The figure shows a P.V graph of the thermodynamic behavious of an ideal gas. Find out form this graph (i) work done by the gas in the process `A to B, B to C, C to D and D to A`. (ii) work done by the gas in complete cycle `A to B to C to D to A`.

Explanation

(i) The work done in a thermodynamic process is equal to the area enclosed between the P-V curve and the volume -exis.
Work done by the gas in the process `A to B` is
`W_1`= area ABB A. `= AB xx A.A`
`=(6.0 =1.0) "litre "xx (12 xx 10^(5))N//m^2`
`= 5. 0 xx 10^(-3)m^3 xx (12 xx 10^5) N//m^2`
`=6000 N-m = 6000` joule.
Work done in the process `B to C` is zero since volume remains constant.
Work done on the gas in the process `C to D` is
`W_2` = area DCB.A.
`=DC xx A.D = (-5.0 xx 10^(-3)) xx (2xx 10^5) = -1000` joule. (Negative sign is taken because volume decreases)
Work `W_1` is positive, while `D to A` is also zero, because volume remains constant.
(ii) Work `W_1` is positive, while `W_2` is negative. Hence the net work done in the whole cycle is
`W = W_1 - W_2 = 6000 - 1000 = 5000` joule
This net work is done by the gas.