1. The document discusses key concepts in thermodynamics including the zeroth law, first law, internal energy, enthalpy, heat capacities, and various thermodynamic processes. 2. It provides mathematical expressions for relating work, heat, internal energy and enthalpy based on the first law of thermodynamics. 3. The different types of thermodynamic processes like isothermal, adiabatic, isochoric and isobaric processes are defined along with their characteristic properties.

1. Dr.P.GOVINDARAJ
Associate Professor & Head , Department of Chemistry
SAIVA BHANU KSHATRIYA COLLEGE
ARUPPUKOTTAI - 626101
Virudhunagar District, Tamil Nadu, India
THERMODYNAMICS

2. THERMODYNAMICS
Zeroth law of thermodynamics
• When two objects are in thermal equilibrium with the third object, then there is
thermal equilibrium between the two objects itself

3. THERMODYNAMICS
Nature of Work and Heat:
• The changes of a system from one state to another is accompanied by change
in energy . The change in energy may appear in the form of heat, work, light, etc.,
• The mathematical relation that relate the mechanical work done (W) and heat
produced (H) is
W  H (or) W = JH ------(1)
Where
J is Joule mechanical equivalent of heat
• When H = 1 Calorie then the equation (1) becomes W = J
i.e., J is the amount of mechanical work required to produce one calorie of heat energy and
its value was calculated as 4.184 Joules

4. THERMODYNAMICS
Internal energy (E)
• The energy associated with a system (matter) by means of its molecular constitution
and the motion of its molecules is called internal energy. It is a state function property
First law of thermodynamics
1. Energy can neither be created nor destroyed but one form of energy can converted to
another form of energy
2. It is impossible to construct a perpetual motion machine. Perpetual machine is a
machine which can produce energy without expenditure of energy
3. The total mass and energy of an isolated system remains unchanged (or) constant

5. THERMODYNAMICS
Mathematical derivation of first law of thermodynamics
• When a system absorbs ‘q’ amount of heat energy and changes from one state (A)
to another state (B)

6. Heat absorbed by the system is used by two ways
i. Increasing the internal energy of the system
i.e., ∆E = EB – EA
ii. Doing work ‘w’ by the system so that the piston can move from A (initial state) to
B (final state)
i.e., Heat absorbed by the system = Increase in internal energy + Work done by the system
(Disappear) (Appear)
q = ∆E + W ------ (1)
THERMODYNAMICS

7. THERMODYNAMICS
• Since w = F x displacement
w = P x area x displacement
w = P x volume change
w = P∆V
• Equation (1) becomes
q = ∆E + P∆V -------(2)
• For small changes
dq = dE + Pdv --------(3)
equation (1), (2) and (3) are the mathematical statements for first law of thermodynamics

8. THERMODYNAMICS
Enthalpy (H)
• It is defined as the total energy stored in the system
Mathematically, H = E + PV
i.e., Enthalpy of a system is obtained by adding the internal energy and the product of
pressure and volume
• When a system changes from state A to state B at constant pressure by absorbing ‘q’
amount of heat energy

9. THERMODYNAMICS
W = P∆V = P (V2 – V1)
• According to First law of thermodynamics
q = ∆E + W -------(1)
q = ∆E + P (V2 – V1)
q = E2 – E1 + PV2 – PV1
q = (E2 + PV2 ) – (E1 + PV1 )
q = H2 – H1
q = ∆H -------(2)

10. THERMODYNAMICS
i.e., The change in enthalpy of a system is the amount of heat absorbed at constant pressure
So equation (1) can become
∆H = ∆E + W
∆H = ∆E + P∆V -------(3)
Note,
Absolute value for E & H cannot be determined but change of internal energy (∆E) and
Enthalpy(∆H) are measured quantities

11. THERMODYNAMICS
• It is defined as the amount of heat energy required to raise the temperature of the
system by 10c
Mathematically
C =
𝑞
𝑇2 − 𝑇1
For small changes
C =
𝑑𝑞
𝑑𝑇
--------(1)
There are two types of heat capacities
1. Heat capacity at constant volume (CV)
2. Heat capacity at constant pressure (CP)
Heat capacity of a system

12. THERMODYNAMICS
Heat capacity at Constant Volume (CV)
• It is defined as the amount of heat energy required to raise the temperature of the system
by 10C at constant volume

13. THERMODYNAMICS
• According to first law of thermodynamics
q = ∆E + P∆V
q = ∆E + 0 (since ∆V = 0 )
q = ∆E
For small changes dq = dE
Equation (1) becomes
C =
𝑑𝑞
𝑑𝑇
=
𝑑𝐸
𝑑𝑇
=
𝜕𝐸
𝜕𝑇 v
i.e., Cv is the increase in internal energy of the system on raising the temperature by 10C
and Cv for one mole gas is called molar heat capacity at constant volume
i.e., Cv =
𝜕𝐸
𝜕𝑇 v

14. THERMODYNAMICS
Heat capacity at Constant Pressure(CP)
• It is defined as the amount of heat energy required to raise the temperature of the system
by 10C at constant pressure

15. • According to first law of thermodynamics
q = ∆E + P∆V = ∆H
For small changes dq = dE + PdV = dH
Equation (1) becomes
C =
𝑑𝑞
𝑑𝑇
=
𝑑𝐸+𝑃𝑑𝑉
𝑑𝑇
=
𝜕𝐻
𝜕𝑇
=
𝜕𝐻
𝜕𝑇 P
i.e., CP is the increase in enthalpy of the system on raising the temperature by 10C
and Cp for one mole gases is called molar heat capacity at constant pressure
i.e., Cp =
𝜕𝐻
𝜕𝑇 p
THERMODYNAMICS
Heat capacity at Constant Pressure(CP)

16. THERMODYNAMICS
Relation between CP & Cv
• Cv is the amount of heat energy required to raise the temperature from T1 to T1+10C
by increasing the internal energy (∆E = E2 – E1) for one of gases
i.e., Cv = ∆E
CV
CP

17. • CP is the amount of heat energy required to raise the temperature from T1 to T1+10C
by increasing the internal energy (∆E) and performing work (P∆V) for one of gases
i.e., CP = ∆E + P∆V
THERMODYNAMICS
• So, CP is greater than Cv by P∆V
i.e., CP > Cv by P∆V and
CP - Cv = P∆V --------(1)
• For one mole of an ideal gas
PV = RT --------(2)
When the temperature is raised by 10C ie., from T →T +1 , so that the volume of the system
is V+∆V , then equation (2) becomes
P(V+∆V) = R(T +1 ) --------(3)

18. THERMODYNAMICS
• Equation (3) – Equation (2), becomes
P∆V = R ------- (4)
• Subtracting Equation (4) in Equation (1) we get
CP – CV = R
1. q = ∆E + W
2. q = ∆E + P∆V
3. H = E + PV
4. ∆H = ∆E + P∆V
5. ∆H = ∆E + W
6. Cv =
𝜕𝐸
𝜕𝑇 v
7. Cp =
𝜕𝐻
𝜕𝑇 p
8. CP – CV = R
Note :

19. THERMODYNAMICS
Process:
Process is an operation by which the system can change from one state to another state
Types of Process :
1. Reversible process
2. Irreversible process
3. Isothermal reversible process
4. Adiabatic reversible process
5. Isochoric process
6. Isobaric process
7. Isothermal irreversible process
8. Adiabatic irreversible process

20. THERMODYNAMICS
Reversible process:
• Reversible process is the process in which the system can change from one state to another
state in a series of small steps by successive small decrements(dP) or increments(dP)
in external pressure that makes successive small increments (dV) or decrements (dV) in
volume

21. THERMODYNAMICS
Irreversible process:
• Irreversible process is the process in which the system can change from one state
to another state suddenly in a single step when the pressure difference between the system
and the surrounding is very high

22. THERMODYNAMICS
Compare reversible and irreversible process:
REVERSIBLE PROCESS IRREVERSIBLE PROCESS
 Slow process  Sudden process
 Pressure difference is infinitesimally
small
 Pressure difference is high
 Involved in many small steps to
complete
 Involved in single step to complete
 Maximum work is obtained  Maximum work is not obtained
 Equilibrium existed between two
successive small steps
 No existence of equilibrium between
initial and final stage

23. THERMODYNAMICS
Isothermal reversible process:
• Isothermal reversible process is the process in which the system can change from one state
to another state in a series of small steps by successive small decrements or increments
in external pressure (dP) that makes successive small increments or decrements in
volume (dV) at constant temperature
Heat conducting material
(so temperature remains constant during the process)

24. THERMODYNAMICS
Change in internal energy and enthalpy for isothermal reversible process:
• The temperature of the isothermal reversible process is constant. Since the internal
energy(E) depends upon the temperature of the system, internal energy (E) also constant
i.e., Change in internal energy (∆E) = 0
• We know that H = E + PV
∆ H = ∆(E + PV) (For 1 mole of gas )
∆ H = ∆(E + RT)
∆ H = ∆E + R ∆T --------(1)
Since, ∆E = 0 & ∆T = 0 at constant temperature equation (1) becomes
∆ H = 0 + R (0)
∆ H = 0

25. THERMODYNAMICS
Heat energy (q) for the isothermal process:
• According to first law of thermodynamics
∆E = q – w -----(1)
Since ∆E = 0 for an isothermal process, equation (1) becomes
0 = q – w
q = w -----(2)
Equation (2) shows that the work(w) is done at the expense of the heat absorbed(q)

26. THERMODYNAMICS
Work done in isothermal reversible expansion of an ideal gas:
• Let dw be the small amount of work done by the system in each step of isothermal
reversible expansion and it is given as,
dw = (P – dP) dV
dw = PdV – dPdV --------(1)
Since dPdV is very small and this value is neglected, then the equation (1) becomes
dw = PdV --------(2)
• The total work (w) is obtained by integrating equation (2) with in the limits V1 & V2
i.e., 𝑑𝑤 = 𝑣1
𝑣2
𝑃𝑑𝑉 --------(3)
Since for an ideal gas PV = RT and P =
𝑅𝑇
𝑉
, then equation (3) becomes

27. 𝑑𝑤 = 𝑣1
𝑣2 𝑅𝑇
𝑉
𝑑𝑉
𝑑𝑤 = 𝑣1
𝑣2
𝑃𝑑𝑉
𝑑𝑤 = 𝑅𝑇 𝑣1
𝑣2 𝑑𝑉
𝑉
𝑤 = 𝑅𝑇 [ ln 𝑉 ] 𝑉1
𝑉2
𝑤 = 𝑅𝑇 [ ln
𝑉2
𝑉1
]
𝑤 = 2.303 𝑅𝑇 l𝑜𝑔
𝑉2
𝑉1
--------(4)
• Since for an ideal gas P1V1 = P2V2 at constant temperature
P1
P2
=
V2
V1
So equation (4) becomes
𝑤 = 2.303 𝑅𝑇 l𝑜𝑔
𝑃1
𝑃2
THERMODYNAMICS
V2
V1

28. THERMODYNAMICS
Adiabatic reversible process:
• Adiabatic reversible process is the process in which the system can change from one state
to another state in a series of small steps by successive small decrements or increments
in external pressure (dP) that makes successive small increments or decrements in
volume (dV) at various temperature
i.e., no heat is enter or leave the system
Heat insulating material
(so temperature changes during the process)

29. THERMODYNAMICS
• Since no heat is enter into the system q = 0
According to first law of thermodynamics ,
q = ∆E + W
0 = ∆E + W
W = - ∆E
i.e., work is done at the expense (decrease) of the internal energy of an ideal gas

30. THERMODYNAMICS
∆E, ∆H and work done for adiabatic reversible process for Enthalpy and work done
We know that
Cv =
𝜕𝐸
𝜕𝑇 v
Cv =
𝑑𝐸
𝑑𝑇
For finite change, equation (1) becomes
∆ 𝐸 = Cv∆𝑇 -----------(2)
Since for adiabatic reversible process W = - ∆E, equation (2) becomes
𝑑𝐸 = Cv 𝑑𝑇 ------------(1)
W = - Cv∆𝑇 -----------(3)

31. For finite change, equation (4) becomes
∆𝐻 = CP∆𝑇 ------------(5)
THERMODYNAMICS
From equation (2), (3) & (5) it is clear that the value for ∆E , W and ∆H depends upon the
value of ∆T i.e., by knowing the initial temperature (T1) and final temperature (T2) for the
adiabatic reversible process, the value for ∆E , W and ∆H are calculated easily
𝑑𝐻 = CP 𝑑𝑇 -----------(4)
CP =
𝑑𝐻
𝑑𝑇
Cp =
𝜕𝐻
𝜕𝑇 P
And also we know that

32. THERMODYNAMICS
Comparison of workdone in isothermal & adiabatic reversible expansion
• Consider the isothermal & adiabatic expansion of ideal gas from initial volume V1 and pressure
P1 to a common final volume Vt
• If Piso and Padia are the final pressure then

33. THERMODYNAMICS
• For isothermal expansion
P1V1 = Piso Vt ---------(1)
• For adiabatic expansion
P1V1
γ = Padia (Vt)γ ---------(3)
Where γ =
𝐶 𝑃
𝐶 𝑉
• From equation (1)
Vt
V1
=
P1
Piso
---------(2)
• From equation (3)
Vt
V1
γ
=
P1
Padia
---------(4)

34. THERMODYNAMICS
• Since for expansion Vt > V1 and for all gases γ >1 and hence
Vt
V1
γ
>
V𝑡
V1
&
P1
Padia
>
P1
Piso
and also Padia < Piso
• Since Padia < Piso the workdone in isothermal expansion (Wiso = Piso ∆Vt) shown by spotted
area ABCD is greater than the workdone in adiabatic expansion (Wadia = Padia ∆V1)
shown by spotted area AECD

35. THERMODYNAMICS
Application of first law of thermodynamics
1. Joule Thomson effect
2. Hess’s law of heat summation
3. Kirchoff’s equation
4. Bond enthalpy

36. “The phenomenon of change of temperature (cooling effect) produced when a gas is
made to expand adiabatically from a region of high pressure to a region of very very low
(vacuum) pressure is called Joule Thomson effect”
THERMODYNAMICS
Joule Thomson effect

37. THERMODYNAMICS
• While adiabatic expansion a part of kinetic energy of the gaseous molecules used to
overcome the vanderwaal’s forces of attraction existed between the molecules. So that the
kinetic energy decreased and the temperature of the gaseous molecule decreased. So
cooling effect takes place.
Reason:
Applications of Joule Thomson effect:

38. THERMODYNAMICS
Let us consider a volume V1 of the gas enclosed between the piston A and the porous plug G
at a pressure P1, is forced slowly through the porous plug by moving the piston A inwards
and is allowed to expand to a volume V2 at a lower pressure P2 by moving the
B outwards, as shown
Derivation of Joule – Thomson coefficient :

39. THERMODYNAMICS
• Work done on the system at the piston A (W1) = - P1V1
• Work done by the system at the piston B (W2) = P2V2
• Net work done by the system W = W1 + W2
W = P2V2 – P1V1 ------------(1)
• Since q=0 for adiabatic expansion and the first law (q =∆E + W) becomes
0 = ∆E + W
W = -∆E -----------(2)
i.e., the work is performed at the expense of internal energy and the internal energy of the
system decreased from E1 to E2 . Substitute equation(1) in equation (2) , we get
P2V2 – P1V1 = -∆E
P2V2 – P1V1 = - (E2 - E1)
P2V2 – P1V1 = E1 - E2
E2 + P2V2 = E1 + P1V1 ------------(3)

40. THERMODYNAMICS
Since E + PV = H, equation (3) becomes
H2 = H1
H2 - H1= 0
∆H = 0 and dH = 0 ----------(4)
i.e., The adiabatic expansion of a gas occurs at constant enthalpy
Since H = f (P,T)
dH =
𝜕𝐻
𝜕𝑃 T dP +
𝜕𝐻
𝜕𝑇 P dT -----------(5)
Since
𝜕𝐻
𝜕𝑇 P = CP equation (5) becomes
0 =
𝜕𝐻
𝜕𝑃 T dP + CP dT ------------(6)

41. THERMODYNAMICS
Rearranging the equation (6) we get
CP dT = -
𝜕𝐻
𝜕𝑃 T dP
dT
dP
= -
𝜕𝐻
𝜕𝑃 T x
1
CP
∂T
∂P H= -
𝜕𝐻
𝜕𝑃 T x
1
CP
-------(7)
where
∂T
∂P H = Joule Thomson coefficient and its symbol μ J.T
Definition:
Joule Thomson coefficient is defined as the decrease of temperature
with pressure at constant enthalpy

42. THERMODYNAMICS
For a finite change equation (7) becomes
∆T
∆P
= -
𝜕𝐻
𝜕𝑃 T x
1
CP
∆T = -
𝜕𝐻
𝜕𝑃 T x
∆P
CP
This equation says that the decrease in temperature (∆T) is directly proportional to the
difference of pressure (∆P)

43. THERMODYNAMICS
We know that
The mathematical statement for μ J.T is
For μ J.T is
μ J.T =
𝜕𝑇
𝜕𝑃 H = -
1
𝐶 𝑃
𝜕𝐻
𝜕𝑃 T --------(1)
Since H = E + PV equation (1) becomes
μ J.T = -
1
𝐶 𝑃
𝜕(𝐸+𝑃𝑉)
𝜕𝑃 T
μ J.T = -
1
𝐶 𝑃
𝜕𝐸
𝜕𝑃 T +
𝜕(𝑃𝑉)
𝜕𝑃 T
μ J.T = -
1
𝐶 𝑃
𝜕𝐸
𝜕𝑉
x
𝜕𝑉
𝜕𝑃 T +
𝜕(𝑃𝑉)
𝜕𝑃 T
μ J.T = -
1
𝐶 𝑃
𝜕𝐸
𝜕𝑉 T
𝜕𝑉
𝜕𝑃 T +
𝜕(𝑃𝑉)
𝜕𝑃 T
Joule Thomson Coefficient (μ J.T ) for an ideal gas

44. THERMODYNAMICS
When an ideal gas adiabatically expands into vacuum
W = 0 and q = 0
So from first law of thermodynamics q = ∆E + W , we get
∆E = 0
i.e., the change of internal energy with volume at constant temperature
𝜕𝐸
𝜕𝑉 𝑇 = 0 ----------(3)
Equation (2) becomes,
μ J.T = -
1
𝐶 𝑃
0 +
𝜕(𝑃𝑉)
𝜕𝑃 T ----------(4)

45. THERMODYNAMICS
Since PV = constant for ideal gases at constant temperature
𝜕(𝑃𝑉)
𝜕𝑃 T = 0 -------(5)
Equation (4) becomes
μ J.T = -
1
𝐶 𝑃
(0 + 0)
i.e., Joule Thomson coefficient for an ideal gas is zero
Joule Thomson Co-efficient for real gases :
μ J.T for real gases can be obtained used Vander Waals equation
PV = RT -
𝑎
𝑣
+ bp -
𝑎𝑏
𝑣2 -------(1)

46. THERMODYNAMICS
Since both a and b are small, the term ab/v2 can be neglected then equation (1) becomes
V =
𝑅𝑇
𝑃
-
𝑎
𝑃𝑉
+ b ---------(2)
Since PV = RT, then the equation (2) becomes
V =
𝑅𝑇
𝑃
-
𝑎
𝑅𝑇
+ b ---------(3)
Differentiating with respect to temperature at constant pressure
𝜕𝑉
𝜕𝑇 P=
𝑅
𝑃
+
𝑎
𝑅𝑇2 ---------(4)
Substituting the value of V from equation (3) and
𝜕𝑉
𝜕𝑇 P from equation (4) in the
following thermodynamic equation of state
𝜕𝐻
𝜕𝑃 T = V - T
𝜕𝑉
𝜕𝑇 P we get

47. THERMODYNAMICS
𝜕𝐻
𝜕𝑃 T =
𝑅𝑇
𝑃
-
𝑎
𝑅𝑇
+ b - T
𝑅
𝑃
+
𝑎
𝑅𝑇2
𝜕𝐻
𝜕𝑃 T =
𝑅𝑇
𝑃
-
𝑎
𝑅𝑇
+ b -
𝑇𝑅
𝑃
−
𝑎
𝑅𝑇
𝜕𝐻
𝜕𝑃 T = b -
2𝑎
𝑅𝑇
---------(5)
we know that ,
μ J.T = -
1
𝐶 𝑃
𝜕𝐻
𝜕𝑃 T ----------(6)
Substitute equation (5) in equation (6) we get
μ J.T = -
1
𝐶 𝑃
b -
2𝑎
𝑅𝑇
μ J.T (real) =
1
𝐶 𝑃
2𝑎
𝑅𝑇
- b

48. THERMODYNAMICS
Inversion temperature:
We know that Joule Thomson coefficient for real gas is
When
2𝑎
𝑅𝑇
= b , equation (1) becomes
μ J.T (real) =
1
𝐶 𝑃
(0 - 0)
μ J.T (real) = 0
The temperature at which μ J.T (real) = 0 is called inversion temperature and its value
is obtained from the equation
2𝑎
𝑅𝑇
= b as
Ti =
2𝑎
𝑅𝑏
• Below the inversion temperature μ J.T (real) is positive i.e., cooling effect occurs
• Above the inversion temperature μ J.T (real) is negative i.e., heating effect occurs
• At the inversion temperature μ J.T (real) is zero i.e., neither cooling nor heating
effect occurs
μ J.T (real) =
1
𝐶 𝑃
2𝑎
𝑅𝑇
- b ------(1)

49. THERMODYNAMICS
Hess’s Law of constant Heat summation
Hess’s Law :
“The enthalpy change of a given chemical reaction is the same whether the process
takes place in one or several steps”
Explanation:
Suppose a substance A can be changed to Z in two ways
First way:
Involved in one way
A → Z + Q1
Where Q1 ---- amount of heat evolved i.e., equal to change in enthalpy ∆H

50. THERMODYNAMICS
Second way:
Involved in three steps
1. A → B + q1
2. B → C + q2
3. C → Z + q3
B C
ZA
q1
q2
q3
Q1
INITIAL STATE FINAL STATE
Where
q1 ---- amount of heat evolved in first step (∆H1)
q2 ---- amount of heat evolved in second step (∆H2)
q3 ---- amount of heat evolved in third step (∆H3)

51. THERMODYNAMICS
The total evolution of heat in second way is
q1 + q2 + q3 = Q2
(or)
∆H1 + ∆H2 + ∆H3 = ∆H’
According to Hess’s law :
Q1 = Q2
or
Q1 = q1 + q2 + q3
or
∆H = ∆H’
or
∆H = ∆H1 + ∆H2 + ∆H3

52. THERMODYNAMICS
Example :
The burning of carbon into carbon dioxide involved in 2 ways
First way:
C + O2 → CO2
∆H = -393 KJ
Second way:
i) C + ½ O2 → CO
∆H1 = -110.5 KJ
ii) CO + ½ O2 → CO2
∆H2 = -283.2 KJ
∆H = ∆H1 + ∆H2
-393 KJ = -110.5 KJ – 283.2 KJ
-393 KJ = -393.7 KJ
According to Hess’s law

53. THERMODYNAMICS
Kirchhoff’s equation:
The variation of ∆H with temperature at constant pressure of a given reaction is
represented by Kirchhoff’s equation
Derivation :
Consider a reaction at constant pressure
A → B
The change of enthalpy
∆H = HB – HA -------(1)
Differentiate equation (1) with respect to temperature at constant pressure
𝜕(∆H)
𝜕𝑇
=
𝜕HB
𝜕𝑇 𝑃 −
𝜕HA
𝜕𝑇 P -------(2)

54. THERMODYNAMICS
Since
𝜕HB
𝜕𝑇 𝑃 = (CP)B
𝜕HA
𝜕𝑇 P = (CP)A then equation (2) becomes
𝜕(∆H)
𝜕𝑇 P = (CP)B −(CP)A
𝜕(∆H)
𝜕𝑇 P = ∆CP --------(3)
𝑑(∆H) = ∆CP dT
𝐻1
𝐻2
𝑑(∆H) = ∆CP 𝑇1
𝑇2
dT
∆H2 - ∆H1 = ∆CP (T2 – T1 ) --------(4)
Equation (3) and (4) Kirchhoff's equations where ∆H1 & ∆H2 are the heats of reaction at
constant pressure at temperature T1 & T2

55. THERMODYNAMICS
Bond enthalpy :
Bond enthalpy is defined as the amount of energy required to dissociate one mole
of bonds present in a compound in the gaseous state
For example
H2(g) → 2H ∆H = 433 KJ / mol
That is 433 KJ of energy is required to break the H – H bond in one mole (Avogadro
number) of Hydrogen gas molecule

56. THERMODYNAMICS
Application of the first law of thermodynamics to real gases :
Work done for the isothermal expansion of real gases :
We know that
dw =Pdv ------(1)
For n moles, the vanderwaal’s equation for real gas is
P +
𝑎𝑛2
𝑣2 (v – nb) = nRT
P +
𝑎𝑛2
𝑣2 =
nRT
(v – nb)
P =
nRT
(v – nb)
-
𝑎𝑛2
𝑣2 -------(2)

57. THERMODYNAMICS
Substitute equation (2) in (1), we get
dw =
nRT
(v – nb)
-
𝑎𝑛2
𝑣2 dv --------(3)
Let the system containing real gas expand isothermally from volume v1 to v2 , the work
done for the process is obtained by integrating the equation (3) with limit v1 and v2
𝑑𝑤 = 𝑣1
𝑣2 nRT
(v – nb)
−
𝑎𝑛2
𝑣2 dv
w = 𝑣1
𝑣2 nRT
(v – nb)
dv − 𝑣1
𝑣2 𝑎𝑛2
𝑣2 dv
w = nRT 𝑣1
𝑣2 dv
(v – nb)
− 𝑎𝑛2
𝑣1
𝑣2 dv
𝑣2

58. THERMODYNAMICS
w = nRT 𝑣1
𝑣2 dv
(v – nb)
− 𝑎𝑛2
𝑣1
𝑣2 dv
𝑣2
w = nRT [ln(𝑣 − 𝑛𝑏)] 𝑣1
𝑣2 - 𝑎𝑛2 [−
1
𝑣
] 𝑣1
𝑣2
w = nRT[ln (v2 - nb) – ln (v1 - nb)] – an2 [
1
𝑣1
−
1
𝑣2
]
w = nRT ln
v2 − nb
v1 − nb
+ an2 [
1
𝑣2
−
1
𝑣1
]

59. THERMODYNAMICS
Change in internal energy for isothermal expansion of real gases :
In Vanderwaal’s equation for real gases, the factor
𝑎𝑛2
𝑣2 is representing the internal pressure
and also
𝜕𝐸
𝜕𝑣 T is also corresponding to internal pressure
i.e.,
𝜕𝐸
𝜕𝑣 T =
𝑎𝑛2
𝑣2
dE =
𝑎𝑛2
𝑣2 dv --------(1)
Integrating the equation (1) with in the limit
𝐸1
𝐸2
𝑑𝐸 = 𝑣1
𝑣2 𝑎𝑛2
𝑣2 dv
E2 – E1 = 𝑎𝑛2
𝑣1
𝑣2 dv
𝑣2

60. THERMODYNAMICS
E2 – E1 = 𝑎𝑛2
𝑣1
𝑣2 dv
𝑣2
∆E = 𝑎𝑛2 [−
1
𝑣
] 𝑣1
𝑣2
∆E = an2 −
1
𝑣2
+
1
𝑣1
∆E = an2 [
1
𝑣1
−
1
𝑣2
]
∆E = - an2 1
𝑣2
−
1
𝑣1
This is the Change in internal energy for isothermal expansion of real gases

61. THERMODYNAMICS
Heat absorbed for isothermal expansion of real gases :
From the first law of thermodynamics
q = ∆E + w ---------(1)
We know that
∆E = - an2 1
𝑣2
−
1
𝑣1
---------(2)
w = nRT ln
v2 − nb
v1 − nb
+ an2 [
1
𝑣2
−
1
𝑣1
] ---------(3)
Substituting (2) & (3) in (1) we get
q = - an2 1
𝑣2
−
1
𝑣1
+ nRT ln
v2 − nb
v1 − nb
+ an2 [
1
𝑣2
−
1
𝑣1
]
q = nRT ln
v2 − nb
v1 − nb

62. THERMODYNAMICS
Work done for adiabatic expansion of real gases :
From the first law of thermodynamics
q = ∆E + w ---------(1)
For adiabatic process, q = 0 so the equation (1) becomes
0 = ∆E + w
∆E = -w ---------(2)
Since E is a state function of T and V
i.e., E = f (T,V)
dE =
𝜕𝐸
𝜕𝑇 V dT +
𝜕𝐸
𝜕𝑉 T dV ---------(3)

63. THERMODYNAMICS
For a vanderwaal’s equation,
P +
𝑎𝑛2
𝑣2 (v – nb) = nRT
𝜕𝐸
𝜕𝑇 V = n Cv -------(2)
𝜕𝐸
𝜕𝑉 T =
𝑎𝑛2
𝑣2 --------(3)
Substitute (2) and (3) in (1) we get
dE = n Cv dT +
𝑎𝑛2
𝑣2 dV
Integrating the equation within the limit
𝐸1
𝐸2
𝑑𝐸 = n Cv 𝑇1
𝑇2
𝑑𝑇 + 𝑎𝑛2
𝑣1
𝑣2 dv
𝑣2
E2 – E1 = n Cv (T2 – T1) - an2 1
𝑣2
−
1
𝑣1
∆E = n Cv ∆T - an2 1
𝑣
−
1
𝑣

64. THERMODYNAMICS
Limitations of first law of thermodynamics:
First law of thermodynamics does not tell
• The direction of flow of heat
• Whether a process occur spontaneously i.e., whether it is feasible
• Heat energy cannot be completely converted into an equivalent of work without
producing some changes elsewhere

65. THANK YOU