Calculate the work done in open vessel at `300K`, when `92g Na` reacts with water. Assume ideal gas neture.

Explanation

`2Na +2H_(2)O rarr 2NaOH +H_(2)`
Moles of `Na = (92)/(23) = 4`
Mole of `H_(2)` formed `= (1)/(2) xx` Mole of `Na` used `= (1)/(2) xx 4 = 2`
Work is done in giving out `2mol H_(2)`. Thus,
`W =- P xx V_(H_(2)) = n_(H_(2))RT =- 2 xx 8.314 xx 300`
`=- 4988.4J`
The `H_(2)` liberated pushes the atmospheric gas block and thus, does work in driving back the atmosphere. Note that in the case of closed vessel, `DeltaV = 0`. Therefore, `W = 0`.