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THERMODYNAMICS 1
The First Law and Other Basic Concepts (part 1)
Department of Chemical Engineering, Semarang State University
Dhoni Hartanto S.T., M.T., M.Sc.
Joule’s Experiments
Understanding of heat and its relation to work developed during the last
half of the nineteenth century
James P. Joule carried out the experiment in Manchester, England. His
experiment is simple enough with good accuracy.
He placed known amounts of water, oil, and mercury in an insulated
container and agitated the fluid with a rotating stirrer.
The work done on the fluid by stirrer were accurately measured and the
temperature changes of fluid were carefully noted.
Joule found for each fluid fixed amount of work was required per unit mass
for every degree of temperature rise caused by the stirring, and that the
original temperature of the fluid could be restored by the transfer of heat
through simple contact with a cooler object
Joule concluded that heat is a form of energy
Internal Energy
In Joule’s experiment, we can see that energy is added into a fluid, and it is
transferred from the fluid as heat.
Why ??
Another form called internal energy is contained in the fluid.
The internal energy of a substance does not include energy that it may
possess as a result of its macroscopic position or movement.
Internal energy refers to energy of the molecules internal to the substances
All molecules possess kinetic energy of translation, except monoatomic
molecules which also have kinetic energy of rotation and internal vibration.
The addition of heat or work to a substance increases molecular activity and
then increases internal energy
Internal Energy (cont.)
The internal energy of a substance also includes the potential energy
resulting from intermolecular forces.
On a submolecular scale energy is associated with the electrons and nuclei
of atoms, and with bond energy resulting from the forces holding atoms
together as molecules. This form of energy is named internal to distinguish
it from the kinetic and potential energy associated with a substance because
of its macroscopic position or motion, which can be thought of as external
forms of energy.
Internal energy, has no concise thermodynamic definition. It is a
thermodynamic primitive. It cannot be directly measured; there are no
internal-energy meters. As a result, absolute values are unknown. However,
this is not a disadvantage in thermodynamic analysis, because only changes
in internal energy are required
The First Law of Thermodynamics
“Although energy assumes many forms, the total quantity of energy is
constant, and when energy disappears in one form it appears
simultaneously in other forms”
In application of the first law of the given process, it divided into two parts :
1. System
2. Surrounding
In its most basic form, the first law requires :
∆(Energy of the systems) + ∆(Energy of the surroundings) = 0
The system may change in its internal energy, in its potential or kinetic
energy, and in the potential or kinetic energy of its finite parts.
The First Law of Thermodynamics (cont.)
In the thermodynamic, heat and work refer to energy in transit across the
boundary which divides the system from its surroundings
These forms of energy are not stored, and are never contained in a body or
system
Energy is stored in its potential, kinetic, and internal forms; these reside
with material objects and exist because of the position, configuration, and
motion of matter
Energy Balance for Closed Systems
Closed systems : no transfer of matter between systems in its surroundings
no streams enter or leave in systems
no internal energy is transported accros the boundary of
the systems
surroundings
systems
Closed Systems
surroundings
energy
systems
energy
mass
Open Systems
All energy exchange between a closed system and its surroundings then
appears as heat and work
Total energy change of equals the net energy transferred to or from it as
heat and work
Energy Balance for Closed Systems (cont.)
∆(Energy of the surroundings) = ± Q ± W
Sign positive means transfer into systems from surroundings
The equation become
∆(Energy of the surroundings) = Q + W
The equation means that the total energy change in closed system equals to
the net energy transferred into it as heat and work.
Energy Balance for Closed Systems (cont.)
Closed systems often happened in the systems which no change occurred
other than its internal energy.
∆Ut = Q + W
Where Ut total internal energy of the system in finite changes.
For differential changes, the equations yield
dUt = dQ + dW
Energy Balance for Closed Systems (cont.)
Properties such as Vt and Ut depend on the quantitiy of material in a
systems, it is called extensive properties.
Properties in homogeneous fluid such as temperature and pressure are not
depend on the quantitiy of the materal in a systems, it is called intensive
properties.
Although Vt and Ut for a homogeneous systems of arbitrary size are
extensive properties, specific and molar volume V (or density) and specific
and molar internal energy U are intensive properties.
For closed system of n moles,
∆(nU) = n ∆U = Q + W
d(nU) = n dU = dQ + dW
When n = 1 mole, thus
∆U = Q + W
dU = dQ + dW
Energy Balance for Closed Systems (cont.)
The equation connect the internal energy to measurable quantities.
It does not represent a definition of internal energy nor lead to absolute
values of internal energy, it is just provide calculation of internal energy
changes
Example
Water flows over a waterfall 100 m in height. Take 1 kg of the water as
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 base of the falls?
b) What is the kinetic energy of the water just before it strikes bottom?
c) After 1 kg of water enters the river below the falls, what change has
occured in its state?
Energy Balance for Closed Systems (cont.)
Solution :
No exchange with surroundings
∆(Energy of the system) = 0
Energy : internal, kinetic, and potential
∆U + ∆Ek + ∆Ep = 0
a) Ep = m g z = 1 kg x 100 m x 9.8066 m s-2
= 980.66 kg m2/ s2
= 980.66 N m = 980.66 J
b) No mechanism exist for conversion of potential or kinetic energy into
internal energy during free fall (∆U must be zero)
∆Ek + ∆Ep = Ek2 – Ek1 + Ep2 – Ep1 = 0
Assumtion Ek1 = Ep2 = 0, then
Ek2 = Ep1 = 980.66 J
Energy Balance for Closed Systems (cont.)
Solution (cont.) :
c) When 1 kg of water strikes bottom and mixed with other falling water,
kinetic energy is converted into internal energy. During this process, ∆Ep =
0.
∆U + ∆Ek = 0
or
∆U =Ek2 – Ek3
The river velocity is small, so that Ek3 is negligible. Thus
∆U =Ek2 = 980.66 J
The temperature increase occured.
(Energy required for a temperature rise of 1 oC is 4184 J kg-1). Thus the
temperature increase is
980.66/4184 = 0.234 oC (assumption : no heat transfer occured with
surrounding)
Thermodynamic state and state functions
The internal energy term on the left side of equation reflect changes in the
internal state or the thermodynamic state of the system.
This state is reflected by its thermodynamic properties, among which are
temperature, pressure, and density.
For a homogeneous pure substance, fixing two of these properties
automatically fixes all the others
i.e Nitrogen gas at temperature 300 K and a pressure 105 kPa (1 bar) has a
fixed specific volume or density and fixed molar internal energy.
If this gas is heated or cooled, compressed or expanded, and then returned
to its initial temperature and pressure, its intensive properties are restored
to their initial values.
Thermodynamic state and state functions
(cont.)
Such properties do not depend on the past history of the substance nor on
the means by which it reaches a given state
They depend only on present conditions, and it is called state functions.
State functions such as internal energy is a property that always has a value,
it may be expressed mathematically as a function of other thermodynamics
properties such as temperature and pressure, or temperature and density.
The right side of equation represents heat and work which is not properties.
They account for the energy changes that occur in the surroundings and
appear only when changes occur in a system.
Thermodynamic state and state functions
(cont.)
The differential of a state function represent as infinitesimal change in its
value. In example :

P2
P1
dP  P2  P1  P
V2

and
V1
dV  V2  V1  V
The differential of heat and work are not change, but are infinitesimal amounts.
When integrated, these differentials give not finite changes, but finite amounts. Thus,
 dQ  Q
and

dW  W
For closed system, the amounts of heat and work required differ for different process,
But that the sum of Q +W is the same for all processes.
This is the basis for identification of internal energy as a state function.
Thermodynamic state and state functions
(cont.)
Example :
A gas is confined in a cylinder by a piston. The initial pressure of the gas is 7
bar, and the volume is 0.10 m3. The piston is held in place by latches in the
cylinder wall. The whole apparatus is placed in a total vacuum. What is the
energy change of the apparatus if the restraining latches are removed so
that the gas suddenly expands to double its initial volume, the piston
striking other latches at the end of the process?
Solution :
System : the gas, piston, and cylinder.
No work done during the process (no force external moves to the system, no
heat is transferred)
Q and W are 0
Total energy of the system doesn’t change
It can not explain more about the distribution of energy among the parts of
the system when further information are not available
Thermodynamic state and state functions
(cont.)
Example :
If the process in previous example is repeated, not in vacuum but in air at
atmospheric pressure of 101.3 kPa, what is the energy change of the
appratus?
Assume the rate of heat exchange between the apparatus and the
surrounding air is slow compared with the rate at which the process occurs.
Solution :
System : the gas, piston, and cylinder.
Work is done by the system in pushing back the atmosphere
Work is given by the product of the force exerted by atmospheric pressure
on the back side of the piston and the displacement of the piston.
Thermodynamic state and state functions
(cont.)
Solution (cont.):
Work done by system : F ∆l= Patm ∆Vt
= (101.3 kPa) (0.2-0.1) kPa m3 = 10.13 kNm
Since work done on the system, so that
W = -10.13 kNm = -10.13 J
Heat transfer is possible occur on the system, but the problem is worked for
the instant before heat transfer take place. Thus Q is assumed to be 0.
∆(Energy of the system) = Q + W = 0 – 10.13 = -10.13 J
Total energy of the system has decreased by amount equal to work done on
the surrounding
Thermodynamic state and state functions
(cont.)
Example :
When a system is taken from state (a) to state (b) in Figure below along
path (acb), 100 J of heat flows into the system and the system does 40 J of
work.
a) How much heat flows into the system along path (aeb) if the work done
by the system is 20 J?
b) The system returns from (b) to (a) along path (bda). If the work done on
the system is 30 J, does the system absorb or liberate the heat? How
much?
Thermodynamic state and state functions
(cont.)
Solution:
Assume the system change only in internal energy.
t
For path (acb) U ab
 Qacb  Wacb  100  40  60 J
The internal energy change from (a) to (b) by any path
a) Path (aeb)
t
U ab
 60  Qaeb  Waeb  Qaeb  20
Qaeb  80 J
b) Path (bda)
t
t
U ba
 U ab
 60  Qbda  Wbda  Qbda  30
Qbda  60  30  90 J
Heat is transferred from the system to the surroundings
Thank you