Physics - Heat & Thermo Question with Solution | TestHub

$PV=\mathrm{constant}(\mathrm{isothermal})$
$P{V}^{\gamma }=\mathrm{constant},\gamma >1(\mathrm{adiabatic})$
Take BC as isothermal and CA as adiabatic.
$P_0{V}_0=3P_0{V}_C\Rightarrow V_C=V_0/3$
$3P_0(V_0/3{)}^{\gamma }=(P_0/2)V_0^{\gamma }$
$3^{\gamma }=6\Rightarrow \gamma =\frac{\ln 6}{\ln 3}>1$
Reversing the assignments gives γ = ln 3/ln 6 < 1, impossible for this gas.
Thus AC is adiabatic. Answers A and D.