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