If the above system is an open vessel the work done is

Options

A. `-1800` cal
B. `-600` cal
C. `+1800` cal
D. `+600` cal.

Answer

A

Explanation

`underset("1 mol")(Al_(4)C_(3)(s))+12 H_(2)O(l) rarr4Al(OH)_(3)+underset("3 mol")(3 CH_(4))`
Initial volume, `V_(1)=0`
Final volume `=V_(2)(` due to 3 mol of `CH_(4))`
`W=- P(V_(2)-V_(1))`
`= - P xx V_(2) = - P xx (nRT)/(P)`
`=- nRT`
`=- 3 xx 2 xx 300 =- 1800 cal`