We were discussing “

Today we will see here the concept of Clausius’ theorem with the help of this post.
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As we have seen above that the area below the reversible path i-f must be equal to the area below i-a-b-f or W

Therefore, we will have

We were discussing above single reversible process, now we will analyze here one complete reversible cycle as shown in figure. Let the reversible cycle is divided into large number of strips and these strips will indicate the reversible adiabatic lines as shown in figure. These strips are closed on bottom and tops with the help of reversible isothermal lines.

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*Carnot cycle and its efficiency*” as well as “*efficiency of reversible heat engine*” and “*COP of refrigerator and heat pump*” in our previous posts. We have also seen the concept of “*corollary of Carnot’s theorem*” and simultaneously “*equivalence of Kelvin Planck statement and Clausius statement”*in our recent posts.Today we will see here the concept of Clausius’ theorem with the help of this post.

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**Let
us see here now Clausius’ theorem**

According to Clausius’ theorem a reversible path or
process or line could be replaced by two reversible adiabatic processes or lines
and one reversible isothermal process or line.

Let us see the following figure, a system is at
equilibrium state i.e. at point i and system reaches to another equilibrium
state i.e. at point f by following the reversible path i-f.

According to Clausius’ theorem reversible process i.e.
i-f could be replaced by two reversible adiabatic processes i.e. i-a and b-f and
one reversible isothermal process i.e. a-b.

Process i-f: Reversible process

Processes i-a and b-f: Reversible adiabatic
processes

Process a-b: Reversible isothermal process

The replacement will be done in such a way that the
area below the reversible path i-f must be equal to the area below i-a-b-f. Mathematically
we will say that

Area
below i-f = Area below i-a-b-f

Let us apply the "

*first law of thermodynamics*" for following path or process
Process: i-f

Q

_{if}= W_{if}+ U_{f}-U_{i}
Process: i-a-b-f

Q

_{iabf}= W_{ iabf}+ U_{f}-U_{i}As we have seen above that the area below the reversible path i-f must be equal to the area below i-a-b-f or W

_{if}= W

_{ iabf}

Therefore, we will have

Q

_{if}= Q_{iabf}
Q

_{if}= Q_{ia}+ Q_{ab}+ Q_{bf}
As we know that i-a and b-f are reversible adiabatic
processes and therefore Q

_{ia}= Q_{bf}= 0
Q

_{if}= Q_{ab}
From above expression, we can say that heat
transferred during the reversible process i-f will be equal to the heat
transferred during the reversible isothermal process a-b.

Therefore we can say that any reversible process or
path could be replaced by a reversible zigzag path with same end states and
this reversible zigzag path will have two reversible adiabatic paths and one
reversible isothermal path.

This replacement will be done in such a way that heat transferred during the original reversible process will be equal to the heat transferred during the reversible isothermal process.

This replacement will be done in such a way that heat transferred during the original reversible process will be equal to the heat transferred during the reversible isothermal process.

We were discussing above single reversible process, now we will analyze here one complete reversible cycle as shown in figure. Let the reversible cycle is divided into large number of strips and these strips will indicate the reversible adiabatic lines as shown in figure. These strips are closed on bottom and tops with the help of reversible isothermal lines.

We can see here that original reversible cycle will
be divided here into numbers of small-small Carnot’s cycle as shown in figure.

Let us focus over here the Carnot’s cycle abcd, dQ

_{1}heat is absorbed reversibly at temperature T_{1}and dQ_{2}heat is rejected reversibly at temperature T_{2}.
dQ

_{1}/T_{1 }= dQ_{2}/T_{2}
Let us consider the sign convention and we will take
heat absorption as positive and heat rejection as negative.

dQ

_{1}/T_{1 }+ dQ_{2}/T_{2}= 0
In similar way for Carnot’s cycle efgh, dQ

_{3}heat is absorbed reversibly at temperature T_{3}and dQ_{4}heat is rejected reversibly at temperature T_{4}.
dQ

_{3}/T_{3 }+ dQ_{4}/T_{4}= 0
Similarly for complete reversible cycle, we will
have following equation

The cyclic integral of dQ/T for a reversible process
will be zero and that is the mathematical expression of Clausius theorem. We must
note it here that above equation will be only valid for a reversible cycle.

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We will see another topic "

*Clausius inequality and explain its significance*" in our next post in the category of thermal engineering.###
**Reference:**

Engineering thermodynamics by P. K. Nag

Image Courtesy: Google