A given section of pipe will have a flow equation like:
$\Delta P = \gamma w^2$ where $\Delta P$ is the loss of head pressure over the length of pipe, $\gamma$ is a coefficient related to length, diameter, Reynold's number etc., and $w$ is the flow rate. You can find expressions for these in engineering books for example.
Now imagine we have a pipe and we want to put one small hole in it. We can analyze it by analogy with an electric circuit, where the hole is represented by a large shunt resistance added at that point going to ground. Let $P_0$ represent the fixed pressure driving the flow (analogous to a voltage source), $P_1$ be the pressure at the point of interest near the hole, and let $\gamma_{1,2,3}$ be the flow coefficients for the section of pipe upstream of the hole, downstream of the hole, and the hole itself. Let $w$ be the flow rate upstream of the hole, $w_2$ downstream, and $w_3$ be the flow through the hole itself. Then you have a system of equations you can solve for $P_1$:
$P_0 - P_1 = \gamma_1 w^2$
$P_1 - 0 = \gamma_2 w_2^2$
$P_1 - 0 = \gamma_3 w_3^2$
$w_2+w_3 = w$ (Assuming incompressible flow)
The case of no hole is given by $\gamma_3 = \infty$. Since the hole is small, you can assume $\gamma_3 >> \gamma_2, \gamma_1$ and derive a perturbative expression for $\Delta P_1$ in terms of $\gamma_3^{-1}$. I will leave this as an exercise for the reader. Anyhow, this will tell you the incremental pressure loss for a single hole.
Of course, I have assumed the flow is driven by a fixed idealized source pressure. In reality the source pressure may be a function of flow rate (analogous to the impedance of a voltage source) which would also effect the result, but which could be added into the above equations in a straightforward manner.
It is not clear from your question whether the gas in the bottle starts out as air, or the only substance in the bottle is water, either in liquid or gas form. I'll assume the latter since it's easier to answer.
Once the bottle is closed, ambient temperature doesn't matter. The pressure of the gas part in the bottle will be stricly a function of temperature, with that function being solely a property of water. You can find what that function is in what is commonly referred to as a steam table.
As you heat the bottle, and presumably everything inside gets to the same temperature, the vapor pressure increases. This causes a little more of the liquid to boil and thereby make less liquid and more gas. Eventually for any one temperature, a new equilibrium is reached where the gas is at the pressure listed in the steam table for that temperature.
If you let the system reach equillibrium at 200°C, for example, then open a valve at the bottom, water at the bottom will be forced out under pressure. As the volume of water in the bottle decreases, the volume available to the gas increases, which decreases its pressure. The liquid that was at equillibrium now no longer is, since its pressure is reduced but its temperature (for the short term) remains the same. This will cause the liquid to boil and make more gas. This cools the liquid, and if done slowly enough the liquid will track the ever decreasing boiling temperature for the ever decreasing pressure.
Meanwhile, the gas is expanding, so it will cool. It won't condense because there is no place for the heat to go. We are making the assumption that the bottle is insulated at this point. Boiling liquid will keep the gas "topped off" at whatever temperature it is according to the steam table.
Once all the liquid is gone, the steam will still be at pressure above ambient. If the valve stays open, the steam will vent until its pressure is the same as ambient. If the bottle is truly insulated, nothing more should happen. If heat is lost thru the bottle, then the temperature will decrease and the pressure of the steam decrease, again according to the steam table. Outside air is then sucked into the bottle. If this air is at 20°C, for example, it will significantly cool the steam, which will cause much of it to condense, which creates lower pressure, which sucks in even more cool air, etc. The proceeds until most of the water is condensed to liquid, with the only remaining water gas being whatever maximum partial pressure of water that air at that temperature and pressure can support.
If the valve is opened so that the 200°C water is expelled "quickly", then it gets a lot more complicated because there is not enough time for the system to track equillibrium as pressure is abruptly release. In that case, I think all we can say is that a bunch of super-heated water will be violently released and quickly undergo decompression, which will cause some of it to boil quickly, making a lot of steam, with the left over liquid being at boiling temperature for ambient pressure, which would be 100°C. After the water is gone from the bottle, a rush of steam will come out, condensing in the (relatively) cold ambient air and adding to the already large saturated water vapor cloud. Once enough steam is expelled so that ambient pressure is reached the remaining steam acts as above.
Best Answer
Assumptions:
you can write the rate of change of the volume of the air pocket as a function of pressure:
$$\frac{PV}{T} = const\\ P_1V_1 = P_2V_2$$ Differentiating $PV = constant$: $$P dV + V dP = 0$$ Dividing by $dT$ and rearranging: $$\frac{dV}{dt}=-\frac{V}{P}\frac{dP}{dt}$$
From this you compute the flow rate from the pressure change. As you can see, the smaller the volume $V$, the smaller the volume change $dV$ that you can calculate for a given change in pressure.
This leaves you with the problem of calibrating the air pocket. This is best done by having an air filled capillary somewhere near the top of your system: you will be able to see the liquid rise in this capillary as the system is pressurized, and from the rate at which it drops you can determine the leakage rate immediately - if you know the diameter of the capillary, no other math is required…
Do note that a heat exchanger is likely to expand when pressurized - you should be able to determine how much this affects the result by having a small calibrated plunger (ideally the same size as the capillary) with which you can inject a small known amount of additional liquid into the system. If the liquid plus heat exchanger were truly incompressible (constant volume, constant density) then the air in the capillary should rise as you push the plunger down. When this does not happen (say the capillary rises half as much) then you know what volume of liquid corresponds to what volume change in the capillary, and this gives you the calibration from capillary volume to liquid volume.