The correct option is D Z1=(1−aRTV), Z2=(1+PbRT), Z3=(1+PbRT)
Van der Waals equation is:
(P+aV2)(V−b)=RT (for 1 mol gas)
Case 1: At low pressure, volume in high, hence taking
(V−b)≈V
(P+aV2)V=RT
PV+aV=RT
PVRT+aRTV=1
Z1=PVRT=(1−aRTV)
Case II: At high pressure, aV2 becomes negligible, so
(P+aV2)≈P
P(V−b)=RT
PV−Pb=RT
PVRT−PbRT=1
Z2=PVRT=1+PbRT
Case III: Gases are of low molar masses thus, forces of attraction are very small i.e. a is negligible
Thus, (P+aV2)≈P
As in case II, Z3=1+PbRT