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THERMODYNAMICS
UNIT-I
NEED OF SECOND LAW
The first law of thermodynamics states that one form of energy can change into
another form but the total amount of energy remains the same. Once we specify a
particular process or change, this law helps us to calculate the internal energy, heat
released and work done in the process. But the first law says nothing about whether the
specified process can occur or not and if it occurs in which direction. This question
concerns the second law of thermodynamics.
STATEMENTS OF SECOND LAW OF THERMODYNAMICS
1. A spontaneous change is one way or unidirectional. The spontaneous flow of
heat is unidirectional from a region at high temperature to a region at lower temperature.
For reverse change to occur, work has to be done.
Clausius statement
2. It is impossible to construct a machine, which can transfer heat in a cyclic
process from a cold reservoir to a hot reservoir, unless some external work is done on the
machine.
3. Spontaneous changes tend towards equilibrium.
The heat flows from high temperature to low as the temperature of the two
regions tend to equalize at an intermediate value.
4. It is impossible by a cyclic process to convert heat into work without at the same
time transferring some heat from high temperature to a low temperature.
5.
Kelvin-Planck statement
It impossible to take heat from hot reservoir end convert into work by cyclic
process without transferring a part of heat to a cold reservoir.
6. It is impossible by a cyclic process to transfer heat from a low temperature region
to a high temperature region without at the same time converting same work into
heat.
7. It is impossible to obtain work by cooling a body below the lowest temperature of
the system.
8. There exists a function ‘S’ called entropy which is a state function. The entropy
of the universe remains constant in a reversible process but it increases in an
irreversible or spontaneous process.
ENTROPY
Exothermic processes are spontaneous as they are accompanied with decrease of
energy. But some endothermic processes, which are accompanied with increase of energy
are also spontaneous. To explain this another thermodynamic function ‘Entropy’ is
introduced.
Entropy is a thermodynamic state quantity which is a measure of the randomness or
disorderness of the molecules of the system. It is also considered as a measure of
unavailable form of energy. It is a function which accounts for the irreversibility of the
given changes.
Consider the following changes.
(a) Melting of ice.
(b) Evaporation of water.
(c) Distribution of solute throughout the solvent.
(d) Expansion of gases.
(e) Flow of water from a hill to the ground.
(f) Mixing of gases.
In all the above cases, the system reaches a state of greater disorder. Eventhough the
energy of the system increases (endothermic) during the above changes the processes are
spontaneous because they are accompanied with increase in entropy. Hence entropy is
another driving force, which explains the spontaneity of a process.
ENTROPY CHANGE OF A REVERSIBLE PROCESS
Since entropy is a state function, its absolute value cannot be found out. But the
change in the entropy when a system undergoing a change can be calculated.
For a reversible change taking place at a fixed temperature (T), the change in
entropy (s) is equal to heat energy absorbed or evolved divided by the temperature.
ds = dqrev/T
or ΔS = qrev/T
If heat is absorbed, then ds or ΔS is +ve (increase in entropy)
If heat is evolved, then ds or ΔS is –ve (decrease in entropy)
For a reversible process
the change in entropy = Change in entropy of the system + Change in entropy of the
Surroundings
ΔS = q/T + (-) q/T = 0
UNITS OF ENTROPY
It is measured in entropy units (eu) which are calories per mole  K i.e. cal-1
mol-1 K-1
In SI system, the units are Joules per mole K i.e. J mol-1 K-1
This is represented by EU.
1 eu = 4.184EU
The entropy of pure crystal is zero at absolute zero (III law) Therefore, it is possible to
calculate the actual amount of entropy that a substance possesses at any temperature.
It is referred to as absolute entropy. The absolute entropy of a substance at 25 C(298K)
and one atmospheric pressure is called standard entropy( S).
ENTROPY CHANGES IN IRREVERSIBLE PROCESS
Irreversible reactions (processes) proceed spontaneously. For a reversible process,
ΔS = qrev/T
We know that a reversible process absorbs more heat than the irreversible one. Hence
qrev/T
 qirr/T
But
qrev/T
=
Hence
ΔSsys
(when T is constant)
ΔS sys.
 qirr/T
Now consider that a small quantity of heat dq flows from a system at temperature T 2 to
the surroundings at temperature T1. The total change in entropy is given by
ΔSnet
=
Since T2
dq/T1 – dq/T2

T1
ΔSnet is +ve or ΔSnet  0
Hence, it is clear that in an irreversible isothermal process, the entropy of an isolated
system is greater than zero.
As most of the processes going on in nature are spontaneous and irreversible, it
has been said that “the entropy of the universe always tends towards a maximum”.
ENTROPY CHANGE FOR AN ISOTHERMAL EXPANSION OF IDEAL GAS
In order to calculate the value of ΔS during the expansion of an ideal gas, the
process is assumed to be going on reversibly.
For a process involving one mole of an ideal gas the entropy change depends
upon the values of temperature, volume and pressure of the system at the initial and final
states of the system.
We know
dq = dE + PdV (I Law) -----------(1)
We also know
ds = dq/T
dq = TdS
Substituting the value of dq in equation (1)
TdS = dE + PdV ------------ (2)
Cv = dE / dT
dE= Cv dT
Substituting the value of dE in equation (2)
TdS = Cv dT + PdV ------------ (3)
for 1 mole of ideal gas
PV = RT
P = RT/V
Substituting the value of P in equation (3)
TdS = CvdT + RT/V dV -------------- (4)
Dividing by T
dS = Cv dT/T + R dV/V -------------(5)
For a definite change, the entropy change ΔS can be calculated by integrating the
equation (5) between the initial state1 and final state (2)
S2
T2
V2
∫ dS = Cv ∫ dT/T + R ∫ dV/V
S1
T1
V1
S2-S1 = Cv( ln T2 – ln T1) + R(ln V2 – ln V1)
ΔS = Cv ln T2/T1 + R ln V2/V1
For n moles
ΔS = n Cv ln T2/T1 +n R ln V2/V1
If the process is isothermal
T2 = T1
So the above equation reduces to
ΔS = R ln V2/V1 for 1 mole
ΔS = n R ln V2/V1 for n mole
Problems
1) Calculate the entropy change when 2 moles of an ideal gas expand isothermally
and reversibly from a volume of 1dm3 to a volume of 10dm3 at 300K.
(ΔS)T = n R ln V2/V1
= 2.303 x 2 x 8.314 x log 10/1
= 38.29 JK-1 (eu)
2) Calculate the entropy change when 1 mole of an ideal gas is allowed to expand at
27C from a volume of 2 litres to a volume of 20 litres against a constant pressure
of one atmosphere (R = 2Cal/deg/mole)
3) Calculate the entropy change when one mole of an ideal gas expand isothermally
from 0.1 litre to 1.0 litre at 27C. (R = 1.987Cal/deg/mole)
4) Two moles of an ideal gas undergoes isothermal reversible expansion from 15
litres to 30 litres at 300K.Calculate the change in entropy.
(R = 1.987Cal/deg/mole)
ENTROPY OF PHASE TRANSITIONS
Processes involving phase transitions like fusion, sublimation and vaporization
take place reversibly at melting point, sublimation point and boiling point respectively.
qrev in these cases will be equal to ΔH fusion, ΔH sublimation and ΔH vaporization
respectively.
Substituting these values in the equation
dS = qrev/T, we can calculate the change in entropy.
Problems
1) Calculate the entropy change for fusion of 1 mole of a solid which melts at 300K.
The latent heat of fusion is 2930 J mol-1
qrev = 2930 J mol-1 and T= 300K
dS = qrev/T = 2930/300 = 9.77 JK-1mol-1
= ΔHb/Tf
2) Calculate the entropy change in the evaporation of 1 mole of water at 100C.
Latent heat of evaporation of water is 9650 cals per mole.
qrev = 9650 cal mol-1 T=100+273=373K
dS = qrev/T = ΔHv/Tb = 9650/373 = 25.87 cal mol-1 K-1
3) Calculate the increase in entropy when one gram molecular weight of ice at 0 C
melts to form water. Latent heat of fusion of ice = 80 cal/ gm.
qrev = 80x18 cal/mole
T = 0+273=273K
dS = qrev/T = 80X18/273 = 5.274 cal mol-1 K-1
4) One mole of water at 100C requires 536cal of heat for conversion into vapours at
100C . Calculate ΔS.
5) Calculate the entropy change involved when 1 mole of liquid oxygen is converted
into gaseous oxygen at its boiling point of 90K. Enthalpy of vaporization of liquid
oxygen is 6820 J mol-1.
FREE ENERGY AND WORK FUNCTION
Besides heat content (H), internal energy (E) and entropy (S), there are two more
thermodynamic functions. They are work function (A) and free energy (G).
WORK FUNCTION (A)
Work function A is defined as
A = E – TS
Since E,S and T are state functions, A is also a state function.
Consider an isothermal change at temperature T from the initial state (1) to final state (2)
A1 = E1 – TS1
A2 = E2 – TS2
A2 – A1 =(E2 – E1) – T(S2 – S1)
ΔA = ΔE – TΔS
But ΔS = q/T
Hence ΔA = E – T q/T
ΔA = ΔE – q
ΔE = q – w (I Law)
Substituting the value of ΔE
ΔA = q-w-q
ΔA = -w
-ΔA =w
The decrease in the work function A in any processes at constant temperature gives the
maximum work that can be obtained from the system during any change.
FREE ENERGY (F or G)
The free energy is defined as,
G =H – TS
At constant temperature, if the system changes from the initial state(1) to the final
state(2)
G1 =H1 – TS1
G2 =H2 – TS2
G2 – G1 = H2 – H1 – T(S2 - S1)
ΔG = ΔH – TΔS
But ΔH = ΔE + PΔV (I Law)
Substituting the value of H
ΔG = ΔE + PΔV – TΔS
ΔG = ΔE – TΔS + PΔV
But ΔE – TΔS = ΔA
Substituting the value of ΔE – TΔS
ΔG = ΔA + PΔV
But ΔA = -w
ΔG = -w +PΔV
-ΔG = w – PΔV
PΔV is the work done due to expansion at constant pressure. Therefore, it is evident, that
the decrease in free energy (-ΔG) accompanying a process taking place at constant
temperature and pressure is equal to the maximum work obtained from the system other
than the work of expansion. This quantity of work is referred to as the network.
Network = - ΔF = w – PΔV
Hence change in free energy is a measure of the network, which may be electrical,
chemical or surface work. (Other than the mechanical work of expansion)
GIBBS-HELMHOLTZ EQUATION
We know ΔH – TΔS
ΔS = S2 – S1
-ΔS = - (S2 – S1) ----------------- (1)
G = H – TS
But H = E + PV
G = E + PV – TS
Differentiating the above equation,
dG = dE + PdV + VdP – TdS – SdT ---------------- (2)
But dq = dE + dw (I Law)
dS = dq/T
dS = dE + dw / T
TdS = dE + dw
dw = PdV
TdS = dE + PdV
Substituting the value of TdS in the equation (2)
dG = dE + PdV + VdP – dE – PdV – SdT
dG = VdP – SdT
If the pressure is constant dP=0, VdP =0
dG = – SdTP
(dG/dT)P = - S
-S2 = (dG2/dT)P
-S1 = (dG1/dT)P
Substituting the value of S2 and S1 in the equation 1
-ΔS = (S2-S1) = (dG2/dT)P - (dG1/dT)P
= [d (G2 – G1)/ dT]P = [d(ΔG)/dT]P
Substituting the value of S in the equation,
ΔG = ΔH – TΔS
ΔG = ΔH + T [d(ΔG)/dT]P ------------I
We know
A = E – TS
Differentiating the above equation
dA = dE + TdS – SdT
dS = q/T
TdS = q
dA = dE –q – SdT
but q – dE = w (I Law)
But at constant volume, no mechanical work will be done by the system (w =0)
dA = - SdTV
(dA/dT)V = - S
[d(A)/dT]V = - S
ΔA = A2 – A1
[d(ΔA)/dT]V = (dA2/dT)V - (dA1/dT)V
[d(ΔA)/dT]V = -S2 – (-S1)
= -(S2 - S1)
= - ΔS
A= E – TS
ΔA = ΔE – TΔS
Substituting the value of ΔS, we get
ΔA = ΔE + T[d(ΔA)/dT]V -----------II
Equations I and II are called Gibbs-Helmholtz equations.
APPLICATIONS
By using Gibbs-Helmholtz equation, we can calculate ΔG, ΔH and ΔS of the reactions
occurring in galvanic cells.
ΔG- In galvanic cells
-ΔG =nFE
n = number of electrons involved in the reaction.
F = 96500 coulombs’
E = emf of the cell.
i.e. decrease in free energy = electrical energy
ΔH ΔG = ΔH + T [d(ΔG)/dT]P
-nFE = ΔH + T [d(-nFE)/dT]P
-nFE = ΔH - nFT(dE/dT)P
ΔH = -nFE + nFT (dE/dT)P
ΔH = -nF(E +T (dE/dT)P
(dE/dT)P is called temperature coefficient of the cell reaction.
ΔG = ΔH – TΔS
TΔS = ΔH – ΔG
ΔS = ΔH – ΔG / T
PROBLEMS
1) Free energy change (ΔG) in a process was found to be -138 KJ at 303K and -135.0KJ
at 313K. Calculate the change in enthalpy (ΔH) accompanying the process at 308K.
ΔG1 = - 138.0 KJ
ΔG2 = -135.0 KJ
T1= 303K
T2 = 313K
[d(G)/dT]P = ΔG2 – ΔG1/T2 – T1
= -135.0 – (-138.0)/ 313 -303
= -135.0 + 138/ 10
= 3/10
= 0.3 KJ K-1
ΔG at 308K = G at 303 +ΔG at 313/2
= -135 + (-138)/2
= - 136.5 KJ
ΔG =ΔH + T [d(ΔG)/dT]P
ΔH = ΔG – T [d(ΔG)/dT]P
ΔH = -136.5 – (308X0.3)
= - 136.5 – 92.4
= -228.9 KJ
2) ΔH and ΔS for a reaction at 27ºC are -10 K cals and 20 K cal K-1 respectively. What
is ΔG for the reaction?
ΔG = ΔH – TΔS
ΔH = -10 Kcals or -10.000cals
ΔS = 20 Kcal K-1
T = 27 + 273 = 300K
ΔG = -10.000 – (300X20)
= -10.000 – 6000
= -16.000 cals.
3) ΔG for a reaction at 300K is -16 K cals. ΔH for the reaction is -10K cals. What is the
entropy of the reaction at 300K?
MAXWELL’S RELATIONS
E = f(S,V)
dE = (dE/dS)V dS + (dE/dV)S Dv
dE = q – PdV (I Law)
dS = q/T
q = TdS
dE = TdS - PdV
We know dE = TdS – PdV
Hence (dE/dS)V = T -------------1
(dE/dV)S = -P -------------2
A = f(V,T)
dA = (dA/dV)T dV + (dA/dT)V dT
A= E-TS
dA = dE + TdS – SdT
dE = -TdS = - PdV
So dA = -PdV – SdT
Comparing the above two equations
(dA/dV)T = -P ------------3
(dA/dT)V = -S ------------4
G = f(T,P)
dG = (dG/dT)P dT + (dG/dP)T dP
We know
dG = VdP – SdT
(dG/dT)P = -S ------------5
(dG/dP)T = V -------------6
H =f(P,S)
dH = (dH/dP)S dP + (dH/dS)P dS
H = E+PV
dH = dE + PdV + VdP
dE + PdV = TdS
dH = TdS + VdP
(dH/dP)S = V-------------7
(dH/dS)P = T -------------8
Differentiating equation (1) with respect V at constant S
d2E/dS dV = (dT/dV)S
Differentiating equation (2) with respect S at constant V
d2E/dV dS = -(dP/dS)V
Comparing the above equations
(dT/dV)S = -(dP/dS)V -------------I
Differentiating equation (3) with respect T at constant V
d2A/dV dT = -(dP/dT)V
Differentiating equation (4) with respect V at constant T
d2A/dT dV = -(dS/dV)T
So (dP/dT)V = (dS/dV)T -----------II
Differentiating equation (5) with respect P at constant T
d2G/dT dP = -(dS/dP)T
Differentiating equation (6) with respect T at constant P
d2G/dP dT = (dV/dT)P
Hence,
-(dS/dP)T = (dV/dT)P ------------III
Differentiating equation (7) with respect S at constant P
d2H/dP dS = (dV/dS)P
Differentiating equation (8) with respect P at constant S
d2H/dS dP = (dT/dP)S
Hence,
(dV/dS)P = (dT/dP)S ------------IV
Equations I,II,III and IV are called Maxwell’s relations.
VAN’T HOFF ISOTHERM
The Van’t Hoff isotherm gives the net work that can be obtained from a gaseous
reactant at constant temperature when both the reactants and the products are at suitable
arbitrary pressures. It can be derived by using the equilibrium box which is a theoretical
device.
Consider the gaseous reaction
A(g) + B(g) ↔ C(g) + D(g)
The initial pressure of A and B are pa and pb.The initial pressure of C and D are pc
and pd. The equilibrium pressures are PA,PB,PC and PD. The following theoretical
operations are performed.
1) Change of pressure of A from pa to PA
work done by the gas = RT ln pa / PA
2) Change of pressure of B from pb to PB
work done by the gas = RT ln pb / PB
3) Now introduce 1 mole of A and B at pressures PA and PB. They react to give 1
mole of C and 1 mole of D at pressures PC and PD.No work done by the gas for
the operation (3). since operation is performed at equilibrium processes.
4) Now Change of pressure of C from PC to pc
work done by the gas = RT ln PC/pc
5) Change of pressure of D from PD to pd
work done by the gas = RT ln PD/pd
Total change in free energy is equal to the total work done
ΔG = RT ln pa/PA + RT ln pb/PB + RT ln PC/pc + RT ln PD/pd.
-ΔG = RT ln PC PD/PA PB – RT ln pc pd/ pa pb.
= RT ln Kp – RT ln pc pd/pa pb
If the reaction is started with reactants at a partial pressure of one atom and the
resulting products are also at 1 atom Pressure(standard state)
-ΔGº = RT ln Kp
or
-ΔGº = RT ln K (K- equilibrium)
VAN’T HOFF EQUATION (VAN’T HOFF ISOCHORE)
-ΔGº = RT ln K
ln K = - -ΔGº/RT
Differentiating the equation with respect to temperature at constant pressure
d ln K/dT = - 1/R d/dT ( ΔGº/dT )
d ln K/dT = - 1/R d( ΔGº/dT )/dT ----------(1)
ΔGº = ΔHº + T d( ΔGº/dT )P
ΔGº/T = ΔHº/T + d( ΔGº/dT )P
Differentiating the above equation with to temperature respect at constant P
d( ΔGº/dT )/dT = - ΔHº/T2 ----------2
Comparing equations (1) and (2)
d ln K/dT = 1/R X ΔHº/T2
d ln K/dT = ΔHº/RT2
d ln K = ΔHº/RT2 dT
Differentiating the equation between K1 at T1 and K2 at T2
d ln K = ΔHº/R 1/T2 dT
ln K2/K1 = ΔHº/R [ 1/T2 –(-1/T1)]
ln K2/K1 = ΔHº/R [ 1/T1 –1/T2]
ln K2/K1 = ΔHº/R [ T2 –T1/T1 T2]
APPLICATIONS
1) If equilibrium constant at one temperature is known the equilibrium constant at
any desired temperature can be found out.
2) If equilibrium constants at two temperatures are known, the heat of reaction can
be found out.
Problems
1) Calculate the equilibrium constant of a reaction at 300K if ΔG at this temperature
for the reaction is 29.29 KJ mol-1.
ΔGº = 29.29 KJ mol-1 = 29.29x103 Jmol-1
T = 300K
R = 8.314 JK-1 mol-1
ΔGº = - RT ln Kp
= - 2.303 RT log Kp
29290 = -8.314 x 300 x 2.303 log Kp
log Kp = - 29290/8.314 x 300 x 2.303
= - 5.0991 = -6 + 0.9009
Kp = 7.96 x 10-6
2) For the formation of ammonia the equilibrium constant at 673K and 773K
respectively are 1.64 x 10-4 and 1.44 x 10-5 respectively. Calculate the heat of the
reaction.
K1 = 1.64 x 10-4
K2 = 1.44 x 10-5
T1 = 673K
T2 = 773K
R = 8.314 JK-1 mol-1
ΔH =?
3) The equilibrium constant for the reaction PCl5 ↔ PCl3 + Cl2 is 1.5 X 10-4 at 300K.
Calculate K at 400K, if ΔHº = -83.68 KJ, R = 8.314 JK-1 mol-1.
CHEMICAL POTENTIAL
The equation dG = SdT + VdP is valid only for a closed system. If the system is an open
and huge system with many components the following equation is derived
G = f(T,P,n1,n2,n3…..)
Where n1,n2,n3 etc are the number of moles of the component 1,2,3 etc
dG = (dG/dT)P,N dT + (dG/dP)T,N dP +(dG/dn1)T,P,n2 dn1 +(dG/dn2)T,P,n1 dn2 and so on
where
(dG/dn1)T,P,n2 = µ1
(dG/dn2)T,P,n1 = µ2
µ1, µ2 etc are called chemical potential or partial molal free energy of the components 1
and 2 respectively.
VARIATION OF CHEMICAL POTENTIAL WITH TEMPERATURE
(dG/dni)T,P,N = µi
Differentiating the above equation with respect T at constant P
d/dT (dG/dni)P,N = (dµi/dT)P,N ------------(1)
dG = VdP – SdT
at constant pressure
dG = - SdT
(dG/dT)P = - S
Differentiating the equation with respect ni
d/dni (dG/dT)P = - (dS/dni)P = -Si ----------(2)
Comparing equations (1) and (2)
(dµi/dT)P,N = - Si (partial molal entropy)
VARIATION OF CHEMICAL POTENTIAL WITH PRESSURE
dG = VdP – SdT
at constant temperature
dG = VdP
(dG/dP)T = V
Differentiating the equation with respect ni, we get
d/dni (dG/dP)N = (dV/dni)T,N = Vi ----------1
µi(P,T,ni) = (dG/dni)T,P,ni
Differentiating with respect P, we get
(dµi/dP)T,N = d/dP (dG/dni)T,N ----------2
Comparing equations(1) and (2)
(dµi/dP)T,N = Vi (partial molal volume)
SIGNIFICANCE
Chemical potential is a property of the substance which determines its spontaneous flow
from one region to another. In other words, it determines the escaping tendancy of the
substance.
SUMMARY
1) Various statements of second law of thermodynamics.
2) Entropy is a measurement of randomness or disorderness of a system.
dS = dqrev/T
Entropy remains constant during a reversible process.
Entropy of irreversible process tends to towards maximum
3) Entropy change for an isothermal expansion of ideal gases = Cv ln T2/T1 + R ln
V2/V1
4) Entropy of phase transitions
solid ↔ liquid
ΔS = Hf/Tf
Liquid ↔ vapour
ΔS = Hv/Tb
solid(1) ↔ solid(2)
ΔS = Ht/Tt
5) Decrease in work function at constant T is the maximum work
-ΔA = Wmax
6) Decrease in free energy is a measure of net work
-ΔF = W – PV
7) Gibbs-Helmholtz equation
ΔG = ΔH + T [d(ΔG)/dT] P
ΔA = ΔE + [d(ΔA)/dT]V
8) Maxwell’s relations
(dT/dV)S = -(dP/dS)V
(dP/dT)V = (dS/dV)T
-(dS/dP)T = (dV/dT)P
(dV/dS)P = (dT/dP)S
9) Van’t Hoff isotherm
-ΔGº = RT ln K
10) Van’t Hoff isochore
ln K2/K1 = ΔHº/R [ T2 –T1/T1 T2]
log K2/K1 = ΔHº/2.303R [ T2 –T1/T1 T2]
11) Chemical potential of the component ‘i’ is the change in free energy when one mole
of the component ‘i’ is added to a large open system at constant T,P and total number of
moles.
(dG/dni)T,P,N = µi
12) Variation of chemical potential with temperature
(dµi/dT)P,N = - Si
13) Variation of chemical potential with pressure
(dµi/dP)T,N = Vi