If Al4C31 mol(s)+12H2O(l)⟶4Al(OH)3+3CH43 mol system is in an open vessel, then work done in calories is

Al4C31 mol(s)+12H2O(l)⟶4Al(OH)3+3CH43 molInitial volume , V1 = 0Final volume =V2 due to 3 moles of CH4W=−pV2−V1=−p×V2=−pnRTp=−nRT=−3×2×300=−1800cal