## Essential University Physics: Volume 1 (3rd Edition)

(a) $40KPa$ (b) $83KPa$ (c) $80KJ$
(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$=\frac{W_{AB}}{W_{BC}}+W_{CA}$ $=-194KJ+0+274KJ=80KJ$