Answer
Please see the work below.
Work Step by Step
(a) We know that for the process $A\rightarrow B$
$W_{AB}=P_AV_A\ln(\frac{V_B}{V_A})$
We plug in the known values to obtain:
$W_{AB}=(600KPa)(0.75\times 10^{-3}m^3)\ln(\frac{4.5\times 10^{-3m^3}}{0.75\times 10^{-3}m^3})$
$W_{AB}=806J$
Now $Q_{AB}=\Delta U_{AB}+W_{AB}$
$\implies Q_{AB}=0+806J=806J$
For process $B\rightarrow C$
$Q_{BC}=\frac{5}{2}nR\Delta T$
$\implies Q_{BC}=\frac{5}{2}P\Delta V$
$\implies Q_{BC}=\frac{5}{2}(100KPa)(0.75\times 10^{-3}m^3-4.5\times 10^{-3}m^3)=-938J$
$W_{BC}=P\Delta V$
$W_{BC}=(100\times 10^3Pa)(0.75\times 10^{-3}m^3-4.5\times 10^{-3}m^3)=-375J$
and $\Delta U_{BC}=Q_{BC}-(-W_{BC})$
$\Delta U_{BC}=-938J-(-375J)=-563J$
For process $C\rightarrow A$
$\Delta V=0$
$\implies W_{CA}=P\Delta V$
$W=0$
and $\Delta U_{CA}=-(\Delta U_{AB}+\Delta U_{BC})$
$\Delta U_{CA}=-(0+(-563J))$
$\Delta U_{CA}=563J$
$Q_{CA}=\Delta U_{CA}+W_{CA}$
$Q_{CA}=563J+0=563J$
(b) We can find the efficiency of the engine as follows:
$e=\frac{W}{Q_k}$
We plug in the known values to obtain:
$e=\frac{806J-375J}{806J+563J}$
$e=0.31$
