Question

If $ 7.1 $ mol of $ A{l_4}{C_3}(s) $ reacts with $ {H_2}O $ in a closed vessel at $ {27^ \circ }C $ against atmospheric pressure. Work done is:
(A) $ - 1800cal $
(B) $ - 600cal $
(C) $ + 1800cal $
(D) Zero

Answer 1

Hint : The formula to find work done is $ W = - P\Delta V $ , where W is work done in joules, P is pressure in atm and $ \Delta V $ is change in volume in liters. Find the pressure and change in volume of the reaction and substitute in the formula then convert the units to get the answer.

Complete step by step solution
First let us write down the reaction which is occurring inside the vessel which is,
 $ A{l_4}{C_3}(s) + 12{H_2}O(l) \to 4Al{(OH)_3} + 3C{H_4} $
We have $ 7.1 $ mol of aluminium carbide reacting with water. The temperature given is $ {27^ \circ }C $ . The pressure is $ 1 $ atm.
We know the formula of work done is negative of the product of pressure and change in volume which is, $ W = - P\Delta V $ .
But we are given that the container is closed. That means there will be no change in the volume of the container. So change in volume will be zero which makes $ \Delta V = 0 $ .
The process which we have here is isochoric. Isochoric process is the process in which there is no change of volume of the container. That is why work done in an isochoric process is always zero.
Therefore we will have $ W = 0 $ , which makes the work done zero as there will be no work done on the system.
So option D is the correct answer.

Note
The work done in a reversible process can also be zero as the system returns to its initial state thus making the total change in volume to be zero. Do keep in mind that the work done formula has a negative sign. So if the product of pressure and change in volume is positive then the work done is negative and if the product of pressure and change in volume is negative then the work done is positive.