4.6 Entropy

The second law of thermodynamics is best expressed in terms of a change in the thermodynamic variable known as entropy, which is represented by the symbol S. Entropy, like internal energy, is a state function. This means that when a system makes a transition from one state into another, the change in entropy $\text{Δ}S$ is independent of path and depends only on the thermodynamic variables of the two states.

We first consider $\text{Δ}S$ for a system undergoing a reversible process at a constant temperature. In this case, the change in entropy of the system is given by

where Q is the heat exchanged by the system kept at a temperature T (in kelvin). If the system absorbs heat—that is, with $Q>0$—the entropy of the system increases. As an example, suppose a gas is kept at a constant temperature of 300 K while it absorbs 10 J of heat in a reversible process. Then from Equation 4.8, the entropy change of the gas is

$$\text{Δ}S=\frac{10\text{J}}{300\text{K}}=0.033\text{J/K}.$$

Similarly, if the gas loses 5.0 J of heat; that is, $Q=\mathrm{-5.0}\text{J}$, at temperature $T=200\text{K}$, we have the entropy change of the system given by

$$\text{Δ}S=\frac{\mathrm{-5.0}\text{J}}{200\text{K}}=\mathrm{-0.025}\text{J/K}.$$

The change in entropy of a system for an arbitrary, reversible transition for which the temperature is not necessarily constant is defined by modifying $\text{Δ}S=Q\text{/}T$. Imagine a system making a transition from state A to B in small, discrete steps. The temperatures associated with these states are $T_A$ and $T_B,$ respectively. During each step of the transition, the system exchanges heat $\text{Δ}{Q}_i$ reversibly at a temperature $T_i.$ This can be accomplished experimentally by placing the system in thermal contact with a large number of heat reservoirs of varying temperatures $T_i$, as illustrated in Figure 4.15. The change in entropy for each step is $\text{Δ}{S}_i=Q_i\text{/}{T}_i.$ The net change in entropy of the system for the transition is

$$\text{Δ}S=S_B-S_A={\displaystyle \sum _i\text{Δ}{S}_i}={\displaystyle \sum _i\frac{\text{Δ}{Q}_i}{T_i}}.$$

We now take the limit as $\text{Δ}{Q}_i\to 0$, and the number of steps approaches infinity. Then, replacing the summation by an integral, we obtain

where the integral is taken between the initial state A and the final state B. This equation is valid only if the transition from A to B is reversible.

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 Figure 4.11, no heat exchange takes place, so $\text{Δ}{S}_2=\text{Δ}{S}_4={\displaystyle \int d}Q\text{/}T=0.$ In step 1, the engine absorbs heat $Q_{\text{h}}$ at a temperature $T_{\text{h}},$ so its entropy change is $\text{Δ}{S}_1=Q_{\text{h}}\text{/}{T}_{\text{h}}.$ Similarly, in step 3, $\text{Δ}{S}_3=\text{−}{Q}_{\text{c}}\text{/}{T}_{\text{c}}.$ The net entropy change of the engine in one cycle of operation is then

$$\text{Δ}{S}_E=\text{Δ}{S}_1+\text{Δ}{S}_2+\text{Δ}{S}_3+\text{Δ}{S}_4=\frac{Q_{\text{h}}}{T_{\text{h}}}-\frac{Q_{\text{c}}}{T_{\text{c}}}.$$

However, we know that for a Carnot engine,

$$\frac{Q_{\text{h}}}{T_{\text{h}}}=\frac{Q_{\text{c}}}{T_{\text{c}}},$$

so

$$\text{Δ}{S}_E=0.$$

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

where ${\displaystyle \oint }$ represents the integral over a closed reversible path.

We can use Equation 4.11 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 Figure 4.16. From Equation 4.11, ${\displaystyle \oint dS=0}$ 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

$${\left[{\displaystyle {\int }_A^{B}dS}\right]}_{\text{I}}+{\left[{\displaystyle {\int }_B^{A}dS}\right]}_{\text{II}}=0.$$

Since the process is reversible,

$${\left[{\displaystyle {\int }_A^{B}dS}\right]}_{\text{I}}={\left[{\displaystyle {\int }_A^{B}dS}\right]}_{\text{II}}.$$

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

$$\text{Δ}S\ge 0,$$

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:

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