Which state has a higher entropy — the liquid–gas equilibrium or the gaseous state? I have learned that at equilibrium the entropy is maximised. But I also learned that the gaseous state has the highest disorder (entropy), if the latter is true then why does the equilibrium exist, why doesn’t the liquid transform into the gaseous state at equilibrium completely? What factor is responsible for this?

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    $\begingroup$ Enthalpy. It is positive for liquid-gas transition as bonds need to be broken. $\endgroup$
    – EJC
    Dec 25 '15 at 10:08

Whatever one observes (whether it is a liquid, a gas, or a mixture thereof) is the state of maximum entropy for the particular values of the internal energy and volume the system currently has. As OP noted, this is a consequence of the second law of thermodynamics. According to this law, a system with a fixed internal energy self-adjusts its free parameters in such a way that the entropy is maximized. The free parameter in OP's example is the fraction of molecules in the liquid (or gaseous) phase.

When we say that the entropy of the gaseous phase is larger than that of the liquid phase, we are referring to the specific entropies (i.e., entropy per mole) of these phases when they coexist. When the two phases coexist, they have a common temperature $T$ and a common pressure $P$. Furthermore, their specific Gibbs free energies are the same, that is, \begin{equation} 0 = \Delta g (T,P) = g_{\mathrm{gas}} (T,P) - g_{\mathrm{liquid}} (T,P) = \Delta u(T,P) + P \Delta v(T,P) - T \Delta s(T,P). \end{equation} The gaseous phase has a higher specific internal energy ($\Delta u>0$) and a larger specific volume ($\Delta v>0$) as well as a higher specific entropy ($\Delta s >0$) in such a way that the effects of the three terms ($\Delta u$, $P\Delta v$, and $T\Delta s$) cancel out.

One may still wonder why liquid-gas coexistence is ever possible. As the gaseous phase has a higher specific entropy, why doesn't the system self-adjust to have all its molecules in the gaseous phase? The Answer: What we know for sure is that if the evaporation took place under a constant temperature and pressure, the entropy would increase (because $\Delta s >0$). But this process would also increase the internal energy and the volume ($\Delta u >0$ and $\Delta v >0$). Therefore, to have more molecules in the gaseous phase but stay at the same internal energy and volume, the system has to lower its temperature and raise its pressure. This results in a decrease in the entropy,1 which counteracts its increase that would have happened if the evaporation had taken place at constant temperature and pressure. Therefore, evaporation at fixed internal energy and volume doesn't necessarily lead to the increase of the total entropy. As stated in the first paragraph, if one observes liquid-gas coexistence, this should correspond to the state of maximum entropy for the given values of internal energy and volume.

1 For any (stable) thermodynamics system, the entropy decreases as the temperature is lowered. On the other hand, no such strict rule exists for the relation between the entropy and the pressure. Still, for most systems, entropy decreases as pressure is raised. [Notice the Maxwell relation $(\partial S/\partial P)_{T} = -(\partial V/\partial T)_{P}$, meaning that the entropy decreasing with increasing pressure is equivalent to the volume increasing with increasing temperature.]


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