A certain amount of gas is expanded from a volume of 20 mL to 42.6 mL by the following three processes (i) against zero pressure (ii) against constant external pressure of 1.2 atm (iii) against constant external pressure of 2.0 atm. Calculate the work done in each of the above three processes.

Explanation

(i) w=0
(ii) `w= - P_(ex)(V_(2)-V_(1))= - 1.2 " atm " (42.6 -20)` mL
`=- 2.74 J`
`(iii) w= - P_(ex)(V_(2)-V_(1))= -2` atm (42.6-20) mL = -4.57 J