Energy absorbed by a system in isothermic process in which the phase changes from liquid to gas

Question

A system contains $x$ mole of material, with surface area $S$ is pressed by a mass $M_1$ (no fraction and no any other outer forces involves). Let $\Delta H_\mathrm{l\rightarrow g}$ be the transition enthalpy from liquid to gas. The system is position in $(T,P_1)$ such that the material phase is liquid for those temperature and pressure values ($T$ and $P_1$). Now instantaneously the mass $M_1$ is decreased to to an unknown value which we denote $M_2$ so the pressure decreases. In case that the temperature stays constant until the inner pressure is equal to the outer pressure $(T,P_2:=\frac{M_2g}{S})$ system gets to back to equilibrium, and the material phase in $(T,P_2)$ is gas, I wish to calculate how much energy was absorbed by the system till it gets to pressure equilibrium $(T,P_2)$.

In the question is says that the volume of the liquid can be neglected when compared to gas' volume. And that the volume of the water is independent of the pressure.

What I did so far:

The chemical equilibrium pressure (i.e the pressure in the phase transition point) can be calculated because the exercise gives the triple point $(T_\triangle , P_\triangle)$ so using the transition enthalpy constant and the constant $T$ of the process, one can calculate $(P_\mathrm{l\rightarrow g, eq},T)$ (using Clausius–Clapeyron relation). And by $PV=nRT$ one may calculate $V_\mathrm{eq} = \frac{nRT}{P}$ the volume in the chemical equilibrium point. By the unstruction $V_\mathrm{eq}$ is the total volume of the material (we neglect the liquid volume)

We know the starting inner pressure which is $\frac{M_1g}{S} \text{[Pa]} $ and by the given instruction the pressure doesn't change the liquid volume we learn that until the phase transition point, the volume of the system stays constant $V_\mathrm{eq}$ which we calculated above.

When the system gets to pressure equilibrium point $(T,P_2=\frac{M_2g}{S})$ we don't know the volume or $P_2$ because $P_2$ is unknown how every we not that its larger then $V_\mathrm{eq}$.

I'm pretty much stuck here, I don't see how to translate this information to evaluate the amount of energy absorbed. It seems to me that maybe one more piece of information is needed, but the exercise doesn't give one.

Answer 1

As indicated by @Buck Thorn,the work done by the substance on its surroundings is $P_2(V_2-V_1)$. So, from the first law, $$U_2-U_1=Q-P_2(V_2-V_1)$$Solving for Q then gives: $$Q=(U_2+P_2V_2)-(U_1+P_1V_1)-(P_2-P_1)V_1=H_2-H_1-(P_2-P_1)V_1$$Here, $$H_2=H_1+(H^{sv}_1-H_1)+(H_2-H^{sv}_1)$$where $H_1$ is the enthalpy of saturated liquid in State 1, and $H_2$ is the enthalpy of the superheated vapor that exists in State 2 (i.e., at a pressure lower than the equilibrium vapor pressure at the constant temperature T). For an ideal gas, the change in enthalpy at temperature T from the saturated vapor in State 1 to the superheated vapor in State 2 is zero. So the previous equation reduces to $$H_2=H_1+(H^{sv}_1-H_1)=H_1+x\Delta h^{vap}$$where $\Delta h^{vap}$ is the heat of vaporization at temperature T. Substitution of this into the equation for the heat added gives: $$Q=x\Delta h^{vap}+(P_1-P_2)V_1$$where $V_1$ is the volume of the saturated liquid at T. The second term on the rhs will typically be negligible.