How can we calculate enthalpy change in a closed tank containing methanol adsorbed by activated carbon? Here's the problem in the simplest way:

I have a closed box containing methanol adsorbed by activated carbon. The box is made from copper. A pipe is connected to that box, which has condenser on the other end to return the methanol back to the liquid state.

The box is painted black and put in the sun to absorb heat.

The data I have collected are:


initial and final pressure
initial and final temperature
0.26 kg of methanol was condensed in the other end
0.3744 kg of methanol remained in the tank


How can we calculate enthalpy change in the box (heat absorbed by the box from the sun)?
Enthalpy ($H$) is defined as, $H=U+PV$ where $U$ is the internal energy and $P$ is pressure and $V$ is volume. The change in enthalpy is given by

$$\Delta H = \Delta U + \Delta(PV)$$

And $\Delta U$ is equal to the sum of the heat transferred $q$ and the work done $w=-P\Delta V$.

At constant $V$ (which is what you have in a closed system), $\Delta V =0$ (and thus $w=0$) and $\Delta(PV)=V\Delta P$.

Thus...

$$\Delta H = q+V\Delta P$$

You can determine $q$ based on the amount of methanol that vaporized using the heat of vaporization of methanol. We can assume that the majority of heat absorbed was used to distil methanol. 

$$q_{vap}=m\cdot c_{vap}$$

How can we determine the volume of the system? Or... if the change in pressure is very small, perhaps that term can be ignored.