For an ideal monoatomic gas, ratio of specific heat γ=
5
3
For adiabatic process A→B pAVAγ=pBVBγ Substituting values of pA,pB and VB from graph, ⇒32×105×VAγ=1×105×(1)γ ⇒VAγ=(
1
32
)γ Here, γ=
5
3
⇒(VA)
5
3
=
1
32
or VA=(
1
32
)
3
5
=
1
(25)
3
5
=
1
8
m3 Similarly for adiabatic compression C→D, pCVCγ=pDVDγ We substitute values from graph, ⇒(1×105)×(8)
5
3
=(32×105)(VD)
5
3
⇒VD5∕3=1 ⇒VD=13∕5=1m3 Now, Work done in process A→B→C→D→A =WAB+WBC+WCD+WDA Here we substitute work done for adiabatic process, Wadiabatic=
piVi−pfVf
γ−1
and work done of isobaric process, Wisobaric=p(Vf−Vi) so, work done in given cycle ABCDA =WAB+WBC+WCD+WDA =
(pAVA−pBVB)
γ−1
+pB(VC−VB)+(
pCVC−pDVD
γ−1
)+pD(VA−VD) Substituting values of pressures and volumes, we get WABCDA=(