thermodynamics propeties

2. MUHAMMADNASIR
CHEMICAL ENGINEERING

3. Thermodynamic -1

4. Property relation for
homogeneous phase
According of first law for a close system for n moles
------1
We also know that
together these three equation

5. Conti……….
U,S and V are molar valves of internal energy ,entroply and valume.
Combing effect of both laws 1ST and 2nd
Derived for reversible reaction
But contain the property of the state of the system not process of the system
Constant mass that result of differenital change from one equiriblrium state to
another.
Nature of the system cannot be relaxed

6. Conti……….
We know that
H = U +PV
GIBBS energy and Helmholtz energy equation
A = U- TS----2 G= H-TS ----3
Putting the values on above equation
so this equation becomes

7. Conti……….
When d(nU) replace by equation no 1 equation becomes 4
Same way multiplying equation 2 and 3 by n and takind differtional equations
becomes 5 and 6

8. Conti……….
Equation 5 and 6 are subject of resection of equation 1.
for the case of one mole of homogeneous fluid at constant pressure
These are fundamental equations for homogeneous equaions

9. Another set of equations follow from equation 6 and 7 for exactness for a
differtional expression for a function f(x.y)

10. Conti……….
As result we get new sets of equations

11. Entropy:
Is a measure of disorder or randomness of a system.
An ordered system has low entropy.
A disordered system has high entropy.
Enthalpy:
Is defined as the sum of internal energy of a system and the product of the
pressure and volume of the system.
The change in enthalpy is the sum of the change in the internal energy and
the work done.
Enthalpy and entropy are different quantities.
Enthalpy has the units of heat, joules.
Entropy has the units of heat divided by temperature, joules per kelvin

12. Enthalpy Vs. Entropy
Enthalpy:
It is donated by 'H', refers to the measure of total heat content in a
thermodynamic system under constant pressure.
Enthalpy is calculated ∆H = ∆E + P∆V
(where E is the internal energy). The SI unit of enthalpy is joules (J).
Entropy:
It is denoted by 'S', refers to the measure of the level of disorder in a
thermodynamic system.
Entropy is calculated ∆S = ∆Q/T (where Q is the heat content and T is the
temperature).
It is measured as joules per kelvin (J/K).

13. Relationship between Enthalpy
and Entropy of a Closed System
(T.∆S=∆H)
Here,
T is the absolute temperature,
∆H is the change in enthalpy, and
∆S is the change in entropy.
According to this equation, an increase in the enthalpy of a system causes
an increase in its entropy.

15. Entropy
pathleirreversibforS
pathcyclicreversiblefor
T
dq
S
0
0

 

16. How does entropy change
with pressure?
The entropy of a system decreases with an increase in pressure.
Entropy is a measure of how much the energy of atoms and molecules
become more spread out in a process.
If we increase the pressure on the system, the volume decreases. The
energies of the particles are in a smaller space, so they are less spread
out. The entropy decreases.
If we decrease the pressure on the system, the volume increases. The
energies of the particles are in a bigger space, so they are more spread
out. The entropy increases.

17. Pressure Dependence of Entropy
For solids and liquids entropy change with respect to pressure is
negligible on an isothermal path. This is because the work done by the
surroundings on liquids and solids is miniscule owing to very small
change in volume. For ideal gas we can readily calculate the entropy
dependence on the pressure as follows












1
2
ln
.
0)(
0
.
P
P
nRS
P
dP
nR
T
dPV
dS
dP
P
V
dVVdPPdVPVd
E
T
dVP
T
dw
T
dq
dS rev


18. Temperature dependence of Entropy
Using the usual conditions such as isobaric or isochoric paths we can
see that:
Just as in case of ΔH the above formulae apply as long as system
remains in single phase. On the other hand if system undergoes a
phase transition, at constant temperature and pressure.
pathisobaric
T
T
C
T
dT
C
T
dTC
T
dq
S
pathisochroic
T
T
C
T
dT
C
T
dTC
T
dq
S
PP
P
vv
v
 
 
























1
2
1
2
ln
ln

19. Enthalpy & Entropy as function of
Temp & pressure
The most useful property relation for the Enthalpy and Entropy of a
homogenous phase result when these properties expressed as function of
T & P
We need to know how H & S vary with Temperature and Pressure.
Consider First the Temperature derivative. Equation 2.2 divide the heat
capacity at constant pressure.
PT
H








PT
S








TP
H








TP
S








P
P
C
T
H









20. Another Expression for this quantity is obtained by division of Eq. (6.8) by
dT and restriction of the result to Constant P.
Combination of this equation with Eq (2.2) gives
The pressure derivative of the entropy results directly from Eq. (6.16)
The Corresponding derivative for the enthalpy is found by division of Eq.
(6.8) by dP and restriction to constant T.
PP T
S
T
T
H
















T
C
T
S P
P








PT T
V
P
S

















21. As a Result of Equation (6.18) this become
The functional relation chosen here for H & S are
H = H(T , P)
S = S(T , P)
V
P
S
T
P
H
TT
















dP
P
H
dT
T
H
dH
TP

















dP
P
S
dT
T
S
dS
TP

















PT T
V
TV
P
H

















22. The partial derivative are given by Eqs. (2.20) and (6.17) through (6.19)
These are general Equation relating the Enthalpy and Entropy of
homogenous fluid at constant composition to Temperature and
pressure.
dP
T
V
TVdTCdH
P
P 














dP
T
V
T
dT
CdS
P
P 









23. Internal energy as
a function of P

24. Internal energy is given as
U = H – PV
Differentiation yields
As we know
Now by putting this equation in above equation

25. The ideal gas state
As we know ideal gas
By differentiating with respect to T and keeping P constant
Now substituting this equation into following equations

26. We got following equations
Alternative forms for Liquids
in following equations
We got following

27. Following
when
In following equation
We obtain

28. Internal energy and entropy as a function of T & P
As we know dQ= 𝑇𝑑(𝑆)……………….1
dU= 𝑑𝑄 + 𝑑𝑊 … … … .2
Putting the of aqua 1 into aqua 2
dU= 𝑇𝑑 𝑆 − 𝑃𝑑(𝑉)….3
Temperature and volume often serve as more convenient independent
variables hen do temperature pressure. The most useful relation are hen
for infernal energy and entropy. Required derivatives are
(
𝜕𝑈
𝜕𝑇
)v ,(
𝜕𝑈
𝜕𝑉
)t ,(
𝜕𝑆
𝜕𝑇
)v ,(
𝜕𝑆
𝜕𝑉
)t

29. taking he derivative of aqua 2 wish respect to temperature and volume at
constant volume and temperature.
(
𝜕𝑈
𝜕𝑇
)v =T(
𝜕𝑆
𝜕𝑇
)v………………………….4
Cv= T(
𝜕𝑆
𝜕𝑇
)v
CV/T=(
𝜕𝑆
𝜕𝑇
)v
(
𝜕𝑈
𝜕𝑉
)t =T(
𝜕𝑆
𝜕𝑇
)t-P………………….……5
(
𝜕𝑈
𝜕𝑉
)t =T(
𝜕𝑆
𝜕𝑇
)t-P

30. the chosen function here are
U=U(T,V) S=S(T,V)
taking derivative
dU= (
𝜕𝑈
𝜕𝑇
)v dT+ (
𝜕𝑈
𝜕𝑉
)t dV………….6
dS=(
𝜕𝑆
𝜕𝑇
)v dT+ (
𝜕𝑆
𝜕𝑉
)t dV…………….7
let

31. dU= (
𝜕𝑈
𝜕𝑇
)v dT+ (
𝜕𝑈
𝜕𝑉
)t dV
dU= Cv dT+[T (
𝜕𝑆
𝜕𝑉
)-P]dV
dS=(
𝜕𝑆
𝜕𝑇
)v dT+ (
𝜕𝑆
𝜕𝑉
)t dV
dS= CV/TdT+ (
𝜕𝑃
𝜕𝑇
)vdV
As we know
(
𝜕𝑃
𝜕𝑇
)v=
𝛃
𝚔

32. It is applied to a state a constant volume
Alternate form

33. Gibbs free energy
The energy associated with a chemical reaction that can be used to do work. The
free energy of a system is the sum of its enthalpy (H) plus the product of the
temperature (Kelvin) and the entropy (S) of the system:

34. According to the second law of thermodynamics, for systems reacting
at STP (or any other fixed temperature and pressure), there is a general natural
tendency to achieve a minimum of the Gibbs free energy

35. The Gibbs free energy is:
which is the same as:

36. where:
U is the internal energy (SI unit: joule)
p is pressure (SI unit: pascal)
V is volume (SI unit: m3)
T is the temperature (SI unit: kelvin)
S is the entropy (SI unit: joule per kelvin)
H is the enthalpy (SI unit: joule)

37. Derivation
The Gibbs free energy total differential natural variables may be derived
via Legendre transforms of the internal energy.
Where, μi is the chemical potential of the ith chemical component. (SI unit:
joules per particle or joules per mole
Ni is the number of particles (or number of moles) composing the ith chemical
component

38. The definition of G from above is

39. Taking the total differential, we have
Replacing dU with the result from the first law gives

40. Applications of Gibbs Free Energy
Colligative properties of solutions
Boiling point elevation and freezing point depression
The pressure on a liquid affects its volatility
Electron-free energy levels

41. Effect of pressure on a liquid
Applying hydrostatic pressure to a liquid increases the spacing of its
microstates, so that the number of energetically accessible states in the
gas, al though unchanged, is relatively greater— thus increasing the
tendency of molecules to escape into the vapor phase. In terms of free
energy, the higher pressure raises the free energy of the liquid, but does
not affect that of the gas phase.

42. Thermodynamics of rubber bands
Rubber is composed of random-length chains of polymerized isoprene
molecules. The poly(isoprene) chains are held together partly by
weak intermolecular forces, but are joined at irregular intervals by
covalent disulfide bonds so as to form a network..

43. Conti’
The intermolecular forces between the chain fragments tend to curl
them up, but application of a tensile force can cause them to elongate
The disulfide cross-links prevent the chains from slipping apart from one
another, thus maintaining the physical integrity of the material. Without
this cross-linking, the polymer chains would behave more like a pile of
spaghetti.

44. Example
Hold a rubber band (the thicker the better) against your upper lip, and notice
how the temperature changes when the band is stretched, and then again when
it is allowed to contract.
a) Use the results of this observation to determine the signs of ΔH, ΔG and
ΔS for the process
rubberstretched → rubberunstretched
b) How will the tendency of the stretched rubber to contract be changed if the
temperature is raised?

45. Solution
 a) Contraction is obviously a spontaneous process, so ΔG is negative.
You will have observed that heat is given off when the band is stretched,
meaning that contraction is endothermic, so ΔH > 0. Thus according to
ΔG = ΔH – TΔS, ΔS for the contraction process must be positive.
b) Because ΔS > 0, contraction of the rubber becomes more spontaneous as the
temperature is raised.