Answer : The work done is, 
Explanation :
The given balanced chemical reaction is:

When 4 moles of
react with 12 moles of
then it gives 8 moles of 
First we have to calculate the change in moles of gas.
Moles on reactant side = Moles of
+ Moles of 
Moles on reactant side = 4 + 12 = 16 moles
Moles on product side = Moles of 
Moles on reactant side = 8 moles
Change in moles of gas = 16 - 8 = 8 moles
Now we have to calculate the change in volume of gas.
Using ideal gas equation:

where,
P = Pressure of gas = 1.0 atm
V = Volume of gas = ?
n = number of moles of gas = 8 mole
R = Gas constant = 
T = Temperature of gas = 
Putting values in above equation, we get:


As the number of moles of gas decreased. So, the volume also deceased. Thus, the volume of gas will be, -195.7 L
Now we have to calculate the work done.
Formula used :

where,
w = work done
p = pressure of the gas = 1.0 atm
= change in volume = -195.7 L
Now put all the given values in the above formula, we get:



conversion used : (1 L.atm = 101.3 J)
Thus, the work done is, 