There is a huge difference with the usual Joules experiment. Here the system is not isolated during the gas expansion into the evacuated chamber.
Two things enter in competition in principle when a setup as the one considered in the article you mention is considered:
- Whether the gas has to work to expand itself in the larger chamber
- Is there anything from outside that helps the gas expanding?
Having an evacuated chamber deals with point 1 and tells us that the gas does not have to work (and hence loose energy) to expand in the chamber to gain extra room.
The fact that the whole system is connected to a piston in contact with atmospheric pressure tells us that the system is a) not isolated and b) that the atmospheric pressure will "help" the gas expanding in the chamber.
Now, assuming an expansion that is quasi-static, the pressure on the piston has to be balanced at every time during the expansion process. Basically, the piston will work a little bit each time there is a pressure drop on the gas side (due to leaking into the chamber) so as to balance pressures again. If the process is adiabatic, the work provided by the atmosphere has nowhere to go and stays in the gas as an increase in temperature/kinetic energy.
The zeroth law posits the existence of temperature by stating that if A is in equilibrium with B and A is in equilibrium with C, then B is in equilibrium with C. We can then assign an intensive property to A, B and C that we call "temperature". They are in equilibrium == they have the same temperature.
As soon as they are NOT in equilibrium, the zeroth law is silent. Thus, as you observed in your question, we cannot derive an ordering of temperatures based on the zeroth law alone.
Here, the second law comes to the rescue. The formulation I am familiar with states
the entropy of a closed system never decreases
If we have two objects that are not in thermal equilibrium, then when we bring them into contact we expect heat to flow between them. Now according to the second law, if we move heat $\delta q$ from $A$ to $B$ (at temperatures $T_B$ and $T_B$ respectively), the change in entropy is
$$\delta S = T_A \delta q - T_B \delta q\\
= \delta q (T_A - T_B)$$
Now if the entropy of the system cannot decrease, then if $\delta q$ is positive we know that $T_A - T_B$ must be positive.
This is where we find the ordering of temperature: heat travels from hotter to cooler until thermal equilibrium is reached. Thus when we have two objects in unequal states we can tell which is hotter by looking at the direction in which heat flows between them. That direction is always from hotter to colder - and to prove this you need the second law.
There is an amusing (although somewhat dated - 50 years old this year) song by the duo of Flanders and Swann that touches on this topic. See http://youtu.be/VnbiVw_1FNs
Best Answer
Specific enthalpy is $$h=u+pv$$
We know u and v are a function of temperature. So h is a function of both temperature and pressure. You can find it in water vapor table. If table does not have specific enthalpy at the temperature and pressure, you can get it by interpolation.