How can I compute the missing volume by looking at the pressure changes? Disclaimer: I'm not in any shape used to deal with physics and forgive me in advance if this question is either out of scope of wrongly formulated.
I have a very practical problem. I have a container that contains a liquid which is under pressure (the air above the liquid, inside the container, is under pressure). I also have a sensor inside the container that tells me, with very high frequency and precision, what is the current pressure of the air.
Now, imagine this container has a valve through which the liquid is expelled. Considering that there is no extra air being pumped into the container, I can perfectly observe the pressure dropping over time, by reviewing the data provided by the pressure sensor.
My question is: how can I compute the volume of liquid that was expelled by only looking at the pressure over time? I have naively assumed that I could just compute the upper integral of the pressure/time curve during the period the valve was open. Is this remotely near a correct answer?
EDIT: Actually what I am looking for is the flow-rate of the liquid that was expelled, not the volume per se.
EDIT 2: To elaborate better on the problem I've compiled the pictures/diagrams bellow. 
The first image depicts the environment as I tried to describe above, hope it makes it clear.

The second image depicts a sample of readings from the pressure sensor during one opening of the valve. As you can tell it is clear that the pressure drops, my hope is that the rate (slope?) of the drop could indicate the flow rate of the expelled liquid.

Best, 
 A: A practical approximation is to assume adiabatic expansion of the gas. 
$$P_iV_i=P_fV_f$$
$$\frac{P_i}{P_f}V_i=V_f$$
$$\Delta V=V_f-V_i=\frac{P_i}{P_f}V_i-V_i=V_i\left(\frac{P_i}{P_f}-1\right)$$
A: Given a known initial volume of liquid (assume water) in the container, it is known that as water flows out of the container, the compressed air on the outside of the water bladder and inside the container, expands.  Using the ideal gas law, it should be possible to calculate the rate of water flow based on the air pressure in the container.  
Let
$P_1$ = the air pressure in the container at the start of a timing period
$V_1$ = the initial volume of air in the container, before any liquid is removed
$P_2$ = the air pressure in the container as water is flowing, at the end of a timing period
$V_2$ = the volume of air in the container as water is flowing
$\Delta P$ = change in pressure in the container over a timing period
$\Delta V$ = change in volume in the container over a timing period
$\Delta t = t_2 - t_1$, which is the length of a timing period
Assuming that the compressed air temperature remains constant, the following derivation applies:
$P_1V_1=P_2V_2$
$P_2-P_1=P_1\frac{V_1}{V_2}-P_1$
$P_2-P_1=P_1\frac{V_1-V_2}{V_2}$
$\frac{\Delta P}{P_1}=\frac{-\Delta V}{V_2}$
$\frac{-V_2}{P_1}\Delta P = \Delta V$
Looking at these changes in pressure and volume over short time intervals leads to:
$\frac{-V_2}{P_1}\frac{\Delta P}{\Delta t} = \frac{\Delta V}{\Delta t}$
It is known that $V_2$ must follow the ideal gas law, so that law can be used to eliminate $V_2$:
$V_2 = \frac{nRT}{P_2}$
$\frac{-nRT}{P_2}\frac{1}{P_1}\frac{\Delta P}{\Delta t}=\frac{\Delta V}{\Delta t}$
The number of moles of air in the container, "n", can also be eliminated with the ideal gas law:
$n=\frac{P_1V_1}{RT}$
$\frac{-P_1V_1}{RT}\frac{RT}{P_2}\frac{1}{P_1}\frac{\Delta P}{\Delta t}=\frac{\Delta V}{\Delta t}$
Cancellation of terms yields:
$\frac{-V_1}{P_2}\frac{\Delta P}{\Delta t}=\frac{\Delta V}{\Delta t}$
If $V_1$ is known, the flow rate, which is $\frac{\Delta V}{\Delta t}$, can be calculated by measuring the rate of pressure drop in the container.  Note that the value of $P_2$ should probably be an average of container pressure at the start of a pressure measurement and the pressure in the container at the end of a pressure measurement. 
