In the given P - V diagram, a monoatomic gas (γ = (5)/(3)) is first...

For adiabatic process $P{V}^{\gamma} = C$

$⇒ 100\left(0 . 8\right)\dfrac{5}{3} = 300\left({V}_{B}\right)\dfrac{5}{3}$

$⇒ {V}_{B} = \dfrac{0 . 8}{{3}^{3 / 5}} = 0 . 4{m}^{3}$

Work done in an adiabatic process is given by,

$W = \dfrac{{P}_{1}{V}_{1} - {P}_{2}{V}_{2}}{\gamma - 1}$

$⇒ {W}_{AB} = \dfrac{{P}_{A}{V}_{A} - {P}_{B}{V}_{B}}{\dfrac{5}{3} - 1} = \dfrac{80 - 300 \times 0 . 4}{2 / 3}kJ$

$⇒ {W}_{AB} = - 60kJ$

Therefore, option \left(C\right)$is correct.</p><p>$C \to A$is isochoric$⇒ \left(D\right)$is correct.</p><p>For process$BC$</p><p>${W}_{BC} = nRTln\dfrac{{V}_{2}}{{V}_{1}} = PVln\dfrac{{V}_{2}}{{V}_{1}}$</p><p>$⇒ {W}_{BC} = 300 \times 0 . 4\ln \left[\dfrac{0 . 8}{0 . 4}\right]$</p><p>$⇒ {W}_{BC} = 120 \times \ln 2kJ ⇒ {W}_{BC} = 84kJ$</p><p>Therefore, option \left(B\right)$ is correct.

Now,

${W}_{AB} + {W}_{BC} = - 60 + 84 = 24kJ$

Therefore, option \left(A\right) is incorrect.