When all entropy changes occur in a process, the total entropy change of the universe is always greater than or equal to zero. This is the second law of thermodynamics.
In 1850, the scientist Clasius first used the word entropy, which means transformation. The unit of entropy is joule/°K or joule/°C.
Examples
1. Heat transfer from a hot object to a cold object
3. Mixing of gases
5. A hot cup of coffee cooling in a room
6. A sugar cube dissolving in a glass of water
7. A chemical reaction occurring in a closed system
8. The expansion of the universe
Entropy ProcessÂ
(a) Change in Entropy in Isothermal Process
In an isothermal process, the temperature T will remain constant when
Change entropy = dQ/T
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(b) Change in Entropy in Adiabatic ProcessÂ
There will be no exchange of heat in the Adiabatic Process
Change entropy = 0
Thus in the adiabatic process, the entropy of the system remains constant.
(c) Change in Entropy in Reversible Process
If a Carnot cycle is taken, then the complete cycle of the engine can be represented by two isothermal curves AB and CD, and Rughdosham curves DA and BC as shown in the figure.
From A to B, the heat T1 is taken on the heat engine, so the entropy is on the AB curve.
S1 = Q1/T1
Q2 from C to D is removed by the heat engine at temperature T2.
S2 = Q2/T2
So now the change in entropy is coming from curve AB to curve CD.
ds = S1 - S2 .....….(१)Â
Now put the values ​​of S1 and S2 in equation (i).
ds = Q2/T2Â - Q1/T1
For Carnot engine
η = 1 - Q2/Q1 = 1 - T2/T1
or Q2/Q1 = T2/T1 or Q1/Q2 = T1/T2
Q1/T1 = Q2/T2
For Equation (i)Â
ds = Q2/T2 - Q2/T2
ds = 0
That is, the change in entropy remains zero in a reversible process. That is, entropy remains constant in a reversible process.
(d) Change in Entropy in Irreversible Process
In an irreversible engine, if engine Q1 receives heat at temperature T1, then
S1 = Q1/T1 ...........(i)
If Q2 removes heat at temperature T2, then
S2 = Q2/T2 ............(ii)
Change in entropy = S2 - S1
= Q2/T2 - Q1/T1 ........(iii)
the efficiency of this irreversible entropy
ηir = 1 - Q2/Q1 ........(iv)
If the Carnot engine is working between T2 temperature and T2 then
ηr = 1 - T2/T1 .......(v)
ButÂ
ηr > ηir
1 - T2/T1 > 1 - Q2/Q1
S1 - S2 = Q2/T2 - Q1/T1 (+ve)
Or Q2/Q1 < T2/T1
Or Q1/T1 < Q2/T2 .......(vi)
From equation (iii) and (vi)Â
ds = S2 - S1 = Q2/T2 - Q1/T1 (+ve)
That is, there is an increase in entropy in an irreversible process.
Principle of Increase of Entropy
The entropy of the world is increasing, it can be explained in the following way.
When there is a change in the entropy of a body, there is also a change in the entropy of its surroundings. All natural processes like conduction, radiation, etc. If the system is taken from initial state I to final state F by a sequential process, then the change in entropy of the system is Sf - Si.
∆System = Sf - Si
ഽ dQ/T (limit i to f) cannot be determined by, because this formula is only for thermodynamic processes, instead of a true reversible process, we will find the change in entropy Sf − Si for a reversible process that can take the system from state i to f. is a function of the state of the system. Hence the entropy change for a truly irreversible process will also have this value.
Also Read:Â
Ideal Gas Equations | Daltons Law, Amagets Law, and PVT Relationship
