Six moles of an ideal gas performs a cycle shown in figure. If the temperature are `T_(D) = 600 K`, `T_(B) = 800 K`, `T_(C) = 2200 K` and `T_(D) = 1200 K`, the work done per cycle is

Options

A. 20 kJ
B. 30 kJ
C. 40 kJ
D. 60 kJ

Answer

C

Explanation

Processes A to B and C to D are parts of straight-line graphs of the form y = mx
Also `P = ( mu R)/( v) T( mu = 6)`
`implies P prop T.` So volume remains constant for the graphs AB and CD.
So no work is done during processes for A and B and C to D . i.e.,
`W _(AB) = W _(CD) = 0 and `
`W _(BC) = P _(2) ( V _(C) - V _(B))= mu R`
`(T _(C) - T _(B))`
`= 6 R (2200 - 800) = 6 R xx 1400 J `
Also `W _(DA) = P _(1) (V _(A) - V_(D)) = mu R`
`(T _(A) -T _(B))`
`= 6 R ( 6000 -800) = - 6 R xx 600 J`
Hence, work done in a complete cycle,
`W = W _(AB)+ W _(BC) + W _(CD) + W _(DA)`
`= 0 = 6 R xx 1400 + 0 - 6 R xx 600`
`= 6 R xx 900 = 6 xx 8.3 xx 800 ~~ 40 kJ`