Our primary result for how the entropy changes, $dS=dQ/T$, is for a slow quasi-static process in which the system is always essentially in equilibrium. But most real-world processes don’t meet these criteria. For example, in Section 20.7 we looked at the free expansion of a gas; FIGURE 20.23 reminds you of the situation. This irreversible process is fast, not quasi-static, and the gas is not in equilibrium during the expansion. We argued that the entropy of the gas must increase even though its thermal energy and temperature do not. Can we calculate the value of $\text{Δ}{S}_{\text{free expansion}}?$

FIGURE 20.23 The free expansion of a gas.

We can, but not by using Equation 20.51, $dS=dQ/T$, because it applies to only a slow quasi-static process. If we tried, we would erroneously conclude that $\text{Δ}{S}_{\text{free expansion}}=0$ because no heat is transferred during a free expansion.

However, we know that entropy is a state variable. We also know that an irreversible process, such a free expansion, starts from an equilibrium state (before the membrane is broken) and ends in an equilibrium state (after the gas fills the larger container). The value of a state variable depends on only the state of the system, not the process by which that state was reached. As a result, the entropy change of an irreversible process is exactly the same as the entropy change of any slow quasi-static process that connects the initial and final states. This is important because it allows us to find the entropy change of an irreversible process by calculating the entropy change due to a reversible process between the same two states.

For example, instead of breaking the membrane, we could let the gas fill the container in a quasi-static isothermal process in which a piston slowly moves outward while the addition of heat keeps the temperature from changing. This is a reversible process because we could always slowly push the piston back in, while removing heat, to restore the initial state. We can use Equation 20.57 for the entropy change of an ideal gas to write

The first term in Equation 20.57 is zero because $\text{ln}(T_{\text{f}}/T_{\text{i}})=\text{ln}(1)=0$ for an isothermal expansion.

Because the initial and final states are the same, $\text{Δ}{S}_{\text{free expansion}}$ for the irreversible free expansion is exactly the same as $\text{Δ}{S}_{\text{slow expansion}}$; that is,

Thus a free expansion that doubles the volume of a gas increases its entropy by $\text{Δ}{S}_{\text{sys}}=nR\text{ln}2$.

If $\text{Δ}{S}_{\text{free expansion}}=\text{Δ}{S}_{\text{slow expansion}}$, then what’s the difference? Why is one a reversible process but not the other? The answer lies not in the gas but in the environment.

A gas undergoing a free expansion is an isolated system; nothing in the environment changes. A doubling of the volume is a process in which $\text{Δ}{S}_{\text{sys}}=\text{Δ}{S}_{\text{free expansion}}=nR\text{ln}2$ and $\text{Δ}{S}_{\text{env}}=0$, so the entropy change of the universe is $\text{Δ}{S}_{\text{universe}}=\text{Δ}{S}_{\text{sys}}+\text{Δ}{S}_{\text{env}}=nR\text{ln}2>0$. A free expansion increases the entropy of the universe. We can’t reverse the free expansion because a reversal would require a decrease in the entropy of the universe, which the second law doesn’t allow.

We can contrast this with doubling the volume in a slow quasi-static expansion. The system’s entropy change is the same, $\text{Δ}{S}_{\text{sys}}=nR\text{ln}2$, but now the system isn’t the only thing changing. A slow isothermal expansion requires a heat input from the environment to keep the temperature constant, so the environment also has an entropy change. If $Q_{\text{env}}$ is the heat flow relative to the environment and the exchange takes place at a constant temperature, then $\text{Δ}{S}_{\text{env}}=Q_{\text{env}}/T$.

Any heat removed from the environment is added to the system, so $Q_{\text{env}}=-Q_{\text{sys}}$. An isothermal process has $\text{Δ}{E}_{\text{th}}=0$ and thus, from the first law of thermodynamics, $Q_{\text{sys}}=-W$. In Chapter 19 you learned that the work done during an isothermal process is $W=-nRT\text{ln}(V_{\text{f}}/V_{\text{i}})$. Thus $Q_{\text{env}}=-nRT\text{ln}(V_{\text{f}}/V_{\text{i}})=-nRT\text{ln}2$ for a doubling of the volume, and $\text{Δ}{S}_{\text{env}}=Q_{\text{env}}/T=-nR\text{ln}2$, exactly opposite that of the system. As a result, the entropy change of the universe is $\text{Δ}{S}_{\text{universe}}=\text{Δ}{S}_{\text{sys}}+\text{Δ}{S}_{\text{env}}=0$.

Thus the difference between the free expansion and the slow isothermal expansion is not the entropy change of the gas, which is the same in both cases, but the entropy change of the environment. The slow isothermal expansion is reversible because the entropy of the universe is unchanged, but we cannot reverse the free expansion without violating the second law of thermodynamics.

Sealed, but not isolated