process, Thermodynamic process,workdone, relation between pressure volume,first law of thermodynamic,need of second law,statement of second law,carnot heat engine,efficiency,numericals

1. Thermodynamics 1
According to Syllabus of S.G.B.Amravati University
Amravati
Class- B.Sc.Sem I
Mr G.D. Rawate
Department of Chemistry
Shri R.R.Lahoti Science ,College
Morshi

2. THERMODYNAMICS
Thermodynamic is a branch of
science that describes the
behaviour of matter and the
transformation between
different forms of energy on a
macroscopic scale (large scale)

3. Process
It is path or operation by which a system
changes fro one state to another state.

4. .
Thermodynamic process
1) Adiabatic process
2) Isothermal process

6. Adiabatic Process
The process is carried out
under such condition that
there is no exchange of
heat take place between
system and surrounding is
called adiabatic process dq
= 0 .
Example : Reaction
carried out in thermally
insulated apparatus
.

7. Isothermal Process
The process which is
carried out at constant
temperature during each
step is called as isothermal
process
Example : reaction carried
out in thermostat
dT = 0

8. Work done in Isothermal process
Consider n moles of gas enclosed
in cylinder fitted with weightless
and frictionless piston.
The gas expand isothermally and
reversibly from initial volume V1
to final volume V2 at constant
temperature T, expansion will take
place in no. of infinitesimally small
step, According to Boyls law
pressure of gas P will decreases at
the same time volume of gas
increases by dV in a single step
and small wok will done

9. .
At the same time volume is increases by
dV in a single step and small quantity of
work done which is given by dw
dw = -P ext dV ----1
But the expansion is reversible, the
pressure of gas is greater by small
quantity dp than P ext
(p-P ext) = dP
P ext = (p-dp) ------2
Putting eq 2 in eq 1
dw =-(p-dp) dV
dw = (-P dV +dP dV)
Ignoring the product dP dV both are
small.
dw = -P dV
Work done in Isothermal process

10. .
To obtain total work done integrate above equation
dw = - PdV ------------------3
For ideal gas PV = nRT
P = nRT/V --------------------4
Putting 4 in eq 3;
W = - nRT/V dV
W = - nRT dV/V
W = -2.303 nRT log V2/V1-------5
At Constant temperature according to Boyl s law
P1 V1 = P2 V2 (p1/P2 = V2/V1)
Eq 5 becomes
W = - 2.303 nRT log P1/P2
-Ve sign indicate work done by system on
surrounding.

11. Work done in adiabatic reversible
expansion of an ideal gas
In adiabatic change there is no exchange of
heat between the system and surroundings.
i.e. q = 0 applying this condition to FLTD
d U = q + w
d U = w or –w =-∆U
The work is done by gas at expense of its
internal enrgy, internal energy is function of
temperature, so decrease of internal energy
result in lowering of temperature

12. .
The final temperature in adiabatic expansion
of gas where as there is an increase of
temperature for adiabatic compression.
From above Eq (1) w can be calculated if ∆U
is known
Consider relation
U = f (T,V)
dU = (dU/dT)v dT+(dU/dT)T dV
dU = CvdT +(dU/dV)T dV
But for ideal gas (dU/dV)T =0
dU = CvdT

13. .
Integrate the above Eq with in suitable limits
∫ dU = Cv ∫ dT
∆U = (U2-U1) = Cv(T2 – T1)
For n mole of ideal gas
∆U = n Cv(T2 – T1)
Cv is molar heat capacity of gas
From Eq (1) ,
W = ∆U = n Cv(T2 – T1)
U2
U1
T2
T1

14. Relation between pressure and volume
For adiabatic reversible expansion of ideal gas we
have
dU = Cv dT = PdV = - n RT d ln V
CvdT = - n RT dln V ---------------(1)
But for n mole of ideal gas Eq is
PV = n RT on differentiation we get
PdV + V dP = n R dT
dT = (PdV + V dP)/nR
dT = (P/n R)dV +(V/nR)dP [ P=n RT/V &V=n RT/P]

15. .
dT = (n RT/nR)(d V /V) +(n RT/nR)( dP /P)
dT = T dlnV + T dln P
Putting this value in eq (1)
Cv (T dln V+ Td lnP ) = - nR T dln V
Cv T dln V+ Cv TdlnP = -nRTdln V
Cv T dln P =[ -nRTdln V – Cv T dln V]
Cv T dln P =-(Cv +nR)dln V
Cv dln P = - Cp dlnV
dlnP = -(Cp/Cv)dlnV
But Cp/Cv= ɤ
d lnP = - ɤ dlnV

16. .
Integrate above eq in suitable limits
∫ dln P = - ɤ ∫ dlnV
[lnP ]p1 = - ɤ [lnV]v1
lnP2/P1 = - ɤ ln V2/V1
lnP2/P1 = ln(V1/V2) ɤ
P2/P1 = (V1/V2) ɤ
P2/P1 = (V1) ɤ/(V2) ɤ
P1V1ɤ = P2V2ɤ = Constant
p2
P1
V1
V2
V2
P2

17. Relation between temperature and Volume
The energy change and heat capacity of an ideal gas
related as
dU = CvdT
Also dw = -PdV
For adiabatic expression q = 0
According to FLTD dU = dW
dU = CvdT = - PdV
CvdT/n RT = - dV/V
(Cv /n R) dlnT = - dlnV
Integrating above within suitable limit;

18. .
Cv/nR ∫ dln T = dlnV
Cv /n R [ ln T ]T1 =-[lnV]v1
Cv /nRlnT2/T1 = lnV1/V2
But Cv = (C p –nR)
(Cp – nR/nR) ln T2/T1 = lnV1/V2
[ ln(T2/T1)] (Cp -nR/nR) = ln V1/V2
(T2/T1)(Cp-nR /nR) = V1/V2
( T2/T1) = (V1/V2)(nR /Cp-nR)
(T2/T1) = (v1/V2) (Cp -Cv /Cv)
(T2/T1) = (v1/V2)(Cp -Cv /Cv)
(T2/T1) = (V1/V2) (ɤ - 1)
T2V2 (ɤ - 1) = T1V1 (ɤ - 1)
T2V2 (ɤ - 1) = T1V3(ɤ - 1)
T2
V2

19. Relation between Pressure and
Temperature
For adiabatic reversible expansion of gas
dU = - P dV = Cv dT -------------------(1)
But for n mole of ideal gas PV = n RT
Differentiating above Eq.(1)
P dV + V dp = nR dT
P dV = (nR dT –V dP)
Putting this value in Eq 1
dU = - (nR dT – V dP) = Cv dT
dU = -nR dT + V dP = Cv dT

20. .
Cv dT + nR dT = V dP
(Cv +nR) dT = (n RT/P) dP
Cp dT/T = nR dP/P
dln T = nR dln P
Cp dln T = nR d lnP
dln T = nR/Cp dln P
dU = -nR dT = V dP =Cv dT
Cv dT = nR dT = V dP
(Cv +nR )dT = (nR T/P)dP
Cp dT/T =nR dP/P

21. .
Cp dln T = nRdln P
dln T = nR/C p dln P
Integrating above Eq within Suitable limit,
∫ dln T = n R/Cp ∫ d lnP
[ln T ] = nR/C p [lnP]
(ln T2/T1) = n R/ Cp ln P2/P1
(T2/T1) = (P2/P1)n R/Cp
(P2/P1) = (T2/T1)C p / nR
(P2/P1) =(T2/T1)C p /C p - C v
(P2/P1) = (T2/T1) ɤ/ɤ -1
(P2/P1) = ( T2 ɤ/ɤ -1 /T1 ɤ/ɤ -1) ɤ/ɤ -1

22. First Law of Thermodynamics
First law of thermodynamics is also known as
law of conservation of energy.According to
this law the total energy of the universe
remains constant when system changes from
one state to another state. First law stated in
different ways.
STATEMENTS OF FIRST LAW
1.The energy can neither created nor
destroyed; it can be converted from one from
to another form of energy.

23. .
2.The total internal energy of an isolated
system is constant.
3.The total amount of energy of univese is
constant.
4. When ever some part of energy disappears,
an exactly equivalent amount of anopther
form of energy makes it appearance.
5.The total amount of energy of system and
surrounding must remains constant .

24. Mathematical Equation of first law
of thermodynamics.
Suppose q quantity of heat is supplied to the
system and w is amount of work done on the
system by surrounding , the internal energy
increases and given by,
∆U = q + w
Where ∆U = increase in internal energy of the
system
q = amount of heat absorbed
W = work done on system by surrounding

25. Needs of Second Law of Thermodynamics
Needs of Second law of thermodynamics felt due
to following limitation of the first law of
thermodynamics.
1.First law can not put any restriction on the
direction of heat flow. Suppose that heat is
flowing from a hot body to cold body , first law
simply reveals that the quantity of heat lost by
the hot body is equal to heat gained by cold body.
It is never reveals that heat flows from hot to
cold body and not in reverse direction hence need
of second law of thermodynamics.

26. .
II)First law of thermodynamics simply
established equivalence between different
forms of energy. However this law fails to
explain under what condition and what extent
it is possible to bring about conversion of one
form of energy into other hence there is need
of SLTD.
III)The first law of thermodynamics does not
explain why chemical reaction do not
proceeds to completion let us consider the
reaction.
CO(g) + H2O (g) CO2 (g) + H2(g)

27. .
If one mole of CO and one mole of H2O in the
reaction vessel then , If reaction went to
completion one mole of CO2 and one mole of
H2 would be formed. If we examine reaction
vessel at equilibrium, the yield of reaction is
less than 100 % first law fail to explain this.
iv) The first law of thermodynamics is a
qualitative statement which fails to contradict
the existence of 100 % efficient heat engine.

28. Statements of second law of
thermodynamics
Work can completely converted into heat but heat
can not converted into work ,with out leaving a
permanent change in the system or surrounding.
The SLTD has been stated in no. of ways ,but all
statement are logically equivalent to one another.
1.Kelvin Planck Statement : It is impossible to
construct engine which operating in complete cycle
,will absorb heat from single body and convert it
into work (100 % efficiency),without leaving
permanent changes in the working system.

29. .
Clausius Statement : Heat flows from hot to
cold body due to difference of temperature.
But in refrigerator heat flows from at lower
temperature and get rejected to body at
higher temperature, here heat is being flow
through some outer agency.This lead to
Clausius to state SLTD which is It is
impossible to transfer for self acting machine
unaided by any external agency to convert
heat continuously, from one body at lower
temperature to body at a higher temperature.

30. Thomson Statement
In a heat engine, the working substance (system)
takes heat from hot body,a part of heat absorbed
is converted into work while remaining is given
to sink (cold body).No engine has so far been
constructed which continuously takes heat from
single body and changes the whole of its work it
work making any change in the system presence
of cold body is necessary for converting heat into
work. This led Thomson to state that ‶It is
impossible to obtain a continuos supply of work
cooling body to temperature lower than that of
coldest of its surrounding.

31. .
4.Diffussion of gas always take place from a
region of higher pressure to one at lower
pressure until pressure becomes uniform .
5. All natural Processes take place in one
direction only and can not be reversed.

32. Carnot Heat Engine or Carnot Heat Cycle
French scientist Sodi Carnot in 1824 proposed a
hypothetical and ideal heat engine which converts
heat into work to the maximum extent. The carnot
heat engine operating between two different
temperatures can perform series of operation
between these two temperatures so that at the end
of these operations the system can return to the
initial state this cycle of different processes is
known as carnot cycle. A carnot heat consist of
following main three parts.

33. .
1.Hot Reservoir ( Hot body at temperature
T2 OR Source of heat.
2. Cold Reservoir (Cold body at temperature
T1) OR Sink
3. Working Substance

34. Cold Body(T1)
Hot Body (T2)

36. 1.P1,V1
2.P2,V2
3.P3,V3
4.P4,V4

37. Expression For the work done and
efficiency of Carnot heat engine
In Carnot heat engine one mole of an ideal gas is
enclosed in a cylinder which fitted with the frictionless
piston . The cylinder is insulated from all side except
base . The Carnot heat engine operated in four
reversible steps as shown in figure
Step 1. Reversible and Isothermal Expansion
The gas is expands isothermally and reversibly from
point 1 to 2 from initial volume V1 to final volume V2
at temperature T2 due to absorption of heat q2 from
hot reservoir . Therefore work done by system on
surrounding is given by,

38. .
q2 = - w1 = - ∫ P d V
But ideal gas equation for 1 mole of gas at
temperature T2 is
PV = RT2
P = RT2/V
q2 =- w1 = - ∫ RT2/ V x d V
= - RT2 ∫ dV /V
q2 = -w1 = - RT2 lnV2/V1 -------(1)
V2

39. 2.Reversible and adiabatic expansion
The gas is expands adiabatically( q=0) and
reversibly from point 2 to 3 from volume V2 to
V3 so that temperature falls from T2 to T1, The
work done by system on surrounding is given by
-w2 = - dU
But definition of Cv, Cv =(dU/dT)v
- w2 = - ∫ Cv dT
- w2 = - Cv(T1 –T2) --------------(2)
T1
T2

40. Step 3: Reversible and isothermal
Compression
The gas is subjected to an isothermal and
reversible compression at temperature T1
From Volume V3 to V4. If q1 amount of heat
is rejected. The work done by surrounding on
system is given by

41. .
q1 = w3 = - ∫ P d V
But ideal gas equation for 1 mole of gas at
temperature T1 is
PV = RT1
P = RT1/V
q1 =- w3 = ∫ RT1/ V x d V
= - RT1 ∫ dV /V
q1 = w1 = - RT1 lnV4/V3 -------(3)
V3
V4
V3
V4
V4

42. Step 4 Reversible and adiabatic Compression
Finally the gas is compressed adiabatically( q=0) and reversibly
from point 4 to 1 from volume V4 to V1 so that temperature
increases from T1 to T2, The work done in this step is given by
w4 = - dU
But definition of Cv, Cv =(dU/dT)v
w4 = - ∫ Cv dT
w4 = - Cv (T2 –T1)
w4 = Cv (T1 –T2) --------------(4)
T2
T1

43. Net work done in cyclic process
W net = -W1 – W2 + W3 + W4
W net = -RT2(V2/V1) – Cv(T1-T2) –
RT1ln(V4/V3) +Cv(T1-T2)
W net = -RT2ln (V2/V1) + RT1ln(V4/V3)---(5)
We know that for adiabatic process
TV (ɤ - 1) = Constant Thus along the adiabatic curve BC
T2V2 (ɤ - 1) = T1V3(ɤ - 1)
(T2/T1)= (V3/V2) (ϒ-1) ----------------------(6)

44. .
Similarly along the curve AD
T2V1 (ɤ - 1) = T1V4(ɤ - 1)
(T2/T1) = (V4/V1) (ϒ-1) --------------(7)
Thus From Eq 6 and 7
(V3/V2)(ϒ-1) = (V4/V1) (ϒ-1)
(V3/V1) = (V4/V1)
(V2/V1) = (V3/V4)
Equation (5) becomes
- W net = RT2ln(V2/V1)+RT1ln(V1/V2)
- W net = RT2ln(V2/V1) - RT1ln(V2/V1) ------------(8)
- W net = R(T2 - T1)ln(V2/V1)
Net heat absorbed by the system in complete cycle is given by
q = { Heat absorbed in various Step –Heat rejected in various step}
= [ q2 – q1 ] -----------------------------(9)
Q = (q2 –q1)
But net heat absorbed = Net work performed
q = -W net = (q2 – q1) = R(T2-T1)ln(V2/V1)
-Wnet = q = (q2 –q1 = R (T2 –T1) ln (V2/V1) ---------(10)

45. Efficiency of heat engine
η = - W net/q2
η = R (T2-T1)ln(V2/V1)/ RT2ln(V2/V1)
η = (T2 –T1)/T1
If T1 =0 Kelvin then η =1 or efficiency =100%
only when temperature of cold body is absolute zero .In practice
this is impossible.
Therefore efficiency of any heat engine is always less than one