Show that the slope of a adiabatic process is γ times . . .

Question : Show that the slope of a adiabatic process is γ times the slope of an isothermal process.

OR

Show that the slope of a adiabatic curve is γ times the slope of an isothermal curve.

Doubt by Jaskirat 

Solution : 

We know, 
For Isothermal Process

PV=constant

differentiating both sides 

PdV+VdP=0
VdP=-PdV
dP/dV = -P/V 
Slope of Isothermal Curve = -P/V — (1)

For Adiabatic Process

PV^γ=constant

differentiating both sides 

P[γV^γ-1]dV+V^γdP=0
γPV^γ-1dV+V^γdP=0
V^γdP=-γPV^γ-1dV
dP/dV=-γPV^γ-1/V^γ
dP/dV=-γPV^γ-1-γ
dP/dV=-γPV^-1
dP/dV=-γP/V
Slope of Adiabatic Curve = - γP/V
Slope of Adiabatic Curve = -γP/V — (2)

Dividing equation (2) by (1)

[Slope of Adiabatic Curve] / [Slope of Isothermal Curve] = γ

Slope of Adiabatic Curve = γ × Slope of Isothermal Curve

Clearly, the slope of a adiabatic curve is γ times the slope of an isothermal curve.

Hence Proved.