How can we calculate enthalpy change in a closed tank containing methanol adsorbed by activated carbon?
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.