BAKU ENGINEERING UNIVERSITY
ENGINEERING FACULTY
DEPARTMENT OF PHYSICS
Speciality: Electrical engineering
Dual Degree Program
Academic year and term: 2022-2023, Autumn term
Subject: Physics 1
The student’s first name and surname: Ali Safarli
Individual Activity Work
Entropy. Change in entropy.
The teacher: Prof. Dr. Vali Huseynov, Corresponding member of the Azerbaijan
National Academy of Sciences
BAKU-2022
There are many one-way processes that is, processes that can occur only in a
certain sequence (the right way) and never in the reverse sequence (the wrong
way). For example, a glass is dropped onto a floor, bread is baked, a paper is
torn, iron rusts. These processes cannot be reversed by the use of only small
changes in their environment . They are called irreversible processes.
One of goals of physics to understand why one way processes is irreversible
and why time has direction.
We use a quantity which is called entropy to understand why one-way process
cannot be reversed.
torn sheet
wrought iron
broken bottle
Irreversible Processes and Entropy
We would be shocked if these mechanisms started acting improperly on their
own (randomly). However, neither of these events would violate the rule of
energy conservation. For instance, you would be surprised if your hands
became colder as the cup of hot tea warmed up if you wrapped your hands
over it. That is undoubtedly the incorrect method for transferring energy, but
the overall energy of the closed system (the hands holding the tea) would be
the same as the total energy if the procedure had been carried out correctly.
As a result, irreversible processes are not determined by changes in energy
within a closed system. Instead, the system's change in entropy ∆S, which we
will analyze in this chapter, determines that direction. The entropy postulate,
which is typically used to characterize the entropy change of a system, has the
following basic property:
If an irreversible process occurs in a closed system, the entropy S of the
system always increases; it never decreases.
In contrast to energy, entropy is not subject to a conservation law. A closed
system's energy is always constant since it is conserved. A closed system's
entropy constantly rises for irreversible operations. The entropy change is
commonly referred to as "the arrow of time" because of this characteristic. As
an illustration, we link the explosion of a granade to the passage of time and a
rise in entropy. Time moving backward (a videotape played backward) would
be analogous to the exploded granade resembling the initial granade. This
reverse process never takes place since it would cause the entropy to
decrease. The two following definitions of a system's change in entropy are
equivalent:
by counting the possible arrangements of the atoms or molecules that make
up the system, as well as in terms of the system's temperature and the energy
it gains or loses as heat. In this post, we apply the first option.
Change in Entropy
Let's approach this explanation of entropy change by revisiting a procedure
known as the free expansion of an ideal gas. The gas is seen in the image
below being contained by a closed stopcock to the left side of a thermally
insulated container in its initial equilibrium condition i.
initial state i
final state f
Upon opening the gas rushes to fill the entire container and ultimately reaches
the stopcock as the gas The following image displays the equilibrium state f.
This is an irreversible process. All the molecules of the gas will never return to
the left half of the container.
The process's p-V graphic displays the gas's pressure and volume in both its
initial condition I and final state (f). Volume and pressure are state attributes,
meaning they only depend on the gas's current condition and not how it got
there. Temperature and energy are other state characteristics. We now
assume that the gas has an additional state characteristic called entropy. In
addition, we describe a system's change in entropy Sf - Sᵢ during a process that
moves the system from a starting state i to an ending state f as
𝒇
∆𝑺 = 𝑺𝒇 − 𝑺ᵢ = ∫
𝒊
𝒅𝑸
𝑻
i
Here, Q represents the energy that is transported as heat to or from the
system throughout the operation, and T represents the system's temperature
in kelvins. As a result, entropy changes are dependent on both the
temperature at which they occur and the amount of heat energy transported.
The sign of ∆S is the same as that of Q since T is always positive. The joule per
kelvin is the SI unit for entropy and entropy change, as shown by the Equation.
However, there is an issue when using the Equation to calculate the free
expansion. The pressure, temperature, and other variables increase as the gas
rushes to fill the entire The gas's temperature and volume fluctuate erratically.
In other words, during the transitional phases from initial state i to final state f,
they lack a sequence of clearly defined equilibrium values. Consequently, we
are unable to identify a pressure-volume relationship. We are unable to
discover a relationship between Q and T that enables us to integrate as the
equation requires path for the free expansion on the p-V plot.
The difference in entropy between states I and f, however, must only depend
on those states and not at all on how the system changed from one state to
the next if entropy is really a state property. So let's say that we choose a
reversible process to link the states I and f in place of the irreversible free
expansion. We can follow a pressure-volume
path for a reversible process on a p-V plot, and
we can discover a relationship between Q and T
that enables us to use the Equation to calculate
the entropy change.
We observed that an ideal ideal gas's temperature remains constant during a
free expansion: Ti = Tf = T. The p-V plot's points I and f must therefore be on
the same isotherm. Then, a reversible isothermal expansion from state I to
state f, which actually follows that isotherm, is a practical replacement
process. Furthermore, the integral of the Equation is greatly facilitated
because T remains constant throughout a reversible isothermal expansion.
How to create such a reversible isothermal expansion is demonstrated in the
image below.
initial state i
final state f
We contain the gas in an insulated cylinder resting on a thermal reservoir kept
at T. We start by adding just enough lead shot on the moveable piston to
match the gas's initial condition I in the first image's initial pressure and
volume. Once the pressure and volume of the gas match those of the final
state f of the first image, we progressively (and piece by piece) remove the
shot. Because the gas continues to be in thermal contact with the reservoir
throughout the operation, its temperature does not change.
Physically, the irreversible free expansion
and the reversible isothermal expansion
are very unlike. However, because the
beginning and ultimate states of both
processes are the same, the change in
entropy for both must be the same. We
may draw the intermediate states of the
states of the gas on the p-V diagram because they are equilibrium states as a
result of our gradual removal of the lead shot. We take the constant
temperature T outside the integral to first apply to the isothermal expansion,
obtaining:
Given that Q is the total amount of energy transported as heat throughout the
process and ∫dQ=Q we have:
It must have been possible for heat Q to have been transferred from the
reservoir to the gas in order to maintain the gas's temperature T throughout
the isothermal expansion. Q is thus positive, and both the isothermal process
and the first free expansion result in an increase in the gas's entropy.
To summarize:
To find the entropy change for an irreversible process, replace that process
with
any reversible process that connects the same initial and final states.
The entropy change can be large when the temperature change ∆T of a system
is large compared to the temperature (in kelvins) before and after the process
equivalent to:
Entropy as a State Function
We have assumed that entropy is a characteristic of the state of a system,
regardless of how that state is attained, much like pressure, energy, and
temperature. That entropy is a state function that can only be produced by
experiment. However, for the unique and significant scenario in which an ideal
gas is put through a reversible operation, we can demonstrate that it is a state
function.
The procedure is carried out gradually in a succession of short stages, with the
gas in an equilibrium condition at the conclusion of each step, in order to make
it reversible. For each little step, the gas performs work by putting off dQ heat,
which is the energy transferred as heat to or from the gas. is dW, whereas
dEint represents the change in internal energy. The first law of
thermodynamics in differential form connects them.
With the gas in equilibrium states, we can swap out dW for p dV and dEint for
nCv dT since the procedures are reversible. The result of finding dQ is equal to:
We substitute nRT/V in place of p in this equation by applying the ideal gas
law. As a result, we obtain by dividing each term in the resulting equation by T:
Let's now integrate each term of this equation between an arbitrarily chosen
starting point i and a randomly chosen ending point f to obtain:
And final equation is: