Entropy: Definition, Application & Implementation

The burning of wood into ash, smoke, and gas melting of ice into water, dissolution of salt and sugar in water, popping of popcorn, the boiling of water are some examples of a day-to-day arrangement of particles as their state changes. The process behind these phenomena is called entropy.

Entropy is not restricted to physics or chemistry; the concept has got a lot of recognition in the field of biology and its relations with life, cosmology, economics, and the information system. Entropy in chemistry can be defined as the “state of disorder” of the system. Summing up, it is how much the energy of the atoms and molecules is during the process. The higher the entropy goes, the more difficult it is to determine the state of the system or object. The universe poses the highest known entropy in humankind.

As in physics, entropy is defined as the quantitative amount of heat energy that is unavailable to carry work of the system. For a thermodynamic process, entropy is a measurable physical quantity. It can be used to discuss the direction of the reaction vis a vis process whether it is reversible, irreversible, or impossible to occur in nature.

Also, check out the Basics of Thermodynamics, here.

What is Entropy?

The first law of thermodynamics makes us familiar with the concept of Internal energy. Entropy is a concept that was formulated by the second law of thermodynamics. The second law of thermodynamics states that “the state of entropy of the entire universe, as an isolated system, will increase over time”. It also includes that Entropy can never be negative for the universe i.e. dS>0.

It is denoted by S. The SI unit of entropy is joules per Kelvin \((J K^-1 )\) or \(kgm^2s^-2K^-1\). The Dimensional formula of entropy is \( [M^1L^2T^-2K^-1]\).

Entropy in thermodynamics is defined as the ratio of the heat (Q) involved in the process per unit Kelvin temperature (T).

Mathematically,

\( S=\frac{Q}{T}\)

Fig (a)

For a reversible process,

\(dS_{rev}=\left(\frac{dQ}{T}\right)_{rev}\)

For Irreversible process,

\(dS_{irrev}>\frac{dQ}{T}\)

For any process,

\(ΔS_{rev}=ΔS_{irrev}=S_2-S_1=\int_1^2\frac{dq_{rev}}{T}\)

As the statistical interpretation of the entropy, it is defined as proportional to the natural log of the total number of microstates. Mathematically, represented as

\(S=K_b\log Ω\)

Here, S is the entropy, KB is the Boltzmann constant and Ω is the number of the microstates.

In terms of thermodynamic work, entropy is defined as proportional to the amount of work done in the system and surroundings. Mathematically,

\(S=K_b\log W\)

For the work done by the system, its negative S<0.

For the work done on the system, it’s positive S>0.

Entropy Key Points:

• • Principle of entropy: According to this entropy change of the universe which is system plus surrounding is always greater than or equal to zero. This is the principle of increase of entropy or the entropy principle.
  • It is never negative. Keeps on increasing.
  • The reactions depicted on basis of entropy of the universe (i.e. sum of the entropies of system and surroundings);

Irreversible reactions or process \(ΔS_{UNI}>0\)

Reversible reactions or process \(ΔS_{UNI}>0\) =0

Impossible reactions or process \(ΔS_{UNI}>0\) <0

For Reversible Adiabatic reactions or processes, entropy is constant.

Learn more about the laws of thermodynamics.

Entropy Change For a System

The change in entropy is defined as the change in the heat (ΔQ) during a reaction per unit kelvin temperature {T) is called the Entropy change of the system.

Mathematically,

The integration is carried out from the initial state to the final state. It does not depend on the path, it’s a state function.

Entropy is defined as the sum of the entropy due to internal reversibility and the entropy due to external interaction. For the internally irreversible state;

ΔS = change in entropy

dq/T = entropy transfer by external interaction

= entropy generation due to internal interaction

For a reversible {internally} state, = 0

Fig (b)

Heat Exchange	Process	Entropy
Heat is added	Reversibly	Positive
Heat is released	Reversibly	Negative
Insulated system (no heat exchange)	Reversibly	Zero i.e. Isentropic process
Adiabatic system	Irreversibly	dS=SGEN

Entropy Points to Note:

Entropy change of the system can be positive, negative, or zero but the entropy change of the universe can never be negative.

• • If a process is reversible and adiabatic then it is also isentropic.
  • The value of the entropy generation is always greater than zero.
  • If the internal reversibility is compensated by a decrease in external interaction then also we can have a zero change in entropy i.e. isentropic process.
  • All the adiabatic processes are isentropic processes but the reverse may not be true due to the above cases.
  • If a perfect gas undergoes a reversible polytropic process then change in entropy.

Here n= polytropic index

𝛄= Specific Heat Ratio

Implementation of Principle of Entropy

Heat exchange in System (for finite temperature)

Let us consider a system having two reservoirs A & reservoir B having temperature T1 and T2, the exchange in heat is Q and the entropy change is ΔS. The changes in the entropy of the reservoir are ΔSA and ΔSB.

Fig (c) Heat exchange in system

Entropy for two Miscible Liquids:

Heat exchange is mathematically derived as

dQ=mCdT

Entropy can be explained in form of specific heat as

Now, we study the method to find out the entropy change for liquids.

For this, we will consider two different liquids A & B of mass and . The specific heat of the liquids are & and the absolute temperatures are Ta & Tb respectively. They are placed in insulated vessels.

The total entropy of the universe will be

\({l}ΔS_{uni}=ΔS_{sys}+ΔS_{sur}\)

\(ΔS_{sur}=0(InsulatedVessels)\)

\(ΔS_{uni}=ΔS_{sys}\)

\(ΔS_{uni}=m_aC_a\ln\left(\frac{T_f}{T_a}\right)+m_bC_b\ln\left(\frac{T_f}{T_b}\right)\)

Tf is the final temperature of the mixture.

Maximum Amount of Work Done :

The total amount of work done by the bodies at a given finite temperature can be calculated with the help of the entropy. We will learn about how to calculate the maximum obtained work.

Let’s consider the two containers of the temperature of \(T_1\)& \(T_2\). The process is carried out at constant pressure so the specific heat at constant pressure will be CP. The heat transfer is \(Q_1\)for the first process and \(Q_2\)for the other and the final temperature is Tf

The change in the heat is stored as the work done W. The work done W = change in heat (ΔQ)

fig(d) work and entropy relation

Derivation:

The heat exchange in the first process

\(Q_1=C_P(T_1-T_f)\)

The heat exchange in the second process

\(Q_2=C_P(T_2-T_f)\)

Now, work done

\({l}W=Q_2-Q_1\)

\(W=C_P(T_1+T_2-2T_f)\)

\(ΔS_{uni}=C_P\ln\left(\frac{T_f}{T_1}\right)+C_P\ln\left(\frac{T_f}{T_2}\right)=C_P\ln\left(\frac{T_f^2}{T_1T_2}\right)\)

For

\(ΔS_{uni}\ge0\)

This implies work is reversible and entropy \(ΔS_{UNI}>0\)= 0

Solving the above relations we get

\(T_f=\sqrt{T_1T_2}\)

I.e. The final temperature will be the square root of the product of the initial temperatures.

\(W_{\max}=C_P(\sqrt{T}_1-\sqrt{T}_2)^2\)

Application in First and Second Law of Thermodynamics

Entropy can be used to merge the first and second law and derive its combined form.

Let’s assume a closed system, as pressure is kept constant the temperature varies from \(T_i\) to \(T_f\). the volume of the system changes from \(V_i\) to \(V_f\) .So, now from the first law of thermodynamics (a closed system)

Integrating the terms w.r.t itself

\(ΔS=C_V\ln\left(\frac{T_f}{T_i}\right)+R\ln\left(\frac{V_f}{V_i}\right)\)

Where ΔS = change in the entropy

Now,

From second law of thermodynamics;

\({l}H=U+PV\\∴TdS=dH+VdP\)

\(∴TdS=C_pdT-VdP\)

\(ΔS=C_P\ln\left(\frac{T_f}{T_i}\right)-R\ln\left(\frac{P_f}{P_i}\right)\)

\(C_P\) = SPECIFIC HEAT AT CONSTANT PRESSURE

\(C_V\) = SPECIFIC HEAT AT CONSTANT VOLUME

Carnot engines or heat engines are also based on the applications of heat exchange at finite temperatures.

Entropy Points to Note:

Different Processes In Terms Of Entropy

Isobaric Process:

Pressure in these processes is constant i.e. P= constant. So, dP=0

\(\left(\frac{dT}{dS}\right)_{_P}=\left(\frac{T}{C_P}\right)\)

Isochoric Process:

Volume remains constant in this process i.e. V=constant. So, dV=0

\(\left(\frac{dT}{dS}\right)_{_V}=\left(\frac{T}{C_V}\right)\)

Note: Slope of Isochoric process is greater than the Isobaric Process

fig(e) T-S DIAGRAM

Isentropic Process:

Process for which the entropy of the system remains constant i.e. dS=0.

The T-S graph for the process will be a vertical line with an infinite slope.

The slope of the graph= ထ

Isothermal Process:

Process for which the temperature of the system remains constant i.e.dT=0.

The T-S Graph for the process is a horizontal line with zero slope.

Slope of the Isothermal Process =0

Learn more about the thermodynamic process and solved examples.

fig(f)

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