THERMODYNAMICS Introductory lecture 5.ppt

Chemical Engineering Thermodynamics-I
Lecture-4
By
Dr. D. M. Nangare

EQUILIBRIUM
A state in which opposing forces or influences are
balanced.
is the state of a system, in which properties have definite,
unchanged values as long as external conditions are
unchanged
A static condition, the absence of change.

Zeroth law of thermodynamics
Thermal equilibrium :
If two bodies are in contact through a thermally-conducting
boundary for a sufficiently long time, no further observable
changes take place
Two systems, which are individually in thermal equilibrium
with a third are in thermal equilibrium with each other; all
three systems have the same value of the property called
temperature.

Object A is in equilibrium
with both the thermometer
and object B. Then the
thermometer should also be in
equilibrium with object B.
This means all three objects
have the same temperature
Thermodynamic equilibrium

First law of thermodynamics
During a process:
Energy can be transferred and converted from one form to
another, while the total energy remains constant
Δ(Energy of the system) + Δ(Energy of the surroundings) = 0

Internal Energy U
No concise thermodynamic definition
On submolecular scale energy is associated with the
electrons and nucli of atoms, and with bond energy resulting
from the forces holding atoms together as molecules.
Cannot be directly measured
In thermodynamics, only changes in internal energy
are used ΔU=U2-U1

All the energy exchange between a closed system and its
surroundings appears as heat and work, and the total energy
change of the surroundings equals The net energy transferred
to or from it as heat and work.
Energy Balance for a Closed System

Water flows over a waterfall 100 m height. Take 1 kg of
water as the system, and assume that it does not exchange
energy with its surroundings.
a) What is the potential energy of the water at the top of the
falls with respect to the base of the falls?
b) What is the kinetic energy of the water just before it
strikes the bottom
c) After the 1 kg of water enters the stream below the falls,
what change has occurred in its state?

3000 J of heat is added to a system and 2500 J of work is
done by the system. What is the change in internal energy
of the system?
ΔU = Q-W
The sign conventions :
Q is positive if the heat added to the system
W is positive if work is done by the system
Q is negative if heat leaves the system
W is negative if work is done on the system
The change in internal energy of the system :
ΔU = 3000-2500
ΔU = 500 Joule
Internal energy increases by 500 Joule.

2000 J of heat is added to a system and 2500 J of work is
done on the system. What is the change in internal energy
of the system?
ΔU = Q+W
ΔU = 2000+2500
ΔU = 4500 Joule

2000 J of heat leaves the system and 2500 J of work is
done on the system. What is the change in internal energy
of the system?
ΔU = -2000+2500
ΔU = 500 Joule

The Gibbs Phase Rule for Multicomponent Systems
the phase rule may be written either in the form
F=2+C−P
Where
F=the number of degrees of freedom (or variance)
=the maximum number of intensive variables that can be
varied independently while the system remains in an
equilibrium state
C=the number of components
= the minimum number of substances (or fixed-composition
mixtures of substances) that could be used to prepare each
phase individually
P= the number of different phases

Consider a system in an equilibrium state. In this state, the
system has one or more phases; each phase contains one or
more species; and intensive properties such as TT, pp, and the
mole fraction of a species in a phase have definite values.
Starting with the system in this state, we can make changes that
place the system in a new equilibrium state having the same
kinds of phases and the same species, but different values of
some of the intensive properties. The number of different
independent intensive variables that we may change in this way
is the number of degrees of freedom or variance, FF, of the
system.

For pure substances C = 1 so that F = 3 − P. In a single phase
(P = 1) condition of a pure component system, two variables
(F = 2), such as temperature and pressure, can be chosen
independently to be any pair of values consistent with the
phase. However, if the temperature and pressure combination
ranges to a point where the pure component undergoes a
separation into two phases (P = 2), F decreases from 2 to 1.
When the system enters the two-phase region, it becomes no
longer possible to independently control temperature and
pressure.

For binary mixtures of two chemically independent
components, C = 2 so that F = 4 − P. In addition to temperature
and pressure, the other degree of freedom is the composition of
each phase, often expressed as mole fraction or mass fraction of
one component.
Boiling Point Diagram
As an example, consider the system of two completely miscible
liquids such as toluene and benzene, in equilibrium with their
vapours. This system may be described by a
boiling-point diagram which shows the composition (mole
fraction) of the two phases in equilibrium as functions of
temperature (at a fixed pressure).

Liquid water in equilibrium with its vapor
Liquid water in equilibrium with a mixture of water vapor
and nitrogen
A liquid solution of alcohol in water in equilibrium with
the vapor

Constant Volume and Constant Pressure Process
For n moles of a homogenous fluid in a closed system
The energy balance is
d(nU)= dQ + dW (1)
The work for a mechanically reversible process is
d(W)= -Pd(nV) (2)
Considering equation (2) equation (1) become
d(nU)= dQ -Pd(nV) (3)

Constant Volume Process
If the process occurs at constant total volume, the
work will be 0. Equation (3) will be
d(nU) = dQ (4)
Q = n ΔU

Constant Pressure Process
If the process occurs at constant pressure, Equation
(3) will be
dQ =d(nU) + Pd(nV) (4)
dQ =d(nU) + d(nPV) = d [n (U+PV)]

Enthalpy
H = (U+PV)
Where H,U and V are molar or unit mass values

THERMODYNAMICS Introductory lecture 5.ppt

Calculate the heat and work requirements and ΔU
and ΔH of the air for each path.
Air at 1 bar and 298.15K (25 ) is compressed to 5 bar
℃
and 298.15K by a processes.
Cooling at constant pressure followed by heating at
constant volume.
The following heat capacities for air may be assumed
independent of temperature: CV= 20.78 and CP=29.10 J
mol-1
K-1
Assume also for air that PV/T is a constant,
regardless of the changes it undergoes. At 298.15K and
1 bar the molar volume of air is 0.02479 m3
mol-1

For complete process
PV/T=const P1 V1/ T1 = P2 V2/ T2
P1= 1 bar
P2= 5 bar
T1= 298.15 K
T2= 298.15 K
V1=0.02479 m3
V2= ???

During the first step the air is cooled at the constant pressure
of 1 bar until the final volume of 0.004958 m3
is reached.
The temperature of the air at the end of this cooling step is
For first half part
P1= 1 bar
P2= 1 bar
T1= 298.15 K
T’
= ?
V1=0.02479 m3
V2= 0.004958 m3

For constant pressure process:
T1= 298.15 K
T2= 59.63 K

For constant pressure process:
H = (U+PV)
∆H = ∆ U+ ∆(PV)
∆ U =∆H - ∆(PV)
∆ U =∆H - P∆V
∆ U = -6941-(1*105
(0.004958-0.002479)
∆ U= -4958 J
W =- P∆V
W =- (1*105
(0.004958-0.002479)) = 1983J

During the second step the volume is held constant
at while the air is heated to its final state
=4958J
W=0 J

The complete process represents the sum of its steps. Hence,
W=1983 + 0=1983 J

Calculate the heat and work requirements and ΔU
and ΔH of the air for each path.
Air at 1 bar and 298.15K (25 ) is compressed to 5 bar
℃
and 298.15K by a processes.
heating at constant volume followed by Cooling at
constant pressure
The following heat capacities for air may be assumed
independent of temperature: CV= 20.78 and CP=29.10 J
mol-1
K-1
Assume also for air that PV/T is a constant,
regardless of the changes it undergoes. At 298.15K and
1 bar the molar volume of air is 0.02479 m3
mol-1

Phases of a Pure Substance:
A pure substance may exist in different phases. There are three
principal phases solid, liquid, and gas.
A phase: is defined as having a distinct molecular arrangement
that is homogenous throughout and separated from others (if
any) by easily identifiable boundary surfaces. A substance
may have several phases within a principal phase, each with a
different molecular structure. For example, carbon may exist as
graphite or diamond in the solid phase, and ice may exist in
seven different phases at high pressure. Molecular bonds are the
strongest in solids and the weakest in gases.

Solid: the molecules are arranged in a three dimensional
‐
pattern (lattice) throughout the solid. The molecules cannot
move relative to each other; however, they continually
oscillate about their equilibrium position.
Liquid: the molecular spacing in liquid phase is not much
different from that of the solid phase (generally slightly
higher), except the molecules are no longer at fixed positions
relative to each other.
Gas: the molecules are far apart from each other, and a
molecular order does not exist. Gas molecules move
randomly, and continually collide with each other and the
walls of the container they are in. Molecules in the gas phase
are at a considerably higher energy level than they are in
liquids or solid phases.

Phase Change Processes of Pure Substances
‐
Consider a process where a pure substance starts as a solid and
is heated up at constant pressure until it all becomes gas.
Depending on the prevailing pressure, the matter will pass
through various phase transformations. At P0:
1. Solid
2. Mixed phase of liquid and solid
3. Sub cooled or compressed liquid (means it is not about to vaporize)
‐
4. Wet vapor or saturated liquid vapor mixture, the temperature will
‐
stop rising until the liquid is completely vaporized.
5. Superheated vapor (a vapor that is not about to condense).

There are two ways that a substance can pass from solid phase
to vapor phase i) it melts first into a liquid and subsequently
evaporates, ii) it evaporates directly without melting
(sublimation). the sublimation line separates the solid and the

vapor. the vaporization line separates the liquid and vapor

regions the melting or fusion line separates the solid and liquid.

these three lines meet at the triple point. if PPTP ) the
 
substance melts into a liquid and then evaporates. matter (like

CO2) which has a triple point above 1 atm sublimate under
atmospheric conditions (dry ice)
for water (as the most common working fluid) we are mainly
interested in the liquid and vapor regions. Hence, we are mostly
interested in boiling and condensation

At a given pressure, the temperature at which a pure substance starts
boiling is called the saturation temperature, Tsat.
Likewise, at a given temperature, the pressure at which a pure
substance starts boiling is called the saturation pressure, Psat.
During a phase change process, pressure and temperature are
‐
dependent properties, Tsat = f (Psat).
The critical point is the point at which the liquid and vapor phases are
not distinguishable.
The “triple point” is the point at which the liquid, solid, and vapor phases
can exist together. On P v or T v diagrams, these triple phase states
‐ ‐ ‐
form a line called the triple line.

Expansivity and compressibility
We now consider a simple system that can be described by
assigning P,V, and T.
The equation of state is f(P,V,T) = 0
Two quantities are found to be independent. We can, for
example, solve the equation of state for V.
V = V(T,P)

Consider an infinitesimal change from one equilibrium
state to another equilibrium state. The temperature and
pressure, the two variables chosen to be independent, will
generally change and we can then write, for the
infinitesimal change in V.
Each partial can be a function of P and T.
We cannot integrate this equation to obtain the change in volume
when there are temperature and pressure changes because the
partial derivatives are unknown. One way to proceed is to try to
obtain expressions for the partials by creating a model of the
system.
However the approach in macroscopic thermodynamics is to
appeal to experimental measurements.

We define two quantities which can be measured and are often
tabulated:
Volume expansivity
Isothermal compressibility
 
T
V,
P,

   
T
V,
P,

 
 P
T
V
V
1




 T
κ P
V
V
1




P
d
V
T
d
V
V
d 
 


For liquid acetone at 20 0
C and 1 bar
β=1.487 * 10 -3 0
C-1
κ = 62* 10 -6
bar -1
V=1.287 cm3
g-1
a) Find the value of (∂P/∂T)v at 20 0
C and 1 bar
b) Pressure generated by heating at constant V from 20 0
C
and 1 bar to 30 0
C
c) Change in volume for a change from 20 0
C and 1 bar to
0 0
C and 10 bar

P
d
V
T
d
V
V
d 
 

(∂P/∂T)v = β/k = (1.487 * 10 -3
)/ (62* 10 -6
)= 24 bar 0
C-1
ΔP=
β/k ΔT = 24* 10 =240 bar
P2=p1+ ΔP = 1+ 240 = 241 bar
Ln (v2/v1)=
β(T2-T1) – k(P2-P1)
Ln (v2/v1)=
0

 P
d
T
d 


THERMODYNAMICS Introductory lecture 5.ppt

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