For an adiabatic process, PVγ= constant,
Differentiating, we have γPVγ−1+dVdPVγ=0 or dVdP=−VγP
Thus the slope of P−V curve is proportional to γ Now, for a diatomic gas, γ(=7/5) is less than that for a monoatomic gas for which γ=5/3. Therefore, the slope of the P−V curve is less for a diatomic gas than for a monoatomic gas. Hence curve 1 corresponds to diatomic gas and curve 2 to monoatomic gas.