This is an unsteady state process and the answer depends on the precise details of the process. If instead of opening the hatch we vent the gas via a valve, we may treat this process as isenthalpic (throttling). With the further assumption that the contents of the tank undergo isentropic change, which amounts to neglecting irreversibilities due to flow inside the tank, we obtain two equations that fix the state of the system inside the tank:
$$
S(P,T) = S(P_0,T_0),\quad
H(T,P) = H(T',P_\text{ext})
$$
where $(P,T)$ are the conditions inside the tank after the venting, $(P_0,T_0)$ are the conditions before venting, $P_\text{ext}$ is the pressure to which the gas vents and $T'$ is the temperature of the gas immediately after venting. In writing these equations we are assuming that the gas that is vented is at the pressure of the surroundings but at temperature $T'$ determined by the isenthalpic condition. For an ideal gas, $T'=T$.
The bottom line is that one cannot solve this problem without making enough assumptions to deal with the complicated nature of the process.