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Q.
The relation between the slope of isothermal curve and slope of adiabatic curve
Thermodynamics
Solution:
For isothermal process, $PV = $ constant
Differentiating both side
$PdV + VdP = 0$ or $\frac{dP}{dV} = \frac{-P}{V}$
Again for adiabatic process, $PV^\gamma =$ constant
Again differentiating both side
$dPV^\gamma + \gamma V^{\gamma-1} dV \, P = 0$ or $\frac{dP}{dV} = - \frac{P}{V} \times \gamma$
$\therefore $ slope of adiabatic curve $= \gamma \times$ slope of isothermal curve