The correct option is C O2 and He
PVγ = constant (in ~adiabatic)
∴ differentiating,
dp.Vγ+γPVγ−1dV=0
⇒dpdV=−γPV
Hence, from same starting point, slope is greates for gas with higher value of γ. Now, γ=1+2f where f : degress of freedom. Higher the vlaue of f, lower is γ So, 'A' has higher 'f'' compared to 'B' ∴(SO2,Ar) and (O2,He) both are possible