Question

an ideal monatomic gas is taken round the cycle ABCDA as shown. calculate the work done during the cycle

Answer 1

work done is given by area between the pv graph

area=(2p-p) x (2v-v)
area =pv
as it is clockwise
work done is negative

work done=-pv
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Answer 2

Work done from A - B is Zero (isochoric process).

W.d from B-c= p.dv=-2p[dv](upper limit =2v; lower limit =v)

We get- (-2pv)[note- p is negative for clock wise direction]

W.d from c-d =0(isochoric process)

W.d from d-a= pdv= p.[dv](upper limit =v; lower limit=2v.) [Note- p is positive for anticlockwise direction]

By integration-

We get = ( pv)

Than total work done=0+(-2pv)+0+(pv)

=-pv