how they work, first we need to get the
terminology right. Some of the terms may look familiar (as they are used in everyday
language as well)- but their meanings are more ‘technical’and ‘precise’, when used in TD
and hence we should not use them ‘casually’.
 System is region where we focus our attention (Au block in figure). A TD system is a
macroscopic system.
 Surrounding is the rest of the universe (the water bath at constant ‘temperature’).
 Universe = System + Surrounding (the part that is within the dotted line box in the figure below)
 More practically, we can consider the ‘Surrounding’ as the immediate neighborhood of the
system (the part of the universe at large, with which the system ‘effectively’ interacts).
In this scheme of things we can visualize: a system, the surrounding and the universe at
large.
 Things that matter for the surrounding: (i) T, (ii) P, (iii) ability to: do work, transfer heat,
transfer matter, etc. Parameters for the system: (i) Internal energy, (ii) Enthalpy, (iii) T, (iv)
P, (v) mass, etc.
The language of TD
In TD we usually do not worry about
the universe at large!
The surrounding does not change in any way during
any process that the system undergoes (i.e. its T, P,
etc. remain the same). I.e. the surrounding is not
transmutable.

5.  To a thermodynamic system two ‘things’ may be added/removed:
 energy (in the form of heat &/or work)  matter.
 An open system is one to which you can add/remove matter (e.g. a open beaker to which
we can add water). When you add matter- you also end up adding heat (which is contained
in that matter).
 A system to which you cannot add matter is called closed.
Though you cannot add/remove matter to a closed system, you can still add/remove heat
(you can cool a closed water bottle in fridge).
 A system to which neither matter nor heat can be added/removed is called isolated.
A closed vacuum ‘thermos’ flask can be considered as isolated.
Open, closed and isolated systems
Type of boundary Interactions
Open All interactions possible (Mass, Work, Heat)
Closed Matter cannot enter or leave
Semi-permeable Only certain species can enter or leave
Insulated Heat cannot enter or leave
Rigid Mechanical work cannot be done*
Isolated No interactions are possible**
* By or on the system
** Mass, Heat or Work
Mass
Heat
Work
Interactions possible

6.  Matter is easy to understand and includes atoms, ions, electrons, etc.
 Energy may be transferred (‘added’) to the system as heat, electromagnetic radiation etc.
 In TD the two modes of transfer of energy to the system considered are Heat and Work.
 Heat and work are modes of transfer of energy and not ‘energy’ itself.
 Once inside the system, the part which came via work and the part which came via
heat, cannot be distinguished*. More sooner on this!
 Before the start of the process and after the process is completed, the terms heat and
work are not relevant.
 From the above it is clear that, bodies contain internal energy and not heat (nor work!).
 Matter when added to a system brings along with it some energy. The ‘energy density’
(energy per unit mass or energy per unit volume) in the incoming matter may be higher or lower than the
matter already present in the system.
* The analogy usually given is that of depositing a cheque versus a draft in a bank. Once credited to an account, cheque and draft have no
meaning. (Also reiterated later).

7. Variables in a TD system
 The TD state is specified by a set of values of all the TD parameters required for the
description of the system.
 The state of a system is determined by ‘Potentials’, which is analogous to the potential
energy of the block under gravity (which is determined by the centre of gravity (CG) of the
block).
These potentials are the Thermodynamic Potentials (A thermodynamic potential is a Scalar Potential to represent
the thermodynamic state of the system).
 There are 4 important potentials (in some sense of equal stature).
These are: Internal Energy (U or E), Enthalpy (H), Gibbs Free Energy (G), Helmholtz Free
Energy (A or F).
* To be discussed later
 The macroscopic variables defining a state are the State or Thermodynamic Variables (A
state variable is a precisely measurable physical property which characterizes the state of
the system- It does not matter as to how the system reached that state). Pressure (P), Volume
(V), Temperature (T), Entropy (S) are examples of state variables.
Macroscopic and Microscopic Variables
Macroscopic Variables

8.  In addition, we can have microscopic variables associated with a system. I.e. these are
associated with the description of the system, via states of individual particles (like position,
velocity, kinetic energy, etc.). These variable change continuously, even for a system in
equilibrium (and hence are typically not considered in classical thermodynamics).
 The macroscopic variables are defined only under equilibrium conditions (or during a quasi-
static process*). The state variables like T & P are not defined during ‘transients’(transient
states of the system). However, even during transients the microscopic variables are well
defined (but changing in value continuously).
Microscopic Variables
 If the TD state of a system does not change with time, then the system is in TD equilibrium.
Often the term state in TD implies a state in equilibrium. A TD Transformation is a change of the state of a TD system.
Thermodynamic Equilibrium* & Thermodynamic Transformation
 Is a functional relation between the TD parameters of the system in equilibrium.
 If P, V, T are TD parameters of the system, the equation of state can be written as:
Equation of State**
* Much more on the chapter on equilibrium. ** Will learn a lot more about this later.
( , , ) 0
f P V T 
 The existence of a such a relation reduces the number of independent variable by one.
 A state of the system is a point in the P-V-T space.
 The equation of state gives us a surface in the P-V-T space and any point on the surface is state in
equilibrium.

9. Q & A What is meant by microscopic in the context of thermodynamics?
 In TD we often take recourse to macroscopic and microscopic viewpoints.
 E.g. if we think of an gas, we can visualize ‘T’ as the parameter, which drives heat transfer
between two bodies (heat flows from high-T to low-T). This is the macroscopic picture. In the
microscopic picture (which is arises from statistical TD), we visualize energy levels available
for the species to populate and the distribution of species across these levels.
 Similarly, we can think of pressure macroscopically as the causative agent for driving the
piston (direction from high-P to low-P). Microscopically, it is the momentum transferred per
unit area per unit time by the species of the medium (e.g. gas molecules).
 When we talk about entropy, again we invoke the microscopic and macroscopic pictures.
Macroscopically, we sit at the system boundary and track heat transfer (Qrev) & S = Qrev/T.
Microscopically, we ‘worry’ about the species occupying (e.g.) certain configurations (micro-
states).
 Hence, in TD micro-scopic does not concern with a lengthscale, but with the details. In the
microscopic picture, we look at the species comprising of the system, like molecules and track
their configurations, energy states they occupy, vibrations, etc.

10.  If a system is in a equilibrium state, then a TD transformation can be brought about only by
changes in the external conditions (parameters) of the system. I.e., actions of the surrounding
can only bring about the change to a system in TD equilibrium.
 The change from one TD state to another is considered as a process.
Thermodynamic Process

11.  Here is a brief listing of a few kinds of processes, which we will encounter in TD:
 Isothermal process → the process takes place at constant temperature
(e.g. freezing of water to ice at –10C)
 Isobaric → constant pressure
(e.g. heating of water in open air→ under atmospheric pressure)
 Isochoric → constant volume
(e.g. heating of gas in a sealed metal container)
 Quasi-static process → the process occurs so gradually, that the system is in internal
equilibrium throughout the process. The macroscopic state variables (like P & T) are well defined during the process.
(e.g. removal of sand, grain by grain, from a piston loaded by the sand)
 Transient process → the process occurs so fast, that the internal equilibrium is not maintained during the process. The
macroscopic state variables (like P & T) are not well defined during the process. This process is not part of the realm of equilibrium TD.
(e.g. expansion of a gas from one part of a system to another, when the partition is removed)
 Reversible process → the system is close to equilibrium at all times (and infinitesimal
alteration of the conditions can restore the universe (system + surrounding) to the original
state. Most quasi-static processes are reversible.
 Cyclic process → the final and initial states are the same. However, q and w need not be
zero.
 Adiabatic process → dq is zero during the process (no heat is added/removed to/from the
system during the process). A system undergoing an adiabatic process is thermally isolated
by adiabatic walls.
 A combination of the above are also possible: e.g. ‘reversible adiabatic process’.
Different types of Processes in TD We will deal with some of these in detail later on

12. What is the relation between ‘quasi-static’ and ‘reversible’ processes?
Funda Check
 We have noted before that:
 actions of the surrounding can only bring about the change to a system in TD equilibrium,
 the change from one TD state to another occurs via a process &
 during a quasi-static process the system is in internal equilibrium throughout the process.
 Also we can think of a reversible process as follows.
 A process is reversible if the transformation retraces path in time, when the external
conditions retraces its path in time.
 E.g. if the external pressure is increased in steps of P each time, the piston will move in &
pressure will equilibrate after each step. I.e. after the first step the internal pressure (Pint = Pext
= P0 + P). We can conceive a series of similar steps to increase the internal pressure to Pf.
Now, if we decrease the external pressure by P, the piston will move out and an equilibrium
pressure will be established (Pint = Pext = Pf  P). By following such steps the physical and
TD path can be retraced.
 A reversible transformation is quasi-static, but the converse may not be true.
Pext = (P0 +P)
1
2
Pt=0 = P0
 A reversible process can be shown as a
continuous path in the P-V diagram (as we shall
see soon).
 Also, a process which is not reversible
cannot be shown as a continuous path in the
P-V diagram.

13.  In chemistry and physics may processes exist. Some of them are listed below.
Other Processes
 Phase Transitions. In phase transitions the composition does not change. A super set of phase
transitions is Phase Transformations.
General:  α phase → β phase.
 Fusion: Solid → Liquid.
 Vaporization: Liquid → Gas.
 Sublimation: Solid → Gas.
 Mixing. Pure A + Pure B → Mixture.  Solution/dissolution: Solute + Solvent → Solution.
 Reaction. Reactants → Products.  Combustion: Element/Compound + Oxygen → Oxide.
 Formation. Elements → Compound.
 Activation. Reactants → Activated complex.

14.  A property which depends only on the state of the system (as defined by T, P, V etc.) is
called a state function. This does not depend on the path used to reach a particular state.
 Analogy: one is climbing a hill- the potential energy of the person is measured by the
height of his CG from ‘say’ the ground level. If the person is at a height of ‘h’ (at point P),
then his potential energy will be mgh, irrespective of the path used by the person to reach
the height (paths C1 & C2 will give the same increase in potential energy of mgh- in figure
below).
 In TD this state function is the internal energy (U or E). (Every state of the system can be ascribed to a unique U).
 Hence, the work needed to move a system from a state of lower internal energy (=UL) to a
state of higher internal energy (UH) is (UH)  (UL). W = (UH)  (UL)
 The internal energy of an isolated system (which exchages neither heat nor mass) is
constant  this is one formulation of the first law of TD.
 A process for which the final and initial states are same is called a cyclic process. For a
cyclic process change in a state function is zero.
E.g. U(cyclic process) = 0.
State functions in TD

15.  A spontaneous process is one which occurs ‘naturally’, ‘down-hill’
in energy*. I.e. the process does not require input of work in any
form to occur.
 Melting of ice at 50C is a spontaneous process.
 A driven process is one which wherein an external agent takes the
system uphill in energy (usually by doing work on the system).
 Freezing of water at 50C is a driven process (you need a
refrigerator, wherein electric current does work on the system).
 Later on we will note that the entropy of the universe will increase
during a spontaneous change. (I.e. entropy can be used as a single
parameter for characterizing spontaneity).
Spontaneous and
Driven processes
Spontaneous process
(Click to see)
* The kind of ‘energy’ we are talking about depends on the conditions. As in the topic on Equilibrium, at
constant temperature and pressure the relevant TD energy is Gibbs free energy.
Q & A When will a process occur?
 Under equilibrium conditions ‘nothing’ will take place (at least macroscopically).
 For a process to occur there has to be a ‘causative agent’ (typically of a critical magnitude). The common
driving forces are differences in temperature, pressure and chemical potential.
 One of the processes of interest, which we will deal with repeatedly in TD, is the reversible
process; which occurs close to equilibrium conditions. In mechanics these are also referred to
as quasi-static processes.
Continued on the next slide

16. Close System
Ideal Gas
Single phase*
T, P, Material Equilibrium
 Starting with an ideal gas# in a closed system at constant T, P at equilibrium, we can progressively relax the conditions to obtain more
‘realistic’ and complicated systems. A broad picture of these is shown below. We will consider a part of this picture in the current e-book.
 Currently we restrict ourselves to P-V work only, noting that other types of work are possible.
System and Process Related Complexities
* Gases always form single phase system. # We will deal with ideal gases in detail soon.
Close System
Ideal Gas
Single phase*
Near Equilibrium
(Reversible Process)
System level complexity Material level complexity
Close System
Real Gas/Solid/Liquid
Single phase*
Single component
T, P, Material Equilibrium
Close System
Ideal Gas
Single phase*
T, P Equilibrium
Chemical non-equilibrium
(Ir-reversible Process)
Various types of complexities can be combined
Open System
Ideal Gas
Single phase*
T, P Equilibrium
Chemical non-equilibrium
(Ir-reversible Process)
Close System
Real Gass/Solids/Liquids
Single phase*
Multi-component
T, P, Material Equilibrium
Close System
Real Gas/Solid/Liquid
Two phases*
Single component
T, P, Material Equilibrium
Close System
Real Gass/Solids/Liquids
Two phases*
Multi-component
T, P, Material Equilibrium
Open System
Ideal Gas
Single phase*
T, P Equilibrium
No (T, P, Chemical) equilibrium
(Ir-reversible Process)

17. In the current set of notes we will follow the path as below.
Closed system at equilibrium Closed system Reversible process Open System Reversible process
 Single phase
 P-V work only
 Single phase
 P-V work only
Open System Reversible process
 Multi-phase
 P-V work only
Phase Equilibrium Reaction Equilibrium

18. How to understand the ‘macroscopic’ versus ‘microscopic variables?
Funda Check
 Let us consider a gas expanding from a high pressure chamber (at pressure P1) into a chamber
under vacuum (via a nozzle). Let the system be insulated, so that no heat can enter the
system.
 As the gas expands the pressure falls from P1 to a lower value. Let us track the process
starting from t = 0 to infinitesimal times.
 As the gas expands the number of collisions (with each other and a fictitious wall) decreases.
 The position and velocity (the microscopic variables) of each molecule can be
‘measured/known’ (say on plane AB, Fig. below). However, under these transient conditions
macroscopic variables like pressure are not defined (the pressure on the left of AB is different
from that on the right of AB) due to the non-equilibrium conditions.
P1
Vacuum
Nozzle A
B

19.  Though we all have a feel for temperature (‘like when we are feeling hot’); in the context of
TD temperature is technical term with ‘deep meaning’.
 As we know (from a commons sense perspective) that temperature is a measure of the ‘intensity of heat’.
‘Heat flows’ (energy is transferred as heat) from a body at higher temperature to one at lower
temperature. (Like pressure is a measure of the intensity of ‘force applied by matter’→
matter (for now a fluid) flows from region of higher pressure to lower pressure).
 That implies (to reiterate the obvious!) if I connect two bodies (A)-one weighing 100 kg at 10C
and the other (B) weighing 1 kg at 500C, then the ‘heat will flow’ from the hotter body to
the colder body (i.e. the weight or volume of the body does not matter).
 But, temperature comes in two important ‘technical’ contexts in TD:
1 it is a measure of the average kinetic energy (or velocity) of the constituent entities (say molecules)
2 it is the parameter which determines the distribution of species (say molecules) across
various energy states available.
Temperature
500C
Heat flow
direction
10C
A B
 How is constant temperature maintained (isothermal conditions)?
 The systems is in contact with the surroundings via dia-thermal walls (walls which conduct
heat). The surroundings acts like a thermal reservoir (i.e. is so large that input or withdrawal
of heat (Q) does not change its temperature).
Thermal reservoir
System

20.  Let us consider various energy levels available for molecules/species in a system to be
promoted to. Let the system be in thermal equilibrium.
 At low temperatures the lower energy levels are expected to be populated more, as compared
to higher energy levels. (Fig.1). In Fig.1 the energy levels are assumed to be equally spaced for simplicity (this will not be true for an real system).
 As we heat the system, more and more ‘molecules’ will be promoted to higher energy levels.
 The distribution of molecules across these energy levels is given by:
Temperature as a parameter determining the distribution of species across energy levels
1
0 0
( )  kT
P E P e P e


 
 Note that  is the only parameter which determines the
distribution.
 The numerical value of  decreases as ‘environment’ gets
colder.
 Hence, we define ‘T’ which is the inverse of ; such that as the
hotter temperatures have a higher numerical value of a
parameter.
 T could have been just the inverse of , but to keep the
magnitude of 1C equal to 1 K, we introduce a constant k (=
kB), which is the Boltzmann constant. This implies that kB is
not a fundamental constant like many others.
 At 0 K only the ground state is populated, while at infinite
temperature all states are populated equally.
 P(E) is the population of species at an energy level E.
  is the single parameters which controls the distribution across
energy levels.
At 0 K only
the ground
state is filled
With increasing T, progressively the population
of higher energy increases
Fig.1

21.  Celsius (Farenheit, etc.) are relative scales of temperature and zero of these scales do not have
a fundamental significance. Kelvin scale is a absolute scale.
Zero Kelvin and temperatures below that are not obtainable in the classical sense.*
 Classically, at 0 K a perfect crystalline system has zero entropy (i.e. system attains its minimum
entropy state). However, in some cases there could be some residual entropy due to degeneracy
of states (this requires a statistical view point of entropy).
 At 0 K the kinetic energy of the system is not zero in the quantum mechanical picture. There
exists some zero point energy due to fluctuations arising from the Heisenberg uncertainty
principle.
Few points about temperature scales and their properties
* In systems with population inversion, we have a negative Kelvin temperature (which is hotter than infinity, rather than
being colder than zero)!

22.  Pressure* is force per unit area (usually exerted by a fluid on a wall**).
 It is the momentum transferred (say on a flat wall by molecules of a gas) per unit area, per unit time. (In the
case of gas molecules it is the average momentum transferred per unit area per unit time on to the flat wall).
 P = momentum transferred/area/time.
 Pressure is related to momentum, while temperature is related to kinetic energy.
Pressure
Wall
of
a
container
‘Crude schematic’
of particles
impinging on a
wall.
* ‘Normal’pressure is also referred to as hydrostatic pressure.
** Other agents causing pressure could be radiation, macroscopic objects impinging on a wall, etc.
 Pressure is a parameter best suited for gases and liquids. For solids stress is the appropriate parameter (though pressure may also be used).
/
/ /
P force area
momentumtransferred area time


2 2
[ ] [ ] [ ]
[ ][ ] [ ] [ ][ ]
mv Kg L Kg
P
At L s s L s
  
 Can we define pressure inside the container (not on the walls as it is easy to visualize pressure on a wall,
but not inside the container)?
 Pressure is a ‘hydrostatic’, ‘homogeneous’ and ‘isotropic’ quantity$  i.e. it is same in each direction and
throughout the inside of the container (including the walls).
 In the interior it is best visualized by introducing a hypothetical wall and computing the momentum
transferred per unit area, per unit time.
$ for an ideal gas, about which we will talk soon.
p mv

A Area


23. Units of pressure
 Mechanical equilibrium
 Thermal equilibrium
 Reaction Equilibrium
 Phase Equilibrium
Kinds of Equilibrium
Material Equilibrium
 The topic of equilibrium is dealt with in detail elsewhere. A system in complete equilibrium
satisfies 4 types of equilibrium as listed below.

24.  Work (W) in mechanics is displacement (d) against a resisting force (F). W = F  d.
 Work has units of energy (Joule, J).
 Work can be expansion work (PV), electrical work, magnetic work etc. (many sets of
stimuli and their responses).
 Heat as used in TD is a tricky term (yes, it is a very technical term as used in TD).
 The transfer of energy as a result of a temperature difference is called heat.
 “In TD heat is NOT an entity or even a form of energy; heat is a mode of transfer of
energy” [1].
 “Heat is the transfer of energy by virtue of a temperature difference” [1].
 “Heat is the name of a process, not the name of an entity” [1].
 “Bodies contain internal energy (U) and not heat” [2].
 The ‘flow’ of energy down a temperature gradient can be treated mathematically by
considering heat as a mass-less fluid [1] → this does not make heat a fluid!
Heat and Work
[1] Four Laws that Drive the Universe, Peter Atkins, Oxford University Press, Oxford, 2007. [2] Physical Chemistry, Ira N Levine, Tata McGraw Hill Education Pvt. Ltd., New York (2002).
To give an example (inspired by [1])
Assume that you start a rumour that there is ‘lot of’ gold under the class room floor. This rumour ‘may’ spread when persons talk to each other.
The ‘spread of rumor’ with time may be treated mathematically by equations, which have a form similar to the diffusion equations (or heat
transfer equations). This does not make ‘rumour’a fluid!
Expansion work P

25.  Work is coordinated flow of matter.
 Lowering of a weight can do work
 Motion of piston can do work
 Flow of electrons in conductor can do work.
 Heat involves random motion of matter (or the constituent entities of matter).
 Like gas molecules in a gas cylinder
 Water molecules in a cup of water
 Atoms vibrating in a block of Cu.
 Energy may enter the system as heat or work.
 Once inside the system:
 it does not matter how the energy entered the system* (i.e. work and heat are terms
associated with the surrounding and once inside the system there is no ‘memory’ of how
the input was received and
 the energy is stored as potential energy (PE) and kinetic energy (KE).
 This energy can be withdrawn as work or heat from the system.
* As Aktins put it: “money may enter a back as cheque or cash but once inside the bank there is no difference”.
Q) Why is work done at constant ‘P’ equal to PV.
 Work done at constant pressure (isobaric) (leading to a volume change V): W = PV.
 This can be understood easily. Consider a ideal gas in a cylinder with a piston of area
A. Let the gas expand, such that the piston moves by x. The work done: W = F . x.
F
W A x P V
A
    

26. Q & A Give examples of a few types of work.
 Work done should have units of energy.
 Mechanical work could be in 3D, 2D or 1D.
Type Sub-type Formula Comments
Mechanical 3D P.V P is Pext
2D (e.g. surface tension () work) .dA  is the surface tension on a fluid surface
1D (e.g. line tension (f) work) f. dl
Electrical Q.V
Magnetic

27.  ‘Ultimately’, all forms of energy will be converted to heat!!
 One nice example given by Atkins: consider a current through a heating wire of a resistor. There is a net
flow of electrons down the wire (in the direction of the potential gradient)  i.e. work is being done.
Now the electron collisions with various scattering centres leading to heating of the wire  i.e. work
has been converted into heat.
(P+P)
 In a closed system (piston in the example figure below), if infinitesimal pressure increase
causes the volume to decrease by V, then the work done on the system is:
 The system is close to equilibrium during the whole process
thus making the process reversible.
 As V is negative, while the work done is positive (work done on the system is positive,
work done by the system is negative).
If the piston moves outward under influence of P (i.e. ‘P’ and V are in opposite directions,
then work done is negative.
Reversible P-V work on a closed system
reversible
dW PdV
 
1
2
Note that the ‘P’is the pressure inside the container. For the work to be
done reversibly the pressure outside has to be P+P (~P for now). Since
the piston is moving in a direction opposite to the action of P, the work
done by the surrounding is PV (or the work done by the system is PV,
i.e. negative work is done by the system).
P

28. What is the significance of the term ‘reversible’ in the context of work?
Funda Check
 As we shall soon see, maximum work is done in a reversible process.
 Typically, in irreversible process, we are far from equilibrium.
 Example of a irreversible process is the expansion of a gas as in Fig.1.
Here, there is a partition separating two regions, one with gas at ‘P1’ and
another with vacuum and the then the partition vanishes. During the
expansion of the gas (states between S1 and S2), macroscopic variables
like ‘P’ are not defined and hence it is not prudent to use formulae like
PV for work.
reversible
dW PdV
 
T, P1
V1
N
State-1 (S1)
State-2 (S2)
T
V1
0
T
2V1
N
Partition
Fig.1
Heat reservoir
Dia-thermal
walls
Q
What is a thermal bath or surrounding (in general)?
Funda Check
 Surrounding remains unchanged in any process that the system undergoes. The surrounding is
‘un-transmutable’.
 The surrounding can act like a (a) thermal, (b) pressure or (c) chemical reservoir.
 I.e. (a) if heat is transferred (in or out, Q can be +ve or ve) from the surrounding to the system, the
temperature of the surrounding does not change. (b) Similarly, if the surrounding does P-V
work (+ve or ve) on the system, its pressure does not change. (c) If the surrounding gives
some species (one or more) to the system (or equivalently takes species from the system), the concentration* of the
species does not change in the surrounding.
* Or chemical potential (a quantity which we will see later and which is responsible for mass transfer)

29.  A reversible process is one where an infinitesimal change in the conditions of the
surroundings leads to a ‘reversal’ of the process. (The system is very close to equilibrium
and infinitesimal changes can restore the system and surroundings to the original state).
 If a block of material (at T) is in contact with surrounding at (TT), then ‘heat will flow’
into the surrounding. Now if the temperature of the surrounding is increased to (T+T), then
the direction of heat flow will be reversed.
 If a block of material (at 40C) is contact with surrounding at 80C then the ‘heat transfer’
with takes place is not reversible.
 Though the above example uses temperature differences to illustrate the point, the situation
with other stimuli like pressure (differences) is also identical.
 Consider a piston with gas in it a pressure ‘P’. If the external pressure is (P+P), then the
gas (in the piston) will be compressed (slightly). The reverse process will occur if the
external (surrounding pressure is slightly lower).
 Maximum work will be done if the compression (or expansion) is carried out in a reversible
manner.
Reversible process
T
Heat flow
direction
T+T
T
Heat flow
direction
TT
Reversible process
40C
Heat flow
direction
80C
NOT a Reversible process
‘Reversible’ is a technical term (like many others) in the context of TD.

30. Why is the work done maximum in a reversible process?
Funda Check
 Let us consider two cases (Fig.1) with pressure inside the cylinder is 100 bar: (C1) the outside
pressure is lowered infinitesimally to 50 bar and (C2) the outside pressure is constant at 50
bar. The gradual reduction in pressure in C1 can be achieved by removal of sand grains as in Fig.1a. For simplicity we replace the
infinitesimals with small finite quantities (). Let us assume an ideal gas in the cylinder.
 In C1 the initial resisting pressure is (100  ) bar or ~100 bar and the work done is: PV =
100 V. This expansion leads to a drop in the pressure to P2 (< 100 bar) say 99 bar. Further
expansion will lead to a drop in pressure to (99  ) bar and the work done in this step will be
~ 99V. We can carry on this process and the net work will be given by the summation
(integration in reality). Wnet (C1) = 100 V + 99V + 98V + ... + 50 V. (We could have done 99.5V +...)
 In C2 the opposing force to the expansion is 50 bar. Hence the work done is:
W (C2) = Pext V + Pext V + Pext V + ...= 50 V + 50 V + 50 V + ...
 Clearly, WC1 > WC2.
Pext = (100 ) bar
C1
100 bar
Fig.1
(a) (b)
Sand grains
Pext = (50) bar
C2
100 bar
Note: in PdV or PV, P = Pexternal = Presisting is the resisting
pressure (against which work is done)
Warning: thermodynamics cannot be used to calculate work during a
irreversible process. Above is some kind of ‘crude’ rationalization.
Kinetic energy gained by the piston, pressure gradients within the gas
and turbulence in the gas will have to be considered.
2
1
irreversible ext
W P dV
 
If the piston is in equilibrium at the start
and finish then we can calculate:

31.  Let us keep one example in mind as to how we can (sometimes) construct a ‘reversible’
equivalent to a ‘irreversible’ processes.
 Let us consider the example of the freezing of ‘undercooled water’* at –5C (at 1 atm
pressure). This freezing of undercooled water is irreversible (P1 below).
Yes, it is possible to obtain water below its freezing point. This is referred to as ‘under-cooled’ water. In fact freezing
always occurs with some undercooling, due to the existence of a ‘nucleation barrier’.
 We can visualize this process as taking place in three reversible steps  hence making the
entire process reversible (P2 below).
How to visualize a ‘reversible’ equivalent to a ‘irreversible’ processes?
* ‘Undercooled’ implies that the water is held in the liquid state below the bulk freezing point! How is this possible?→ read chapter on phase
transformations
Water at –5C Ice at –5C
Irreversible
Water at –5C
Water at –0C
Reversible
Ice at 0C
Ice at –5C
Heat Cool
P2
P1
Freezing

32.  We will take up this concept here briefly and return to it later. The kinetic theory gives some ‘nice results’, which are to be treated as
approximate in many circumstances.
 When we want to understand any new subject/concept, it is best to start with some
idealizations. In TD some of these idealizations include: ‘reversible’, ‘quasi-static’, etc.
 In an ideal gas:
(i) the particles are point particles (no size) and
(ii) there are NO interactions between the particles.
 The ideal gas obeys the ideal gas law (which is a simplified equation of state*).
 Many real gases (like the noble gases, oxygen, hydrogen, nitrogen, etc.) under certain
conditions of P & T behave close to an ideal gas. All gases tend to behave more like an ideal
gas at high T and low P.
 In an ideal gas ALL the Internal energy is due to the translational motion (velocity) of the
particles (molecules). The velocity is a function of the temperature (T). Higher the temperature,
higher the kinetic energy and higher the internal energy.
 Three kinds of ideal gases are differentiated in the literature: (i) the classical or Maxwell–
Boltzmann ideal gas (which follows the Maxwell-Boltzmann distribution), (ii) the ideal quantum Bose gas
(which is composed of Bosons, Bose-Einstein statistics), (iii) the ideal quantum Fermi gas (composed of Fermions,
Fermi-Dirac statistics).
Ideal Gas Kinetic theory of gases: Pressure, Temperature, Molecular speeds.
* For gases, equation of state relates P, V & T. The equation may have one or more parameters. (More about this later).

33.  The equation of state obeyed by an ideal gas is given by the Boyle’s law. N is no. of molecules.
PV
Constant
N
 The value of the constant depends on the experimental scale of temperature used
 The equation of state of an ideal gas can be used to define a temperature scale the ideal gas
temperature ‘T’.
PV
k T
N
 
23
5
=1.380 10 / /
8.617 10 / /
B (Boltzmann Constant)
J K mole
k k
eV K mole




 
 The value of the Boltzmann constant is determined by the choice of temperature intervals,
usually chosen as 1C. The universality of the scale arises from the universal character of the
ideal gas.
 The procedure to define an ideal gas temperature scale is as
follows. Determine the value of PV/Nk at the freezing (P1)
and boiling (P2) points of water. These points are plotted with
PV.Nk and T as the axes. A straight line is drawn through the
points, which intersects the T axis at T = 0. The interval
between P1 & P2 is divided into 100 divisions (as we are
using the degree Celsius scale). The resulting scale is the
Kelvin scale.
 To measure the T of an unknown system, it is brought into thermal contact with an ideal gas
and PV/Nk is determined for the ideal gas. Then, the T is read off from the plot.

34. 1 Mole of N2
 Typically, we plot P versus V at constant temperature
(isotherms). (Fig.1). The curves are asymptotic to the x
and y axis.
 At a constant volume (say V1) if heat the gas (say from 75C to
100C) then the pressure increases (vertical dashed line).
 Similarly, at constant pressure if we heat the gas, the volume increases
(a horizontal line).
V1
Fig.1a
 As the molecules of a ideal gas do not interact with each other, the internal energy of the
system is expected to be ‘NOT dependent’ on the volume of the system.
 Internal energy (a state function) is normally a function of T & V: U = U(T, V).
 For an ideal gas: U = U(T) only.
0
T
U
V

 

 

 
 The ideal gas law is the combination of the Charles law, the Boyle’s law and the Avogadro's
law and equivalently we can write the equation of state for an ideal gas as PV = nRT.
1 1 2 2
1 2
P V P V
nR
T T
 
This leads to
PV nRT

Ideal gas law
n
R
A
A
N
PV Nk T N k T n RT
N
  
NA is the Avogadro's no. (=6.023  1023 atoms/mole), n is the number of moles, R is the gas constant (= 8.315 J/C)

35.  (Fig.2 & 3). For an ideal gas the plot of PV versus P should be a horizontal line at constant
temperature. However, real gases (like N2) show marked deviation at high P and low T.
 At room temperature (RT), for gases like H2 and He the variation of PV/nRT with P is nearly
linear. The behaviour for other gases (like CO and CH4) is more complicated. We will
consider the details of the behaviour later, when we discuss real gases and the
compressibility factor (Z).
Increasing deviation
from Ideality:
 P
 T
Fig.2
1 Mole of N2
Fig.3

37.  In an idea gas the following assumptions are made:
(i) there is no (negligible) attractive force between the molecules,
(ii) the molecules are point particles
(volume occupied by the molecule is << the volume of the container),
(iii) the collisions are perfectly elastic,
(iv) the duration of the collisions is negligible as compared to the time between the collisions.
Ideal gas
Calculation of pressure
• Let the velocity of a ‘typical’ molecule be ‘c’ and its components along x, y, z be u, v, w,
respectively. This implies: c2 = u2 + v2 + w2. Let the box be a cube of dimensions: L L L.
• For this typical molecule, if we consider the velocity along y-direction and elastic collision with
the wall, the change in momentum is: mv  (mv) = 2 mv.
• The time (t) taken for the molecule to return to this wall is: 2L/v = t. The number of impacts per
second (the rate) on this wall due to this molecule is: v/2L.
• Momentum change due to one molecule per second: (v/2L).(2mv) = mv2/L = Force on +X face.
• Pressure on +X face due to one molecule = F/area = (mv2/L)/L2 = mv2/L3 = P.
• If there are ‘N’ molecules in the chamber, each with a velocity vi (i = 1 to N), the total pressure
due to these ‘N’ molecules is:
Pressure is force/area = change in momentum/area/time.
 
2 2
2
3 3 3
1 1
n n
i i
i i
mv Nm v Nm
P v
L L N L
 
  
 
Continued…
(1)

38. • Since none of the axis is special it is reasonable to assume: 2 2 2
u v w
 
Hence: 2 2 2 2
c u v w
   & 2 2
3
c v
 Hence from (1):
2 2
3 3
1
3 3
Nm c Nm c
P
L L
 
 
 
 
 
• So far we assumed that the molecules do no suffer any collisions and are free to move from end
to end in the container. In reality, the molecules will collide and redistribution of speeds occur.
However, it is reasonable to assume that the average speed of the molecules does not change
with time.
Root mean square (RMS) speed
2
c is the mean square speed
The pressure can be written in terms of density:
2 2
1
3
3
1
3

Nm
P c c
L
 
 
 
 
The square root of the mean square speed can be written in terms of macroscopically measurable
quantities P and :
2 3

P
c 
For H2 at STP if we substitute the values of P and  we get: 2
1840 /
RMS speed c m s
 
Incredible speeds!
Of the same order as that of speed of sound.
• James Prescott Joule (1818–1889) had first done this calculation in 1848. This incredible speed
is if the same order of magnitude as the speed of sound in in H2 at 0C (~1.3 km/s).
• The equation implies that for as the density of the gas increases RMS speed decreases. O2 (with
a molecular mass of 32 amu has a RMS speed given by:
(2)
(3)
(4)
2 2
1840 460 /
32
Oxygen
c m s
   
 
 
Continued…
 
2
3
Nm
P v
L


39. Temperature
3 2
1
3
PL Nm c

From (3):
2
1
3
PV Nm c

According to the ideal gas equation PV = nRT:
2
1
3
nRT Nm c

• This is an important result, that for an ideal gas the T in Kelvin is proportional to the mean
square velocity of the molecules. I.e. the average K.E. associated with the translation of a
molecule of the gas is proportional to the absolute temperature (T).
2
3 1
2 2
R
n T m c
N

If we consider an Avogadro no. of gas molecules (N = NA & n = 1), then R/NA = k is the gas
constant per molecule and is the Boltzmann constant (k = kB).
2 3 3
1
2 2 2
A
R
m c T kT
N
 
2 3
1
2 2
. .
K E permolecule m c kT
 
NA = 6.023  1023
R (Molar gas constant) = 8.31 J/mol/K
k (Boltzmann constant) = 1.38  1023 J/K
We had noted before that:
(5)
(6)
2 2 2 2
3 ( 3 3 )
c v u w
  
Hence, the kinetic energy associated with one ‘degree of freedom’ (DOF) is:
2
1 1
2 2
K.E. per moleculeperdegree of freedom m v kT
 
Continued…
2 2
1
3
3
1
3

Nm
P c c
L
 
 
 
 
(3)

40. Degrees of freedom associated with a molecule (& contributions to the kinetic energy)
• In the discussions so far, the molecule was mono-atomic (like a hard ‘point-like’ sphere). Hence,
the only contribution to the kinetic energy is the translational motion. This true for molecules
like Ar & Xe. Such molecules do not have rotational DOF.
• Molecules like H2 and H2O have additional degrees of freedom at the molecular level, which can
contribute to the kinetic energy.
• These molecules can vibrate and rotate. Hence, contributions to the K.E. and hence the internal
energy (U) arise from these DOF. I.e. the internal energy of a polyatomic molecule is shared
between translation, vibration and rotation. We will ignore vibration for now.
Contributions to the K.E. (& hence Internal Energy (U) Translation + Vibration + Rotation
• Let us consider H2 (a liner molecule) and H2O (an angular molecule).
• Let us further assume (due to Maxwell) that the K.E. per degree of freedom per molecule is ½kT.
This is referred to as the principle of equi-partition of energy. This principle is approximately
valid at ‘high’ temperatures.
• DOF of H2 (lying along the x-axis): rotation along y-axis & z-axis.
• DOF of H2O rotations along all 3 axis.
  3
1
2 2
K.E. monoatomic molecule (3 DOF) 3 kT kT
 
3 5
2
2 2 2
K.E. atomic molecule (3 tranlation+2 rotation DOF)
Translation Rotation
di kT kT kT
  
3 3 6
2 2 2
K.E. atomic molecule (3 tranlation+3 rotation DOF)
Translation Rotation
poly kT kT kT
  
Continued…

41. Internal Energy (U)
3 3
2 2
( ) ( . . / )
Monoatomic A A
Internal Energy U N K E molecule N kT RT
    
• The kinetic theory of gases, assuming an ideal gas, gave us the kinetic energy (K.E.) of
monoatomic, diatomic and polyatomic molecules.
• The internal energy of an ideal gas is the K.E. of a mole of gas molecules (as there are no other
contributions to U*).
* Internal Energy (U) in other systems can have contributions from the following.
 Molecular translational energy  Molecular rotational energy  Molecular vibrational energy
 Energy due to electronic states  Interaction energy between molecules
 Relativistic rest- mass energy (mc2) arising from electrons and nucleons.
5 5
2 2
( ) ( . . / )
Diatomic A A
Internal Energy U N K E molecule N kT RT
    
6
2
( ) ( . . / ) 3
Polyatomic A A
Internal Energy U N K E molecule N kT RT
    
Note again that these are under the assumptions' of the kinetic theory of gases
Molar heat capacities (CV & CP)
• The molar heat capacity at constant volume (CV) is the heat required to increase the internal
energy of 1 mole of a gas through 1 K.
3
2
Monoatomic
U RT

5
2
Diatomic
U RT

3
Polyatomic
U RT

  3
2
Monoatomic
V Monoatomic
U
C R
T

 

 
P V
C C R
    3 5
2 2
P Monoatomic
C R R R
  
• The ratio of molar heat capacities (CP/CV ) is given the symbol  (known as the adiabatic index
or the isentropic expansion factor).
5
2
3
2
5
3
 P
Monoatomic
V
C R
C R
 
  
 
 

42. Q & A Why is CP greater than CV ?
 We have noted that: (which implies that CP > CV).
 This can be understood as follows.
 The amount of heat supplied at constant pressure is utilized for: (i) increasing the internal
energy (and hence temperature since internal energy is a function of temperature) and (ii) for
doing work.
 In contrast at constant volume, no work can be done. This implies that the heat supplied at
constant volume is utilized only for increasing the internal energy.
 Hence, more heat has to be supplied at constant pressure and CP > CV.
 
P V
C C R
 

43. Mean Free Path (MFP, )
• The ‘average’ distance travelled by a gas molecule before suffering a collision is called the MFP
().
• Let: (i) there be ‘nV’ molecules of gas per unit volume, (ii) the radius of the molecule be ‘r’
(diameter ‘d’); then the MFP is given by the formula below (we do not derive it here).
• If ‘n’ is the number of moles of the gas, then:
2 2
4 1
2 2

 
V V
n r n d
 
• This implies that MPF decreases with increasing ‘n’ or the pressure.
• Using data for H2, r = 3  1010 m*, at STP (1 bar, 0C) 6.023  1023 molecules occupy 22.4 L,
we get:
  
3
8
23 20
22.4 10
9 10 90
2 6.023 10 9 10


m nm




   
  An extremely short distance!
• We have seen earlier that the mean speed is 1840 m/s (~2000 m/s) and the number of collisions
per second is:
10
8
2000
Collisionspersecond = 10
9 10

mean speed

 

A really large number!
* Noting that hydrogen molecule is not spherical!
For an ideal gas 2
1
2

 A
RT
N P
d

A
V
n N
n
V


44. Distribution of molecular speeds
• The distribution of the no. of molecules with speed ‘c’ (N(c)) as a function of ‘c’ is given by the
Maxwell-Boltzmann distribution (1).
• The no. of molecules N with speeds in the range c & (c + c) is the area of the shaded portion
of the curve. (Fig.1). N = N(c) c.
• The following quantities are marked in the figure: (i) the most probable speed (c0), (ii) the mean
speed (cm) & (iii) RMS speed (cr).
• For a Maxwell-Boltzmann distribution: c0 : cm : cr = 1.00 : 1.13 :1.23.
• The RMS speed is related to macroscopic gas properties like pressure and specific heat capacity.
The mean speed is related to macroscopic gas properties like diffusion through porous partitions.
Continued…
2
3
2
2
2
( ) 4
2


mc
RT
m
N c c e
RT
 

 
 
 
 
  
 
0
2RT
c
m

8

m
RT
c
m

3
r
RT
c
m

(1)

45. • We had noted earlier (eq.(4)) that RMS speed is an inverse function of the density (and hence the
atomic mass). Hence, molecules of lighter gases have (crudely speaking) a higher speed on the
average (i.e. the Maxwell-Boltzmann distribution is shifted to higher speeds).
• However, the pressure exerted (P1) by n1 moles of these gases at a given T1 is same in a constant
volume (V1) container.
2 3

P
c  (4)
T = RT
1 1
1
1
n RT
P
V

What is the relevance of the fact that there is a distribution of speeds?
Funda Check
 There are many important implications of the fact that there are molecules “hotter”* than the average molecule (and similarly “colder”*).
 If all molecules had the same KE, then we have to heat (say water) above the boiling point to take water to the vapour state. However, given the
distribution, some of the molecules have a higher speed to escape the water surface and hence evaporation can occur at lower temperatures
(than the boiling point).
* We already know that hot and cold cannot be defined for ‘some/few molecules’ here hotter implies “faster” molecules.

46.  Case-A. Isobaric process. Constant Pressure Process. Let an ideal gas at P1 & T1 be enclosed
in a chamber with a frictionless movable piston (of initial volume V1). Let the piston be
loaded with sand giving rise to an external pressure Pext = P1. This is initial state of the sytem
in equilibrium (S1).
 Let us heat the system to a temperature T2 (very gradually), such that the gas expands inside
the chamber to a new volume V2. In this new state (S2) the Pext = Pint = P1 (due to
equilibrium).
 Due to the expansion the system does mechanical work on the surrounding, which is the area
under the P-V curve (= P(V2  V1) = PV). As the temperature has changed the internal
energy of the system has changed (by U).
PV diagrams
Closed
System
Ideal
gas
Pext = P1
1 1 1 2
1 2
P V P V
T T
 1 2
2
1
V T
V
T

U Q W Q P V
     
State-1
(S1)
State-2
(S2)
N, T1, P1, V1
Closed
System
Ideal gas
Pext = P1
N, T2, P1, V2
 In the current case (as we go from
S1 to S2) the work is done In the
current case (as we go from S1 to
S2) the work is done by the
system on the surrounding. Hence,
as far as the system goes, it is
negative.
 If the gas is compressed (i.e. we
go from S2 to S1), then work is
done on the system and PV term
is positive.
As work is done by the system
Many important concepts and processes in TD can be understood using P-V (PV) diagrams.
N, T1, P1, V1
N, T2, P1, V2
Heat reservoir
Dia-thermal
walls
Q Q is given throughout
the process and not in
state-2

47. N, T1, P1, V1
N, T2, P2, V2
 Case-B. For an arbitrary process in a closed system from S1 to S2, the work done by the
system during the process can be computed by dividing the area under the curve into small
parts, with the assumption that during the ‘small’ change in volume by V, the P remains
constant. Hence, the work for each segment is given by:
Wi = Pi V.
1
n
i
i
W P V

 

The net work is the sum of all the rectangular areas:
2
1
V
V
W PdV
 
For infinitesimal areas the summation is
replaced by integration:

48.  We had written the first law as:
 We used a positive (+) sign for W. In this sign convention, ‘anything’ (Q or W) which goes on
to increase the internal energy of the system is given a positive sign.
 The opposite sign convention for work is also found in literature.
 From S1 to S2 (expansion of the system), work is being done by the system (the sign of W is
negative, i.e. W is a negative quantity), while if we go from S2 to S1 (compression of the
system), work is being done on the system and hence W has a +ve sign.
Work done by the system versus work done on the system.
Funda Check
U Q W
  
The colour coding is w.r.t to the effect on the internal energy
0
2RT
c
m

Let the pressure inside the cylinder be:
P = PInternal  PExternal = PResisting
Let the decrease in the volume be V
(i.e. V is a negative quantity).
.Work done on the system is:
WReversible = P.V
This is a positive quantity as V is negative.

49.  Case-C1a. Isothermal process. Let us consider a system at constant temperature (T) with N
molecules (particles) of an ideal gas. Let the container (the system) have two chambers: C1
and C2 (each of V1 (Fig.1)). All the ideal gas is in C1 and C2 is under vacuum in state-1 (S1).
Now let the partition ‘vanish’, such that the gas expands into C2 also (with total volume
2V1).
 In the new state (S2) the volume of the system is 2V1 (for simplicity we have assumed 2V1  in general can be
any V2), the temperature is T (which is maintained via contact with a heat reservoir heat is transferred from the
reservoir into the system). As the number of molecules (N) & T of the system is constant, the
kinetic energy (velocity) of the molecules is constant. However, since the volume has
doubled, the molecules will hit the walls of the container less frequently and hence a
pressure would be lower.
 Note that we cannot draw a path from S1 to S2, as the system is not under equilibrium
between S1 and S2 (Fig.2). Macroscopic thermodynamic variables like P & T are not defined (and so too V), during the expansion.
T
V1
N
State-1 (S1) State-2 (S2)
T
V1
0
T
2V1
N
Partition
Fig.1
Fig.2
1 1 2 2 2 1
2
P V P V P V
T T T
 
1
2
2
P
P

Between S1 and S2 (after the partition vanishes) the gas is
expanding, the P is falling and the T is tending to fall; but in
parallel, heat is flowing into the system in an attempt to maintain
the temperature. The system is in transient condition from S1 to S2.
N, T1, P1, V1
N, T1, P2, 2V1
Heat reservoir
Dia-thermal
walls
Q
Continued…
PV = nR T = Constant
C
P
V

This is like y = 1/x
(asymptotic to both
x & y axis)

50. Case C1: isothermal process
During the isothermal process
0
T
  0
U
 
( ) 0
PV
 
0
U Q W
   
We can generalize the process for expansion from V1 to V2
 
into the system work done by the system
Q = W
Closed
System
Ideal
gas
Pext = P1
State-1
(S1)
State-2
(S2)
N, T1, P1, V1
Closed
System
Ideal gas
Pext = P2
N, T1, P2, V2
Heat reservoir
Dia-thermal
walls
Q
 Case-C1b. The system previously considered can be visualized as that with a piston. S1: N, T1,
P1, V1. S2: N, T1, P2, V2. We go from S1 to S2 via removal of sand grains gradually, such that
the system is always under equilibrium. The temperature of the system will tend to fall due to
the expansion, but heat transfer from the heat reservoir (via the dia-thermal walls) will
maintain the temperature. In this process there are no transients and hence macroscopic TD
variables like P, T & V are defined throughout the process. (Fig.2).
 As T is constant, U is constant and U = 0. The work done can be calculated using the first law
as below, which is the area under the PV curve.
Gradually remove
sand grains
(PV) = nR T = Constant
isothermal
N, T1, P1, V1
N, T1, P2, V2
2
1
V
V
W PdV
 
1 1 2 2
1 1
P V P V
T T
 PV nRT
 nRT
P
V

2
1
V
V
nRT
W dV
V
 
  2
1
ln( )
V
V
W nRT V

2
1
ln
Isothermal
reversible
process
V
W nRT
V
 
 
  
 
 
 
Hence we can use
the ideal gas equation
at each point on the curve
Fig.1
Fig.2
Q W

This is the work done by the
system (a negative quantity
as far as the system goes)

51.  Case-C2. Isothermal process. Let us consider a system at constant temperature (T). If volume
of a container (the system) reduces (say ‘magically’ to half its original volume, i.e. from V1 to V2 (= V1/2)), the
pressure will increase. As the number of molecules (N) & T of the system is constant, the
kinetic energy (velocity) of the molecules is constant. However, since the volume has
reduced, the molecules will hit the walls of the container more frequently and hence a higher
pressure. This is similar to the previous case, but going from S2 to S1.
 Again in this case, between S1 and S2 the system is in transients and hence macroscopic
thermodynamic variables like P & T are not defined (and so too V).
T, V1, N
T
V1/2
N
State-1 State-2
 Case-C3. Isothermal process. Let us consider two
isothermal processes at two different temperatures.

52.  Case-D. Isochoric process. Let us consider a system at a temperature T1 with N molecules
(particles) of an ideal gas. Let the container (the system) have a volume V1 which is constant
throughout the process (Fig.1)).
 Let heat Q be transferred gradually to the chamber, such that the system is under equilibrium
throughout. This will lead to an increase in the temperature (gradually to T2). The molecules
of the gas will impinge with higher velocity on the walls, which will increase the internal
pressure. The system will try to expand this will have to be countered (to maintain constant volume) via
gradual addition of sand grains (to increase the external pressure and maintain equilibrium).
 The states are as follows. S1: N, T1, P1, V1. S2: N, T2, P2, V1.
 The area under the curve (Fig.2) is zero, hence NO PV work is done during the process. All
the heat Q goes into an increase in the internal energy (as seen by an increase in the temperature of the system).
 Another way to visualize the process is to introduce Q gradually into a rigid container with dia-thermal walls.
Closed
System
Ideal
gas
Pext = P1
State-1
(S1)
N, T1, P1, V1
Heat reservoir (at T2)
Dia-thermal
walls
Q
Gradually add
sand grains
Closed
System
Ideal
gas
Pext = P2
N, T2, P2, V1
State-2
(S2)
U Q W Q P V
      1 1 2 1
1 2
P V P V
T T

U Q
 
Fig.1
Fig.2
N, T2, P2, V1
N, T1, P1, V1
1 2
1 2
P P
T T


53.  Case-E. Adabatic process. Let us consider an insulated & closed system with an ideal gas at
S1: N, T1, P1, V1. (Fig.1)). The walls are insulated.
 Let us increase the volume of the system by moving the piston (by removing grains of sand) to expand the
gas to a volume V2. Since the walls are insulated, no heat can enter the system to equilibrate
the temperature (Q = 0). This implies that molecules will impinge on the walls less
frequently and the pressure will drop. The system has done work in the expansion, which
will lead to a decrease in the internal energy (and the T). Hence, U is negative. The work
done by the system is the area under the PV curve. S2: N, T2, P2, V2.
 If this is compared with an isothermal expansion to V2, it will be seen that P2 (adiabatic) > P2
(isothermal). (Fig.2)
 The adiabatic process can be visualized as jumping from one isotherm to another. And as the T is changing during the
process, we cannot determine the work using the relation PV = C. (Fig.3).
U Q W
  
Closed
System
Ideal
gas
Pext = P1
State-1
(S1)
State-2
(S2)
N, T1, P1, V1
Closed
System
Ideal gas
Pext = P2
N, T2, P2, V2
Gradually remove
sand grains
Insulated walls
N, T1, P1, V1
N, T2, P2, V2
Fig.1
Fig.2
Fig.3
Continued…
U W
 

54. 2
1
V
V
W PdV
 
Work done in an adiabatic process
1 1 2 2
Constant = C
  
PV P V P V
  
2
1

V
V
C
W dV
V
 
2
1
(1 )
(1 )


V
V
V
W C

 
  

 
(1 ) (1 )
2 1
(1 )
 

V V
W C
 
 

  

 
(1 ) (1 )
2 1
(1 )
 

CV CV
W
 
 

  

 
   
(1 ) (1 )
2 2 2 1 1 1
(1 )
   

P V V PV V
W
 
 

  

 
 
   
2 2 1 1
(1 )

Adibatic
P V PV
W

 
  

 
 Work done is the area under the PV curve. During the expansion this is the work done by the
system, leading to the reduction in the internal energy.
For an adiabatic process the equation of state is:
Here  is ratio of the specific heats γ = CP /CV. It is a factor which determines the speed of
sound in a gas. (γ = 1.66 for an ideal monoatomic gas and γ = 1.4 for air, which is
predominantly a diatomic gas).
Using Eq.(1), with two different combinations of PV:
Eq.(1)
U W
  

55. A comparison of the 4 processes (isobaric, isothermal, isochoric &adiabatic) on the PV diagram
P = 0
U = 0
T = 0 (PV) = 0
V = 0
Q = 0
2
1
ln
Isothermal
V
W nRT
V
 
 
  
 
 
 
   
2 2 1 1
(1 )

Adibatic
P V PV
W

 
  

 
0
Isochoric
W 
Isobaric
W P V
 
 P
V
C
C

U Q
 
U W
 
Q W

U Q P V
   
Conditions
Work done
First Law
Q
S
T
 
(S1) N, T1, P1, V1
N, T2, P2, V2
N, T1, P2, V2
N, T2, P2, V1
N, T2, P1, V2
dU TdS PdV
 
(PV) = Constant
PV = Constant
 The adiabatic curve (adiabat) lies below the isotherm; as, the adiabat follows PV (=C), while
the isotherm follows PV (=C). [  > 1, hence the adiabat falls more steeply as compared to the isotherm].
 Suppose that the volume doubles in an (i) isothermal and (ii) adiabatic process. Let  = 5/3 and let the inital pressure be 100 bar. Then:
2 2 1 1
( ) :
ii Adiabatic P V P V
 

2 2 1 1
( ) :
i Isothermal P V P V

5/3
2
1
100 32
2
Adiabatic
P bar
 
 
 
 
2
1
100 50
2
Isothermal
P bar
 
 
 
 

56.  18
T-S diagrams

57.  The Carnot cycle is a conceptual ideal thermodynamic cycle (due to Nicolas Léonard Sadi
Carnot). The cycle effectively does one of the following:
(M1) use transfer of heat from a hot source to a cold source to produce work (act like a work
engine) or
(M2) use work as an input to transfer heat from a cold source to a hot source (act like a heat
engine of refrigerator) and gives the upper limit on the efficiency that any classical
thermodynamic engine can achieve.
Carnot Cycle
Fig.1
Details are in the upcoming slides
 The cycle can be represented in a PV diagram
and consists of four steps (Fig.1); such that
start and finish states are the same. In the ‘work
engine mode’a complete cycle consists of:
(P1) Isothermal Expansion,
(P2) Adiabatic Expansion,
(P3) Isothermal Compression,
(P4) Isothermal Compression.
 At the end of the cycle the internal energy (U)
remains unchanged (as it is a state variable and we are back
to the same P1 & V1).

58.  In M1 (clock-wise operation), the combined effect of the 4 processes (P1 to P4) is to transfer heat
from a hot source (Q1, +ve magnitude) to a cold sink (Q2, ve magnitude) and to produce work.
The amount of work produced is given by the area enclosed by the curve.
 In P1 and P2 work is done by the system on the surrounding (W1 & W2), while in P3 and P4 work is
done on the system (W3 & W4) by the surrounding.
 In the clockwise cycle, working as a work engine. As |[(W1) + (W2)]| > | (W3 + W4)|. The net effect is that the
system does (produces) work during the clockwise operation.
 The net effect of the cycle is to covert (Q1  Q2) amount of heat into [W1 W2 + W3 + W4]
amount of work.

59.  (P1) Isothermal expansion at T1 (= Th). The system takes in Q1 amount of heat and does work
on the surrounding of W1. Being an isothermal process, the internal energy remains unchanged
(U = 0). The heat input (Q1) can be conceived as coming from a hot bath at T1.
 (P2) Adiabatic expansion. No heat is exchanged with the surrounding. The system does work
(W2) on the surrounding and this comes at an expense of the internal energy (decreases, U <
0).
 (P3) Isothermal compression at T2 (= Tc). The system rejects Q2 amount of heat and the
surrounding does work on the system of W3. Being an isothermal process, the internal energy
remains unchanged (U = 0). The rejection of heat (Q2) can thought of going to a cold bath at
T2.
 (P4) Adiabatic compression. No heat is exchanged with the surrounding. The surrounding does
work on the system (W4) and this leads to an increase in the internal energy (U > 0).
P1 & P2 are expansions, while P3 & P4 are compressions
 The net effect of the cycle is the following.
 Take heat Q1 from a hot source and reject Q2 to a cold sink.
 Produce [W1 W2 + W3 + W4] amount of work.
 Lose internal energy in P2 (adiabatic expansion) and gain the
same in P4 (adiabatic compression), such that the net change in
internal energy is zero.
 Gain entropy during the isothermal expansion and lose an equal
amount of entropy during the isothermal compression.

60. P1
Q1
U- W1
P2
U
S1
P3
Q2
U-
P2
U
W4
S2
S3
S4
Q1 > Q2
P1
Q1
U- W1
Hot source
Cold Sink
Work done at each step is the area under each one of these curves
P1
P2
P3
P4
P3
Q2
U- W3
Hot source
Cold Sink
W2
W3
W4
W1
W2
W3
P2
U
W2
P4
U
W4
• The source and sink are both reservoirs, whose temperature does not change
by the addition or removal of Q.
• The internal energy lost in P2 is the same as that gained in P4. 2 4
P P
U U
  
• Since work is involved in all the steps, this implies that we must have some
kind of piston connected to the system (not shown).

61. Carnot Cycle: T-S diagrams
 The Carnot cycle can be drawn with T & S as axes. Clearly, the isothermal sections are
horizontal.
S1 S2
S3
S4
 The interesting point to note is that the adiabatic
processes are isentropic and hence vertical.
 During the adiabatic process, the T is changing. If we
consider an infinitesimal section where T is constant (say
Ti), then Si = Qrev/Ti. Qrev = 0, for the entire adiabatic
process and for each section. Hence, Si = 0. Hence, the
adiabatic processes are vertical in the T-S diagram.
1
1 2
1
S S
Q
S
T

 
During the isothermal
expansion (S1S2):
U = 0 Q W

Q W T S
  
Which is the area under the curve
Note: symbol S is being used for
both state and entropy (pl do not
get confused)
S1 S2
1 2 1 2 1
( )
W T S S
  
From S3S4 the work done is the
area as shown
S3
S4
Negative quantity
Negative quantity
2
3 4
2
S S
Q
S
T

 
2
1
1 2 3 4
1 2
S S S S
Q
Q
S S
T T
 
    

62. The Carnot cycle (operating as a work engine) can be compact drawn as below (Fig.1).
The cycle operating as a heat engine can be drawn as Fig.2. The heat engine is also called a heat pump. Terms body
and reservoir are used interchangeably (essential the properties, including T of the reservoir does not change).
Heat reservoir
Heat Q1
Work (W)
Cold Reservoir
Heat Q2
Sink
Source
Cyclic engine
Hot body
Heat Q1
Work (W)
Cold Body
Heat Q2
Cyclic engine
‘Heat engine’
Like a refrigerator
‘Work engine’
Like a steam engine
Fig.1
Fig.2

63.  The efficiency of a heat engine is the amount of work output divided by the amount of heat
input.
 The efficiency can be calculated using the equation of the first law applied to the whole
cycle.
The efficiency of a heat engine
Nicolas Léonard Sadi Carnot in 1824.
“Reflections on the Motive Power of Fire”, Chapman & Hall ltd, London, 1897.
* Not shown here.
1
 net
Carnot engine
Output W
Input Q
  0
cycle net net
U Q W
   
By definition 2 1
net net
Q Q Q W
  
2
1 2
1 1 1
1
 net
Carnot engine
Q
W Q Q
Q Q Q

   
2
1
ln
Isothermal
V
W nRT
V
 
 
  
 
 
 
As seen earlier, the work done during an isothermal process is:
3
2
4
2
1
1
ln
1
ln
Carnot engine
V
RT
V
V
RT
V
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
3 2
4 1
V V
V V

Now*: Hence: 2
1
1
Carnot engine
T
T
 
   
 
max sink
source
1
heat engine
T
T

 
  
 
output
heat engine
input
w
q
 
Continued…

64.  This is surprising as:
 there is no mention of the medium of the system (or its properties),
 the formula has only temperatures and
 the temperature of the sink seems to play a major role (as the presence of the sink is
usually not intentional or obvious→ in a steam engine sink is the air around the engine and
source is the hot steam).
Important message Sink (characterized by its temperature) is as important as the source.
 To increase the maximum possible efficiency of a heat engine, either the temperature of the
source has to be increased on the temperature of the sink has to be decreased.
2
1
1
Carnot engine
T
T
 
   
 
max sink
source
1
heat engine
T
T

 
  
 
The efficiency of a heat engine

65.  Let us assume there exists an Engine E2 which is of higher efficiency (Fig.2) than the Carnot
Work Engine (CWE) (Fig.1). This implies that this engine can produce more work (say
W(1+x)) than the CWE.
 Let us couple this higher efficiency E2 with a CHE* (reverse of CWE), which needs W(1+x)
amount of work as input (Fig.3). In this engine all the variables are scaled up by (1+x).
 The net effect of such an coupled system (Fig.4) is to transfer heat from a cold source to a hot
source (by an amount Q1 x) without any other influence. This violates the second law.
The Carnot engine is the most efficient engine
Hot
Q1
W
Cold
Q2 = Q1  W
CWE
Fig.1
Hot
Q1
W(1+x)
Cold
Q2 = Q1  (W(1+x))
E2
Hot
Q1 (1+x)
W(1+x)
Cold
Q2 = (Q1  W)(1+x))
CHE
Fig.2 Fig.3 Fig.4
Hot
Q1
W(1+x)
Cold
Q2 = Q1  (W(1+x))
E2
Hot
Q1 (1+x)
Cold
Q2 = (Q1  W)(1+x))
CHE
Fig.4
Hot
Cold
Q1 x
E2+CHE
Q1 x
Hence, an engine with an efficiency higher
than the Carnot engine is not possible.
* Carnot Heat Engine (CHE)

66. What is the work done for an arbitrary closed cycle?
Funda Check
 Let us consider an arbitrary path from S1 to S2 and
back to S1, forming a closed cycle.
 The work done for the path from S1 to S2 is the area
under the curve (Eq.1).
 The work done during the cyclic process is the area
enclosed by the curve.
2
1
1 2 ( )
V
S S
V
W P V dV
   (1)
What is the entropy change for an arbitrary closed cycle?
 Entropy is a state function and hence for a close cycle (like the Carnot cycle) the entropy
change is ZERO.
0
rev
closed cycle
closed cycle
dQ
S
T
 


68.  Heat capacity is the amount of heat (measured in Joules or Calories) needed to raise an unit
amount of substance (measured in grams or moles) by an unit in temperature (measured in
C or K). As mentioned before bodies (systems) contain internal energy and not heat.
 This ‘heating’ (addition of energy) can be carried out at constant volume or constant
pressure. At constant pressure, some of the heat supplied goes into doing work of expansion
and less is available with the system (to raise it temperature).
 Heat capacity at constant Volume (CV):
It is the slope of the plot of internal energy with temperature.
 Heat capacity at constant Pressure (CP):
It is the slope of the plot of enthalpy with temperature.
 Units: Joules/Kelvin/mole, J/K/mole, J/C/mole, J/C/g.
 Heat capacity is an extensive property (depends on ‘amount of matter’)
 If a substance has higher heat capacity, then more heat has to be added to raise its
temperature. Water with a high heat capacity (of CP = 4186 J/K/mole =1 Cal/C/Kg) heats
up slowly as compared to air (with a heat capacity, CP = 29.07J/K/mole)  this implies that
oceans will heat up slowly as compared to the atomosphere.
 As T0K, the heat capacity tends to zero. I.e near zero Kelvin very little heat is required to
raise the temperature of a sample. (This automatically implies that very little heat has to
added to raise the temperature of a material close to 0K.
This is of course bad news for cooling to very low temperatures small leakages of heat will lead to drastic increase in temperature).
Heat Capacity
V
V
E
C
T

 
  

 
P
P
H
C
T

 
  

 

70. The Laws of Thermodynamics
 There are 4 laws of thermodynamics. Like any of the laws, there are generally found to be
true (i.e. no violation of the laws have been found so far); but, there is no proof for any of the
laws.
 The laws are numbered from the 0th to the 3rd (for historical reasons).
 The second law is considered as one of the most (if not the most) profound laws of nature. it
introduces the concept of Entropy.
 The second law further gives us the direction for the spontaneity of a process. Applied to the
universe at large, it gives us the arrow of time and Cosmological concepts like the ‘heat-death
theory’ (of the Universe).
 The third law sets the scale for entropy.

71.  The internal energy of an isolated system is constant. A closed system may exchange energy
as heat or work. Let us consider a close system at rest without external fields.
 There exists a state function U such that for any process in a closed system:
U = Q + W [1] (For an infinitesimal change: dU = (U2  U1) = dQ + dW)
 Q → heat flow into the system.
 W → work done on the system (work done by the system is negative of above- this is just ‘one’ sign convention).
 U is the internal energy. Being a state function for a process U depends only of the final
and initial state of the system. U = Ufinal – Uinitial. Hence, for an infinitesimal process it can be written as dU.
 In contrast to U, Q & W are NOT state functions (i.e. depend on the path followed).
 q and w have to be evaluated based on a path dependent integral.
 For an infinitesimal process eq. [1] can be written as: dU = dQ + dW.
 The change in U of the surrounding will be opposite in sign, such that:
Usystem + Usurrounding = 0
 Actually, it should be E above and not U {however, in many cases K and V are zero (e.g.
a system at rest considered above) and the above is valid- as discussed elsewhere}.
 It is to be noted that in ‘w’ work done by one part of the system on another part is not included.
 The experimental foundation of the first law is the Joule’s demonstration of the equivalence between heat and mechanical energy.
The First Law
* Depending on the sign convention used there are other ways of writing the first law:
dU = dq  dW, dU = dq + dW

72.  Previously, we had dealt with the concept of state functions and path functions. Internal energy
(U) (& H, A, G, S) are state functions and the change between two states is not dependent on
the path of the process. E.g. if we go from state ‘1’ to state ‘2’, the change in the internal
energy (U) is given by:
A note on exact and in-exact differentials
2
1
U dU
   And we refer to dU as an exact differential*
* An exact differential is an infinitesimal quantity which upon integration gives a result, that is independent of the path taken.
2
1,path
Q dQ
 
 In contrast, the Q and W for a process are path dependent.
 
2 1
Q Q Q
 
Here, dQ as an in-exact differential and hence the LHS is Q and not Q
2
1,path
dQ Q
 


73.  It is impossible to build a cyclic machine* that converts heat into work with 100%
efficiency  Kelvin’s statement of the second law.
 Another way of viewing the same:
it is impossible to construct a cyclic machine** that completely (with 100% efficiency)
converts heat, which is energy of random molecular motion, to mechanical work, which is
ordered motion.
 The unavailable work is due to the role of Entropy in the process.
The Second Law
Heat reservoir Cyclic engine
Heat q
Work (w)
100%
Not possible
Heat reservoir Cyclic engine
Heat q
Work (w)
Cold Reservoir
Heat q’


Kelvin’s
statement of the
second law
The second law comes in many equivalent forms
* For now we are ‘building’‘conceptual machines’!
** These ‘engines’ which use heat and try to produce work are called heat engines.
Called the sink
Possible
Called the source G

T S

H


74. Hot body
Cold body
 Heat does not ‘flow*’ from a colder body to a hotter body, without an concomitant change
outside of the two bodies Clausius’s statement of the second law.(a)
 This automatically implies that the spontaneous direction of the ‘flow of heat*’ is from a
hotter body to a colder body.(b)
 The Kelvin’s and Clausius’s statements of the second law are equivalent. I.e. if we violate
Kelvin’s statement, then we will automatically violate the Clausius’s statement of the
second law (and vice-versa).
Another statement of the second law → the Clausius statement
* Used here in the ‘common usage sense’.
(b) is obvious, but not (a) → though they represent the same fact.
Not possible
Hot body
Cold body
Spontaneous flow not possible
Work
G

T S

H

 Heat cannot spontaneously flow from a cold (low temperature) body to a hot body.
 To make heat flow from a cold body to a hot body, there must be accompanying change elsewhere (work has to be done to achieve this).

75.  The entropy* of a closed system will increase during any spontaneous change/process.
If we consider the Universe to be a closed system (without proof!!)**, then,
 the entropy of the universe will increase during any spontaneous change (process).
A combined (Kelvin + Clausius) statement of the II Law
* Soon we will get down to this mysterious quantity.
** For all we know the Universe could be ‘leaky’ with wormholes to other parallel Universes!
You may want to jump to chapter on
equilibrium to know about Entropy first
Entropy sets the direction for the arrow of time !
 The universe can be thought of as an isolated system. The entropy of an isolated system will
increase during an spontaneous process.

76. Heat reservoir
Heat q
Work (w)
Cold Reservoir
Heat q’
Sink
Source
Q & A What is the difference between ‘heat engine’ and ‘work engine’?
 Actually both the engines we are going to describe here are usually known as heat engines.
 We are differentiating two types of engines to see which one produces work and which one
actually transfers heat.
 In the heat engine as the temperature of the cold body tends to zero Kelvin, more and more
work has to be done to transfer the heat from the cold body to the hot body.
Cyclic engine
‘Work engine’
Hot body
Heat q
Work (w)
Cold Body
Heat q’
Cyclic engine
‘Heat engine’
The main objective here
is to produce work
The main objective here is
to transfer heat from a
cold body to a hot body
Like a steam engine Like a refrigerator

77.  For substances in internal equilibrium, undergoing an isothermal process, the entropy
change goes to zero as T (in K) goes to zero.
The Third Law
0
lim 0
T
S

 
 The law is valid for pure substances and mixtures.
 Close to Zero Kelvin, the molecular motions have to be treated using quantum mechanics
→ still it is found that quantum ideal gases obey the third law.
Phenomenological description of the third law.
 There does not exist any finite sequence of cyclical process, which can cool a body to zero
Kelvin (absolute zero).
Other statements of the third law.
 For a closed system in thermodynamic equilibrium, the entropy of the system approaches a
constant value as the temperature goes to absolute zero.
 If there is a unique ground state with minimum energy at zero Kelvin, then the entropy at
zero Kelvin is ZERO. However, if there is a degeneracy with respect to the number of
microstates at absolute zero, then there will be some Residual Entropy.

78.  The first law says: “you cannot win”.
 The second law says: “you can at best break even- that too at zero Kelvin”.
 Third law says: “zero Kelvin is unattainable”.
Humorous look at
the three laws
Q & A What is the difference in the ‘status’ of quantities like T, U, S, H, A & G?
 T, U & S are fundamental quantities of thermodynamics.
 H, A & G do not give us new fundamental concepts, but are for better ‘accounting’ in
thermodynamics.

80. Chemical Thermodynamics
Some ‘thingamajigs’ we will encounter:
 Material Equilibrium
 Phase equilibrium
 Reaction equilibrium
o Gibbs Equations
 Non-equilibrium systems
 Chemical Potential
* Etymology. [Ergon (Greek) = Work] + [Hodos (Greek) = Way] = [Ergoden (German)] + [ic (English)] = [Ergodic]  a dynamical systems
term
Equilibrium and Non-equilibrium Conditions

81.  Nothing will happen if a system is in equilibrium. This implies that non-equilibrium
conditions are required for a process to occur.
 The first step is to consider processes in closed systems, wherein the system is close to
equilibrium conditions these are the reversible processes. Also, we restrict ourselves to P-V
work only.
 Later on we can consider open systems and non-equilibrium processes which are not
reversible (i.e. the process occurs irreversibly and hence is not close to equilibrium).
Equilibrium and Non-equilibrium Systems
 First, using the first law and the definition of entropy we derive the equation for dU.
 The using definitions for H, A, G, CV and CP we get set of 6 basic equations.
 Reversible composition changes are included in the processes (e.g. when we heat a two phase
mixture and the proportions of the two phases changes).
 The closed system condition precludes the introduction of additional species, leading to a
change in the composition.
 Reaction of species, leading to a composition change is also not included in the above. This is
because chemical reactions occurs spontaneously, which implies that the system is not under
equilibrium.
Reversible processes passing through equilibrium states in a closed system
 Next, akin to that for U (i.e. dU), we will derive expressions for dH, dA and dG. Collectively
these are known as the Gibbs equations.

82.  A system undergoing a reversible process passes through equilibrium states (only).
Thermodynamic Relations for a System in Equilibrium
dU dQ dW
 
The first law for a closed system is:
rev
dW dW P V
   
For only P-V work done reversibly
Again, for a reversible process (second law)
rev
dQ dQ TdS
 
From (1-3) (combining the first and second laws) dU TdS P V
  
(1)
(2)
(3)
(e1)
Definition of H H U PV
  (e2)
Definition of A A U TS
  (e3)
Definition of G G H TS
  (e4)
Definition of CV V
V
U
c
T

 
  

 
(e5) Close system in equilibrium, P-V work only
Definition of CP P
P
H
c
T

 
  

 
(e6) Close system in equilibrium, P-V work only
We are assuming here that work is done by
the system & hence the ve sign
 We have considered reversible processes, i.e. system is passing through equilibrium states. Reversible composition changes are included in
the processes (e.g. when we heat a two phase mixture and the proportions of the two phases changes).
 The closed system condition precludes the introduction of additional species, leading to a change in the composition.
 Reaction of species, leading to a composition change is also not included in the above. This is because chemical reactions occurs
spontaneously, which implies that the system is not under equilibrium.
(e1) to (e6) are basic
thermodynamic equations
of a closed system in
equilibrium

83.  We had derived the (e1) in differential form for U.
 Similar equations can be derived for dH, dA & dG.
 Collectively, e1, e7, e8, e9 are known as the Gibbs equations.
The Gibbs equations
dU TdS P V
   (e1)
( )
TdS PdV
dH d U PV PdV VdP TdS VdP
dU

       dH TdS VdP
 
( )
TdS PdV
dA d U TS TdS SdT SdT PdV
dU

        dA SdT PdV
  
( )
TdS VdP
dG d H TS TdS SdT SdT VdP
dH

        dG SdT VdP
  
(e7)
(e8)
(e9)

84.  The primary question is: “to what extent can we use concepts of equilibrium TD to non-
equilibrium systems?” or equivalently “to what extent can we assign definite values of TD
properties to non-equilibrium systems?”.
 In general systems, the non-equilibrium can arise from variations in (i) time or (ii) space
(location within the system) or both time and space. The variations could be in one or more of
the following.
(a) T (thermal non-equilibrium),
(b) P (mechanical non-equilibrium),
(c) Phase fractions (system not in phase equilibrium) or
(d) Chemical species (system not in chemical equilibrium).
 We can achieve the understanding of non-equilibrium systems (at least partial) in the
following steps.
C1 Assume that there is no spatial variation in quantities within a phase (i.e. any variations
are only in time).
C2 System is in T and P equilibrium, but not in phase equilibrium.
C3 System is in T and P equilibrium, but not chemical equilibrium.
C4 Consider system with spatial variations in quantities as well. (E.g. T, P, composition is
varying spatially within the system).
Non-equilibrium Systems

85. Q & A How can the composition of a system change?
 The Gibbs equations are not valid for open systems, wherein the composition can change due
to entry of matter.
 Composition can change due to any of the following.
P1 Entry of matter (open system).
P2 Inter-phase transport of matter (within the system).
P3 Chemical reaction.
 P1 above is easy to understand. E.g. if we introduce alcohol into a beaker with water.
 P2. Let us consider a two phase mixture in the micrograph below (Pb-Sn eutectic system*).
The phase in light contrast is a Pb rich phase (labelled as the -phase, with Sn dissolved in it).
The dark phase is Sn rich phase (labelled as the -phase, with Pb dissolved in it). Now if Pb
diffuses from dark phase to the light phase, then the transport of matter is within the system;
i.e. inter-phase transport of matter.
 P3. In a chamber with H2 and O2, if we introduce a catalyst, then the reaction will begin to
give rise to H2O, which is composition change due to chemical reaction.
* We will learn about this in the chapter on phase diagrams.
( )
dU TdS PdV terms
  
If P1, P2, P3 occur, then equations we considered
before have to be modified with additional term(s)

86.  Let us start with conditions C1 & C2.
 Consider a system under thermal and mechanical equilibrium (i.e. chemical reactions do not
occur at fast rate (explosively)). Assume that the composition within each phase is uniform
(i.e. the diffusion rate within each phase is rapid compared to the rate of transport of
components from one phase to another).
System NaCl Solution
U U U
  System NaCl Solution
S S S
 
 Let the partition vanish ‘magically’ (or equivalently is
removed reversibly and adiabatically, such that Q and
W are zero for the process, along with U and S).
 Once the partition is removed the system is not in phase equilibrium anymore and salt starts to
dissolve in the solution. (Assume that saturation of the solution does not occur during our discussions!).
 We consider first a system which is not in phase equilibrium (Fig.1). In this system a remove
able partition separates NaCl (crystal) from water (or unsaturated solution of NaCl in water).
 The internal energy and entropy of the system, which are extensive quantities, can be
obtained by adding the values for the two phases (salt and salt solution).
System Not in Phase Equilibrium
Fig.1
C1 & C2
C1 Assume that there is no spatial variation in quantities within a phase (i.e. any variations are only in time).
C2 System is in T and P equilibrium, but not in phase equilibrium.
C3 System is in T and P equilibrium, but not chemical equilibrium.
C4 Consider system with spatial variations in quantities as well. (E.g. T, P, composition is varying spatially within the system).

87.  In spite of the fact that the system is not in equilibrium, at any stage of dissolution we can
define USystem and SSystem. This can be understood as follows.
 At any point in time* (or any stage in the dissolution) we can freeze a frame (or equivalently
re-insert the thin partition in the ‘ideal’ way we removed it) and measure the amount of solid
and solution and hence compute U and S. (Noting that the solution has changed, while the solid is the same).
 This implies that, in spite of the non-equilibrium situation, we can use the concepts and
parameters of equilibrium TD (proviso we are able to visualize a sequence of equilibrated
steps).
 We consider first a system which is not in reaction equilibrium (Fig.2). In this system there is
mix of H2, O2 and H2O gases. The relative fractions of these gases will not change with time
at RT as H2 will not react with O2. We can introduce a catalyst to start the reaction, which will
lead to an increase in the amount of H2O at the expense of H2 and O2. (I.e. the composition of
the mixture will change with time).
 The quantities U and S can be calculated at the initial state. At any stage of the reaction if the
catalyst vanishes (magically) /is removed then the reaction can be stopped and U and S for
the system can be calculated for the composition at that ‘time’.
System Not in Chemical Equilibrium
Fig.2
* Unfortunately, there is no time in thermodynamics (it is the domain of kinetics)!
C1 & C3

88.  The two examples we considered (systems in phase and reaction non-equilibrium conditions),
serve to illustrate an important point.
 For systems in thermal and mechanical equilibrium (with well defined P, V & T) and such that
the composition of each phase is uniform, we can assign/determine U and S values. This is in
spite of the fact that the system is not in material equilibrium.
 For such systems, first we have to take an instant of time (freeze a frame of the video).
 Then, we will have to locate a small region (maybe a thin slice) where we can assume the T to
be constant. For the case of reactions, we have to assume that composition is constant in that
region of the system.
 We can then assign U and S values to these regions and then sum up the quantities to obtain S
and U for the whole system.
What about systems with varying T and composition?
Region-1 Region-2 Region-3
System
U =U +U U ...

C4
Region-1 Region-2 Region-3
System
G =G +G ...
G


89.  Let us consider a single phase system in thermal and mechanical equilibrium (but not material
equilibrium).
 If the system is made of ‘k’ components with ni moles of each component (i goes from 1 to
k), then the TD state of the system can be defined by a set of values for T, P and ni.
 As discussed before, in spite of the system not being in material equilibrium, we can assign
values to U and S. Further, since T, P, S and U are defined, so are other TD state functions (H,
A, G).
 The state functions U, H, A & G can be expressed as functions of T, P, ni.
Gibbs equations for non-equilibrium systems
( , , )
i
G G T P n

At any instant during the progress of a chemical process
(a ‘frame’ of the movie), we can write:
 Let T, P, ni (‘k’ values) change by infinitesimal amounts due to an irreversible process
(chemical reaction or transport of matter into the system). The changes will be: dT, dP, dn1,
dn2, ..., dnk.
 The aim is to compute dG for this irreversible process. However, keeping in view that G is
state function, we substitute the irreversible process with a reversible one and calculate dG for
this reversible process.
   
2 1
State State
Substitute
Irreversible Reversible
G G
dG dG
 



 Assume that the process occurring is a chemical reaction. To perform the process reversibly, first we stop the chemical reaction (say by
removing the catalyst). Then, we add dn1 moles of species ‘1’, dn2 moles of species ‘2’ and so forth. Further, we change the T by dT and P by
dP.
State-2 is only slightly different from State-1

90. 1
1
, , 1 , , , ,
...
i i j j k
Rev k
P n T n k
T P n T P n
G G G G
dG dT dP dn dn
T P n n
 
 
 
   
   
      
 
   
   
       
The change in G can be written as:
P & Composition is held constant T & Composition is held constant T. P & amount of all species except ‘1’ is held constant
We already know that for a reversible process where no change in composition occurs:
dG SdT VdP
  
, i
P n
G
S
T

 
 
 

 

, i
T n
G
V
P

 

 

 
1 , , j i
k
i
i i T P n
G
dG SdT VdP dn
n


 

     

 

Single phase
Open system
Reversible process
P-V work only
The term in the brackets has a special meaning it is referred to as the Chemical Potential.
Chemical Potential (i)
, , j i
i
i T P n
G
n


 

  

 
(1)
(2)
(3)
Using i Eq. (3) can be written as:
1
k
i i
i
dG SdT VdP dn


    
(4)
 The chemical potential is an important quantity in TD.
 It is the change in the Gibbs free energy of a system, when a small amount of a given species
(say ‘1’ or ‘A’) is added into the system. I.e. it is the slope of the G vs composition curve at a
given composition. (When we say added, we also can considered removal of a given species from the system).
 If i is positive (G increases on addition of ‘A’ to the system), it implies that the system does “not like” to
accommodate ‘A’.
(5)
, , j i
i
i T P n
G
Single Phase System
n


 

  

 

91. 1
k
i i
i
dG SdT VdP dn


     (5)
 The equation we saw before (eq. (5)) is an important one which is applicable to open systems
(single phase). The equation is applicable to systems in thermal and mechanical equilibrium
but not in material equilibrium. The equation is valid during irreversible chemical reaction
taking place in the system and during transport of matter into (or out of) the system.
 Similarly, we can write down equations for the other thermodynamic potentials (U, H, A).
These are the Gibbs equations for open systems.
Single phase
Open system
Reversible process
P-V work only
i i
i
dU TdS P V dn

    
i i
i
dH TdS VdP dn

   
i i
i
dA SdT PdV dn

    
i i
i
dG SdT VdP dn

    
(e1a)
(e7a)
(e8a)
(e9a)
Single phase system
Mechanical & Thermal equilibrium
P-V work only
Not in Chemical equilibrium
(dni arises due to irreversible chemical reaction or
irreversible transport of matter into the system)
 In the above discussions, the system was open, but was a single phase system. We now need to
generalize the equation(s) to multi-phase systems. Gibbs free energy is s state function which is
extensive and hence we can add the G’ for individual phases to obtain that for the whole
system.
Multi-phase Phase-1 Phase-2 Phase-3
G =G G G ....
   (6)
1
k
i i

 

92. Generalization of the Gibbs equation for a single phase open system to a multi-phase system
 Let the phases be labelled , , , etc. Then the total ‘G’ for the system can be written as below.
( has been used as the label/name for a single phase and also as an index for the summation
over all the phases).
...
Total
G G G G G
   

     
Given that the differential of a sum is the sum of the differentials (i.e. d(u + v)= du + dv), we can
write:
1
k
i i
i
dG S dT V dP dn
    


     (8)
  ...
Total
dG d G dG dG dG dG
    
 
     
 
Eq. (5), the formula for dG can be written for a given phase () as follows (the equation is basically the
same, but written for the -phase).
(6a)
(7)
From (7) & (8)
1
k
Total i i
i
dG S dT V dP dn
   
  


 
     
 
   
S is the entropy of the -phase
V is the volume of the -phase
ni
 is the no. of moles of the ‘ith’ species in the -phase
i
 is the chemical potential of the ‘ith’ species in the -phase
The term i
 is:
, , j i
i
i T P n
G
n 





 

  

  (9)
(10)