Answer
(a) $40KPa$
(b) $83KPa$
(c) $80KJ$
Work Step by Step
(a) We know that for adiabatic expansion
$\frac{P_B}{P_A}=(\frac{V_A}{V_B})^{\gamma}$
We plug in the known values to obtain:
$P_B=250(\frac{1}{3})^{1.67}=40KPa$
(b) As $P_CV_C=P_AV_A$
$\implies P_C=\frac{P_AV_A}{V_C}$
We plug in the known values to obtain:
$P_C=250\times \frac{1}{3}=83KPa$
(c) As $W_{AB}=\frac{P_BV_B-P_AV_A}{\gamma-1}$
$W_{AB}=\frac{-250\times 1+40\times 3}{1.67-1}=-194KJ$
Now $W_{CA}=-nRT\ln\frac{V_A}{V_C}=-P_CV_C\ln{\frac{V_A}{V_C}}$
$W_{AB}=-83\times 3\ln(\frac{a}{3})-274KJ$
Thus, the total work done on gas$=\frac{W_{AB}}{W_{BC}}+W_{CA}$
$=-194KJ+0+274KJ=80KJ$
