For an isothermal change, pV=K Differentiating both side, we get
p⋅dV+Vdp=0⇒Vdp=−pdV
∴ Slope of an isothermal curve (‌
dp
dV
)‌iso ‌=−‌
p
V
For an adiabatic change, pVγ=K′ Differentiating both side, we get p⋅γVγ−1⋅dV+Vγ⋅dp=0 ⇒‌‌γpdV+Vdp=0‌‌ or ‌‌Vdp=−γ⋅pdV ∴ Slope of an adiabatic curve, (‌