One mole of a monoatomic gas is carried along process ABCDEA as shown in the diagram. Find the net work done by gas :-

Answer

`10.45`

Explanation

The work performed by the gas is area bounded by the curve and V-axis. From A to B total area is the area of triangle ABC, and rectangle below CB.
`W_(AB) =1/2 [(5.00xx10^(-3)) – (2.00xx10^(-3))][600 – 300]+[(5xx10^(-3))-(2xx10^(-3))](300) =1.35 J`
Work done from B to C is negative because the gas is being compressed, volume is decreasing. BC is isobaric process. Thus work done is `W_(BC) = PDeltaV = P_B (V_C - V_B) = (300)[(2.00xx10^(-3))- (5.00xx10^(-3))] = -0.90 J`
The negative work shows that work is done on the system. Change in volume, `DeltaV`, is always `V_f - V_i`
The net work done by the system is `W_("total") = W_(AB) + W_(BC) = 1.35 J +(0.90J) = 0.45 J`.