2. Content
1. (a) State and Explain second law of thermodynamics.
(b) Zinc melts at 693K and its entropy at 298K is 9.95 cal/deg/mol. Calculate the
standard entropy of zinc at 750°C.
Given: Heat of fusion of Zn at the melting point
ΔΗf = 1.74 Kcal/mol
Cp <Zn>= 5.35 +2.40 X 10³T cal/deg/mol
Cp {Zn} = 7.50 cal/deg/mol.
2. (a) Explain the concept of free energy. What are the equilibrium conditions of a
given process in terms of the free energy change.
(b) Vapour pressure of liquid Titanium at 2500K is 1.053mm. Hg. The heat of
vaporization at the boiling point of Titanium is 104 Kcal/mol. Calculate its boiling
point.
3.(a) From the Claussius Clapeyron derive equation for liquid vapour equilibrium.
(b) Explain the practical importance of Claussius Clapeyron equation.
3. 1.(a)
Second law of Thermodynamics:
Statement: Heat absorbed at any one temperature cannot be
completely transformed into work without leaving some change in the
system or its surroundings.
Spontaneous processes are not thermodynamically reversible. The
entropy of an isolated system always increases
A spontaneous process occurs without external intervention of any
kind.
Examples : Flow of heat from higher to longer temperature, diffusion of
a species from higher concentration to lower concentration, mixing,
and acid-base reactions
In a refrigerator, heat is pumped out from lower to higher temperature
by an artificial device. This device consumes electrical energy
provided from outside.
4. Explanation: Entropy is a measure of the systems disorder or
randomness. The second law implies that natural processes tend
to move towards a state of greater disorder.
In a reversible process, when a substance absorbs infinitesimal amount
of heat 𝑞, in a reversible manner, at an absolute temperature, 𝑇 the
entropy of the substance increases by 𝑑𝑆, such that,
𝑑𝑆 =
𝛿𝑞𝑟𝑒𝑣
𝑇
Where,
𝑑𝑆 = change of entropy, J/K/mol
𝛿𝑞𝑟𝑒𝑣 = heat absorbed by the substance in a reversible manner
𝑇 = temperature of the substance in K
6. 1.(b) Given,
Melting Point of Zn (Tm) = 693K
T₁ = 298K ; T₂ = 750°C = 1023K
S°₂₉₈{Zn} = 9.95 cal/deg/mol ; S°₁₀₂₃{Zn}= ?
ΔΗf = 1.74 kcal/mol
Cp <Zn>= 5.35 +2.40 X 10³T cal/deg/mol
Cp {Zn} = 7.50 cal/deg/mol
There is a phase change at 693K, to calculate the standard entropy of zinc at
1023K,There will be latent ΔΗf of melting point at 693K should be consider
Therefore,
S°₁₀₂₃ = S°₂₉₈ + ꭍ Cp <Zn>/ T . dT + ΔΗf /693 + ꭍ Cp {Zn}/ T . dT
Here,
S°₁₀₂₃ = 9.95 + 5.35 ( In 693 - In 298) + 2.40 x 10³ (693 - 298) + 2.51
+7.50 ( In 1023 - In 693)
S°₁₀₂₃ = 20.84 cal/deg/mole
693
298
1023
693
7. 2.(a)
Free Energy
In thermodynamics, the thermodynamic free energy is one of the state
functions of a thermodynamic system (the others being internal energy,
enthalpy, entropy, etc.)
Free Energy is the available amount in a system that can be utilize to do
some work
The change in the free energy is the maximum amount of work that the
system can perform in a process at constant temperature, and its sign
indicates whether the process is thermodynamically favorable or
forbidden.
By Equating first and second law of thermodynamics
we get,
Maximum work : (δW) = – (dE – TdS ) …..(2.1)
8. We know that,
Maximum Work = Mechanical Work + Non-Mechanical Work
where,
Mechanical work = PdV
non-mechanical work = chemical work in chemical reaction
or
electrical work in electrochemical cell.
There are two kinds of free energies:
1. Helmholtz Free Energy: A = E-TS
2. Gibbs Free Energy : G = H-TS
9. Helmholtz Free Energy(A)
Its change (dA) is equal to the amount of reversible work done on, or
obtainable from, a system at constant T.
Therefore,
dA = dE-TdS …..(2.2)
By Comparing eqn. 2.1 &2.2 we get
δW = – dA
Since it makes no reference to any quantities involved in work Thus, it
decreases is the maximum amount of work which can be done by a
system at constant temperature, and it can increase at most by the
amount of work done on a system isothermally
Where, δW = Maximum Work
dE = Change in Internal Energy
dS = Change in Entropy
10. Gibbs Free Energy(G)
Gibbs Free Energy is the most useful for processes involving a system at
constant pressure P and temperature T, because, in addition to
subsuming any entropy change due merely to heat, a change in G(δG)
also excludes the PdV work needed to "make space for additional
molecules" produced by various processes
Gibbs free energy change therefore equals work not associated with
system expansion or compression, at constant temperature and pressure
Therefore,
δW – PdV = – (dE – TdS ) – PdV
δW – PdV = – (dE + PdV – TdS ) = – (dH – TdS)
In Reversible Non-Mechanical work:
δW – PdV = – δG
i.e δG = dH - TdS
11. Free Energy as Criteria of Equilibrium
It is measure of energy that is available for the system to do Non
Mechanical work
It serves as Criterion for equilibrium ,because ΔE and ΔH are available
A system of substances reacting chemically attains equilibrium state
when free energy attains the minimum value. Thus, free energy
becomes a criterion of chemical or phase equilibrium.
For a closed system undergoing, a irreversible change i.e., spontaneous,
the free energy decreases because entropy increases.
ΔG = – ve, since ΔS= + ve
For a closed system undergoing, a reversible change i.e., equilibrium,
free energy remains unchanged and entropy change is also zero.
ΔG = 0, since ΔS = 0
12. 2.(b)
Given,
vapour pressure of liquid(P₁) =1.503 mm Hg
Temperature of liquid(Tₐ) = 2500K
Heat of vaporization(Hᵥ) = 104 kcal/mole=104000 cal/mole
Standard vapour pressure(P₂) = 1atm = 760 mm Hg
Boiling Point at Temperature(Tb) =
According to,
Clausius-Clapeyron Equation to Phase Changes in Liquid to vapour
The equation will be
log
𝐏₂
𝐏₁
=
−𝐇ᵥ
𝟐.𝟑𝟎𝟑 𝐑
x
1
Tb
−
1
Tₐ
14. 3.(a)
Liquid — Vapour Equilibrium
When some liquid is present in a container some vapour will form at a given
temperature. After some time liquid and vapour will come to equilibrium(as
shown in the below figure)
To a liquid - vapour equilibrium:
𝑑𝑃
𝑑𝑇
=
∆𝐻
𝑇.∆𝑉
𝑑𝑃
𝑑𝑇
=
∆𝐻𝑉
𝑇 . 𝑉𝑉𝑎𝑝𝑜𝑢𝑟 − 𝑉𝐿𝑖𝑞𝑢𝑖𝑑
…..(3.1)
Where ΔHV is the heat of vaporization or latent heat of evaporation, Vvap is the
molar volume of vapour, and Vliq is the molar volume of liquid.
Fig 3.1: liquid — vapour equilibrium
15. Since Vvapour >> Vliquid , so value of Vliquid is negligible with respect to
value of Vvapour
Hence, eq 3.1 becomes:
𝑑𝑃
𝑑𝑇
=
∆𝐻𝑉
𝑇 . 𝑉𝑣𝑎𝑝𝑜𝑢𝑟
. . . (3.2)
Assuming that the vapour behaves as an ideal gas, the volume Vvapour
may be related to:
𝑉𝑉𝑎𝑝𝑜𝑢𝑟 =
𝑅𝑇
𝑃
. . . (3.3)
16. Substituting this value of Vvap (eq 3.3) to eq (3.2), we get:
𝑑𝑃
𝑑𝑇
=
𝑃 . ∆𝐻𝑉
𝑅𝑇2 . . . (3.4)
𝑑𝑃
𝑑𝑇
1
𝑃
=
∆𝐻𝑉
𝑅𝑇2 Or
𝑑(𝑙𝑛𝑃)
𝑑𝑇
=
∆𝐻𝑉
𝑅𝑇2 . . . (3.5)
This eq (3.5) is another form of Clausius-Clapeyron equation for
Liquid – Vapour Equilibrium
17. 3.(b)
Clausius-Clapeyron equation
The Clapeyron equation (also called the Clausius-Clapeyron
equation) relates the slope of a reaction line on a phase diagram
to fundamental thermodynamic properties.
The form of the Clapeyron equation most often used is as
𝑑𝑃
𝑑𝑇
=
𝛥𝑆
𝛥𝑉
The Clapeyron equation helps us determine thermodynamic values
for reactions or phases. When combined with volume data, we can
use the slope of an experimentally-determined reaction to
calculate the Δ S of the reaction, and to calculate the entropy of
formation (ΔSf) of a particular phase. Often the volumes of phases
are very well-known, but the entropy data may have
large uncertainties.
18. Practical Importance of Clapeyron equation
The Clausius-Clapeyron equation is applicable to any phase
change-fusion, vaporization, sublimation, allotropy transformation
Vapor Pressure Estimation: Explain how the equation helps
estimate vapor pressure at different temperatures.
Boiling Point Elevation and Freezing Point Depression:
Discuss its role in understanding these phenomena.
Showcase its importance in studying phase transitions in
materials.
