The ratio of the slopes of adiabatic and isothermal curves is

\frac{1}{γ}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{γ^{3}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

γ^{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

γ

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $γ$
For an isothermal process $P V = c o n s t a n t$
Differentiating we get
$P d V + V d P = 0$
or
$\frac{d P}{d V} = - \frac{P}{V}$
or
$tan θ = \frac{d P}{d V} = - \frac{P}{V}$ -----------------------(i)
For a adiabatic process we have
$P V^{γ} = c o n s t a n t$
Differentiating we get
$d P V γ + P γ V^{γ - 1} = 0$
or
$\frac{d P}{d V} = - γ \frac{P V^{γ - 1}}{V^{γ}} = - γ (\frac{P}{V})$
Thus slope is $tan θ = \frac{d P}{d V} = - γ \frac{P}{V}$ ................(ii)
Thus we get the ratio of the slopes of adiabatic and isothermal curves from (i) and (ii) as $γ$

Suggest Corrections

Similar questions

γ

γ

p - V

α, β, γ

x^{3} - 2 x + 3 = 0

\frac{1}{α^{3} + β^{3} + 6} + \frac{1}{β^{3} + γ^{3} + 6} + \frac{1}{γ^{3} + α^{3} + 6}

{(\frac{α}{γ})}^{3}

{(\frac{β}{γ})}^{3}

α, β, γ

x^{3} - 8 = 0

α, β, γ

x^{3} + x^{2} + x + 1 = 0

α^{3} + β^{3} + γ^{3} =

Join BYJU'S Learning Program