The short answer is, it doesn't really mean much of anything, because it's not meaningful to compare the Gibbs free energy of two systems unless they're at the same temperature. To show you why, I'll quickly run you through the derivation of the Gibbs free energy and point out where the constant temperature assumption comes in.
The second law of thermodynamics tells us that the total entropy of a system, plus the entropy of its surroundings, must be non-decreasing. I'll write this as $\Delta S_\text{total} = \Delta S + \Delta S_\text{surroundings} \ge 0$.
Suppose we have a system (of any kind) in contact with an environment that remains at constant temperature and pressure. This system undergoes a process (of any kind) that changes its internal energy by $\Delta U$ (with positive sign meaning the system's energy increases) and its volume by $\Delta V$ (positive sign meaning it does work on the environment). We know that the system has done an amount of work $p\Delta V$ on the environment, so we can say that $\Delta U = Q - p\Delta V$, with $Q$ an amount of heat transferred from the surroundings to the system (or in the opposite direction if it's negative).
Since the surroundings are at a constant temperature $T$, they must have lost an amount of entropy equal to $Q/T = \frac{\Delta U}{T} + \frac{p\Delta V}{T}$. So we have
$$\Delta S_\text{total} = \Delta S - \frac{\Delta U}{T} - \frac{p\Delta V}{T}.\qquad\qquad(i)$$
This quantity, as we know from the second law, must always seek a maximum, as long as the temperature and pressure remain constant.
Now, up to this point the assumption of constant temperature hasn't really been necessary. If we assume that $\Delta U$ etc. are small we can replace them with differentials to get $dS_\text{total} = dS - pdV/T - dU/T$, which we can integrate if we know how $p$ and $T$ depend on $U$ and $V$.
However, for reasons that always struck me as somewhat arbitrary, the tradition in physics is to multiply equation $(i)$ by $-T$ to get
$$\Delta G \stackrel{\textit{def}}{=} -T\Delta S_\text{total} = \Delta U + p\Delta V - T\Delta S,$$
which must now be minimised because of the change of sign. I suppose this is done in order to put the quantity into energy units, which physicists and chemists tend to find more intuitive than entropy units. But it comes at a price - the transformation from maximising $\Delta S_\text{total}$ to minimising $-T\Delta S_\text{total}$ only works if $T$ is a constant, and if you violate this assumption then $\Delta G$ can either be positive or negative for a spontaneous process, so its sign no longer really means anything.
To work out $\Delta S_\text{total}$ for your system you would have to integrate $dS_\text{total}$ as described above, and to do that you would have to know the heat capacity and thermal expansion coefficient of steam (as functions of $T$ and $p$, though I'd guess they're pretty constant over that range) in addition to the data you posted. This would certainly come out positive, and a lower value would mean a more efficient turbine, with $\Delta S_\text{total}$ approaching zero in the reversible limit.
Expanding Liquid: the expansion per energy unit is tiny compared with the expansion at the phase transition from liquid to gas.
Expanding Solid: even worse than expanding liquid
Bimetallic strips: very little energy transfer capability and long cool-down cycle.
Ice: see "expanding solid" above.
The next best choice after conversion of (any) liquid to gas is direct gaseous expansion as is done in internal combustion engines; the latter is far less efficient.
As to the choice of water: it's easily available, relatively nontoxic , and boils at a reasonable temperature: not terribly hot, but not so low that it boils at, say, room temperatures.
Best Answer
Enthalpy(H) is defined as H=U + PV. U is internal energy and P and V are pressure and volume respectively.
I do not know much about steam power plants, so I looked up steam boilers on the net. Since they are closed vessels, if the pressure falls, the temperature also falls. The internal energy is directly related to temperature, so it decreases. Moreover, the PV term also decreases with temperature ( for both ideal and real gases). Hence, equating with U and PV, H(enthalpy) should also decrease with pressure in this particular situation.