In the figure shown, same monoatomic working substance $\left(\gamma =\frac{5}{3}\right)$ is taken through an isothermal and an adiabatic, starting from state-A, and the curves are as shown. The slope of isothermal at A is S. The slope of the adiabat at B is $\left(2^{1.6}=3.03\right)$

For isotherm:

pV = constant$\Rightarrow$pdV + Vdp = 0

slope $\mathrm{S}=\frac{\mathrm{dp}}{\mathrm{dV}}=-\frac{{\mathrm{p}}_0}{{\mathrm{V}}_{\mathrm{A}}}$

For adiabatic $p{V}^{\gamma }=$constant

Slope S  (adiabat) at $\mathrm{A}=\gamma \left(-\frac{\mathrm{p}}{\mathrm{V}}\right)=\gamma \mathrm{S}$

For adiabat${\mathrm{pV}}^{\gamma }=$ constant $\Rightarrow$ Hence for state at A and B

${\mathrm{P}}_0{\mathrm{V}}^{\gamma }=\frac{{\mathrm{P}}_0}{2}{\mathrm{V}}_{\mathrm{B}}^{\gamma }\Rightarrow {\mathrm{V}}_{\mathrm{B}}={\mathrm{V}}_{\mathrm{A}}{2}^{1/\gamma }={\mathrm{V}}_{\mathrm{A}}{2}^{(3/5)}$

$\therefore$slope of the adiabat at B :

${\mathrm{S}}_{\mathrm{B}\text{ (adiabat) })}=\gamma \frac{{\mathrm{P}}_{\mathrm{B}}}{{\mathrm{V}}_{\mathrm{B}}}=\frac{5}{3}\cdot \frac{{\mathrm{p}}_0}{2}\cdot \frac{1}{{\mathrm{V}}_{\mathrm{A}}{ }^{(3/5)}}=\frac{5}{3}\cdot \frac{1}{2^{8/5}}\cdot \frac{{\mathrm{p}}_0}{{\mathrm{V}}_{\mathrm{A}}}$

$=\frac{5}{3}\frac{1}{3.03}\mathrm{S}\simeq \frac{5}{9.1}\mathrm{S}$