The document discusses key concepts in thermodynamics including: 1. Thermodynamic states are characterized by macroscopic properties like temperature, pressure, and volume that determine a system's internal state and interaction with external bodies. 2. Thermal equilibrium exists when temperature is uniform throughout a system, as stated by the zeroth law of thermodynamics. 3. Internal energy (U) is the energy associated with the random, disordered motion of molecules within a system.

3. Thermodynamic State: It is the state in
which a thermodynamic system exists.
Thermodynamic states are characterized by a
set of macroscopic properties, which determine
the internal properties of a system in that state
and the interaction of the system with external
bodies. These properties include temperature,
pressure, volume, electric polarization, and
magnetization.
Thermal Equilibrium: When the
temperature throughout a system is uniform,
the system is in thermal equilibrium.

4. Zeroth law of thermodynamics: The zeroth law of
thermodynamics states that if two thermodynamic
systems are each in thermal equilibrium with a third,
then all three are in thermal equilibrium with each
other.
If two systems A and B are separately in thermal
equilibrium with a third system C, then the three
systems are in thermal equilibrium with each other.
Zeroth law of thermodynamics states that two systems
which are individually in thermal equilibrium with a
third one, are also in thermal equilibrium with each
other. This Zeroth law was stated by Flower much later
than both first and second laws of thermodynamics.
This law helps us to define temperature in a more
rigorous manner.

5. Internal Energy - U is the most common
symbol used for internal energy.
Internal energy is defined as the energy
associated with the random, disordered motion
of molecules.

6. Concept of Heat
• Heat may be defined as energy in transit.
• Word heat is used only if there is a transfer of energy from
one
thermodynamic system to the another.
• When two systems at different temperatures are kept in
contact with each other then after some time temperatures of
both the systems become equal and this phenomenon can be
described by saying that energy has flown from one system to
another.
• This flow of energy from one system to another on account of
temperature difference is called heat transfer.
• Flow of heat is a non-mechanical mode of energy transfer.
• Heat flow depends not only on initial and find states but also
on path it's.

8. P-V Indicator Digram
• Only two thermodynamic variables are sufficient to describe a system
because third vaiable can be calculated from equation of state of the
system.
• P-V Indicator Digram is just a graph between pressure and volume of a
system undergoing an operation.
• When a system undergoes an expansion from state A (P1 V1) to a state
B (P2V2) its indicator diagram is shown as follows.

9. In case of compression system at state A(P1 V1) goes to a state
B(P2V2) its indicator digram is as follows.

11.  In isothermal process temperature of the system
remains constant throughout the process.
 For an iso-thermal process equation connecting P, V and
T
gives PV = constant
i.e., pressure of given mass of gas varies inversly with its
volume this is nothing but the Boyle's law.
 In isothermal process there is no change in temperature,
since internal energy for an ideal gas depends only on
temperature hence in iso thermal process there is no
change in internal energy.
Thus, ΔU=0
therefore, ΔQ =ΔW
 Thus during isothermal process
Heat added (or substacted) from the system = wok done
by
(or on) the system

13. • Process in which no heat enters or leaves a system is called an
adiabatic process (Temperature doesn’t remain constant)
• For every adiabatic process Q=0
• Prevention of heat flow can be accomplished by surrounding system
with a thick layer of heat insulating material like cork, asbestos etc.
• Flow of heat requires finite time so if a process is performed very
quickly then process will be practically adiabatic.
• On applying first law to adiabatic process we get
ΔU=U2 - U1= +ΔW (adiabatic expansion)
Here, internal energy of systems decreases resulting a drop in
temperature. But, the work done is positive.
• On applying first law to adiabatic compression we get
ΔU=U2 - U1= - ΔW (adiabatic compression)
Here, internal energy of systems increases resulting an increase in
temperature. Bur, the work done is negative.
.

15.  A process taking place at constant
pressure is called isobaric process.
 we see that work done
in isobaric process is
W = P(V2 - V1) = nR (T2-T1)
where pressure is kept constant.
 Here in this process the amount of heat
given to the system is partly used in
increasing temperature and partly used in
doing work.

17. • In an isochoric process volume of the
system remain uncharged throughout i.e.
ΔV = O.
• When volume does not change no work is
done ; ΔW = 0 and therefore from first
law U2 - U1 = ΔU =ΔQ
• All the heat given to the system has been
used to increase the intenal energy of the
system.

18. Workdone in an isothermal expansion
Consider one mole of an ideal gas enclosed in a cylinder
with perfectly conducting walls and fitted with a
perfectly frictionless and conducting piston. Let P1, V1
and T be the initial pressure, volume and temperature of
the gas. Let the gas expand to a volume V2 when
pressurereduces to P2, at constant temperature T. At any
instant during expansion let the pressure of the gas be P.
If A is the area of cross section of the piston, then force
F = P × A.
Let us assume that the pressure of the gas remains
constant during an infinitesimally small outward
displacement dx of the piston.

19. Work done
dW = Fdx = PAdx = PdV
Total work done by the gas in expansion from initial volume V1 to final volume V2 is

20. Work done in an adiabatic expansion
Consider one mole of an ideal gas enclosed in a
cylinder with perfectly non conducting walls and
fitted with a perfectly frictionless, non conducting
piston. Let P1, V1 and T1 be the initial pressure,
volume and temperature of the gas. If A is the area of
cross section of the piston, then force exerted by the
gas on the piston is F = P × A, where P is pressure of
the gas at any instant during expansion. If we assume
that pressure of the gas remains constant during an
infinitesimally small outward displacement dx of the
piston,
then work done dW = F × dx = P × A dx

23. Adiabatic relations of system for perfect gas
Consider 1 gram of the working substance (ideal gas) perfectly insulated from
the surroundings. Let the external work done by the gas be δW.
Applying the first law of thermodynamics
δH = dU + δW
But δH = 0
and δW = P.dV
Therefore,
Where P is the pressure of the gas and dV is the change in Volume.
0= dU + P dV

24. As the external work is done by the gas at the cost of its internal energy, there
is fall in temperature by dT.

25. Let r = Cp - Cv
CvP dV + Cv V dP + Cp P dV − Cv P dV = 0
Cv V dP + Cp P dV = 0
Dividing by Cv PV,
𝑑�
�
+
��
��
𝑑�
�
= 0
Substitute
��
��
= �
𝑑�
�
+ �
𝑑�
�
= 0

26. Integrating,
log P + � log V = Constant
log P + log �� = Constant
log P��
= Constant
or
P�� = Constant …………………..(4)
This is the equation connecting pressure and volume
during an adiabatic process.
Taking PV = r T
Or P =
𝑟�
�
Substitute in eqn (4)
𝑟�
�
��
= Constant
rT �� -1
= Constant
or
T �� -1
= Constant …………..(5)

27. Taking PV = r T
Or V =
𝑟�
�
Substitute in eqn (4)
P (
𝑟�
�
)�
= Constant
𝑟� ��
��−1 = Constant
Or
��−1
�� = Constant …………………(6)
Thus during adiabatic process,
(i) P��
= Constant
(ii) T �� -1
= Constant
(iii)
��−1
�� = Constant

28. Reversible process
A thermodynamic process is said to be reversible when (i) the various stages of an
operation to which it is subjected can be reversed in the opposite direction and in
the reverse order and (ii) in every part of the process, the amount of energy
transferred in the form of heat or work is the same in magnitude in either
direction. At every stage of the process there is no loss of energy due to friction,
inelasticity, resistance, viscosity etc. The heat losses to the surroundings by
conduction, convection or radiation are also zero.
Condition for reversible process
(i) The process must be infinitely slow.
(ii) The system should remain in thermal equilibrium (i.e) system and surrounding
should remain at the same temperature.
Examples
(a) Let a gas be compressed isothermally so that heat generated is conducted away
to the surrounding. When it is allowed to expand in the same small equal steps,
the temperature falls but the system takes up the heat from the surrounding and
maintains its temperature.
(b) Electrolysis can be regarded as reversible process, provided there is no internal
resistance.

29. Irreversible process
An irreversible process is one which cannot
be reversed back.
Examples : diffusion of gases and liquids,
passage of electric current through a wire,
and heat energy lost due to friction. As an
irreversible process is generally a very rapid
one, temperature adjustments are not
possible. Most of the chemical reactions are
irreversible.

30. Efficiency of Carnot’s cycle
 is independent of the working substance,
but depends upon the temperatures of heat
source and sink.
 Efficiency of Carnot’s cycle will be 100% if
T1 = ∞ or T2 = 0 K.
 As neither the temperature of heat source
can be made infinite, nor the temperature
of the sink can be made 0 K, the inference
is that the Carnot heat engine working on
the reversible cycle cannot have 100%

31. Relation between Cp and Cv (Meyer’s relation)
Let us consider one mole of an ideal gas enclosed
in a cylinder provided with a frictionless piston of
area A. Let P, V and T be the pressure, volume
and absolute temperature of gas respectively
(Fig.). A quantity of heat dQ is supplied to the gas.
To keep the volume of the gas constant, a small
weight is placed over the piston. The pressure and
the temperature of the gas increase to P + dP and
T + dT respectively. This heat energy dQ is used
to increase the internal energy dU of the gas. But
the gas does not do any work (dW = 0).
∴ dQ = dU = 1 × Cv × dT ... (1)

33. The additional weight is now removed from the
piston. The piston now moves upwards through a
distance dx, such that the pressure of the enclosed
gas is equal to the atmospheric pressure P. The
temperature of the gas decreases due to the
expansion of the gas. Now a quantity of heat dQ’ is
supplied to the gas till its temperature becomes
T + dT. This heat energy is not only used to
increase the internal energy dU of the gas but also
to do external work dW in moving the piston
upwards.