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The figure shows four heat reservoirs. The first reservoir at temperature T has gas with volume V, the second reservoir at T plus delta T has gas V plus delta V, the third reservoir at T plus 2 delta T has gas V plus 2 delta V and the final reservoir at T plus n delta T has gas at V plus n delta V.
The gas expands at constant pressure as its temperature is increased in small steps through the use of a series of heat reservoirs.

As an example, let us determine the net entropy change of a reversible engine while it undergoes a single Carnot cycle. In the adiabatic steps 2 and 4 of the cycle shown in [link] , no heat exchange takes place, so Δ S 2 = Δ S 4 = d Q / T = 0 . In step 1, the engine absorbs heat Q h at a temperature T h , so its entropy change is Δ S 1 = Q h / T h . Similarly, in step 3, Δ S 3 = Q c / T c . The net entropy change of the engine in one cycle of operation is then

Δ S E = Δ S 1 + Δ S 2 + Δ S 3 + Δ S 4 = Q h T h Q c T c .

However, we know that for a Carnot engine,

Q h T h = Q c T c ,

so

Δ S E = 0 .

There is no net change in the entropy of the Carnot engine over a complete cycle. Although this result was obtained for a particular case, its validity can be shown to be far more general: There is no net change in the entropy of a system undergoing any complete reversible cyclic process. Mathematically, we write this statement as

d S = d Q T = 0

where represents the integral over a closed reversible path .

We can use [link] to show that the entropy change of a system undergoing a reversible process between two given states is path independent. An arbitrary, closed path for a reversible cycle that passes through the states A and B is shown in [link] . From [link] , d S = 0 for this closed path. We may split this integral into two segments, one along I, which leads from A to B , the other along II, which leads from B to A . Then

[ A B d S ] I + [ B A d S ] II = 0 .

Since the process is reversible,

[ A B d S ] I = [ A B d S ] II .
The figure shows a a pear shaped closed loop graph with x-axis V and y-axis p.
The closed loop passing through states A and B represents a reversible cycle.

Hence, the entropy change in going from A to B is the same for paths I and II. Since paths I and II are arbitrary, reversible paths, the entropy change in a transition between two equilibrium states is the same for all the reversible processes joining these states. Entropy, like internal energy, is therefore a state function.

What happens if the process is irreversible? When the process is irreversible, we expect the entropy of a closed system, or the system and its environment (the universe), to increase. Therefore we can rewrite this expression as

Δ S 0 ,

where S is the total entropy of the closed system or the entire universe, and the equal sign is for a reversible process. The fact is the entropy statement of the second law of thermodynamics    :

Second law of thermodynamics (entropy statement)

The entropy of a closed system and the entire universe never decreases.

We can show that this statement is consistent with the Kelvin statement, the Clausius statement, and the Carnot principle.

Entropy change of a system during an isobaric process

Determine the entropy change of an object of mass m and specific heat c that is cooled rapidly (and irreversibly) at constant pressure from T h to T c .

Strategy

The process is clearly stated as an irreversible process; therefore, we cannot simply calculate the entropy change from the actual process. However, because entropy of a system is a function of state, we can imagine a reversible process that starts from the same initial state and ends at the given final state. Then, the entropy change of the system is given by [link] , Δ S = A B d Q / T .

Solution

To replace this rapid cooling with a process that proceeds reversibly, we imagine that the hot object is put into thermal contact with successively cooler heat reservoirs whose temperatures range from T h to T c . Throughout the substitute transition, the object loses infinitesimal amounts of heat dQ , so we have

Δ S = T h T c d Q T .

From the definition of heat capacity, an infinitesimal exchange dQ for the object is related to its temperature change dT by

d Q = m c d T .

Substituting this dQ into the expression for Δ S , we obtain the entropy change of the object as it is cooled at constant pressure from T h to T c :

Δ S = T h T c m c d T T = m c ln T c T h .

Note that Δ S < 0 here because T c < T h . In other words, the object has lost some entropy. But if we count whatever is used to remove the heat from the object, we would still end up with Δ S universe > 0 because the process is irreversible.

Significance

If the temperature changes during the heat flow, you must keep it inside the integral to solve for the change in entropy. If, however, the temperature is constant, you can simply calculate the entropy change as the heat flow divided by the temperature.

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Practice Key Terms 2

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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