A thermodynamic system is taken from an original state A to an intermediate state B by a linear process as shown in the figure. Its volume is then reduced to the original value from B to C by an isobaric process. The total work done by the gas from A to B and B to C would be :

Step-by-step solution

Work done during the path A → B can be written as W AB = 1 2 ( 8000 + 4000 ) dyne cm - 2 × 1 N 10 5 dyne × 1 cm 2 10 - 4 m 2 × 7 - 3 m 3 = 2400 J The work done during the process B → C is given by W BC = 4000 dyne cm - 2 × 1 N 10 5 dyne × 1 cm 2 10 - 4 m 2 × 3 - 7 m 3 = - 1600 J Hence, the total work done is given by W = W AB + W BC = 2400 J - 1600 J = 800 J