(3)

First, isothermal expansion

$PV=P\text{'}\left(2V\right)\Rightarrow P\text{'}=\frac{P}{2}$

Then, adiabatic expansion, $P\text{'}{\left(2V\right)}^{\gamma }=P_f{\left(16V\right)}^{\gamma }$

(For adiabatic process, $P{V}^{\gamma }$ = constant)

         $\frac{P}{2}{\left(2V\right)}^{5/3}$$=P_f{\left(16V\right)}^{5/3}$

$P_f=\frac{P}{2}{\left(\frac{2V}{16V}\right)}^{5/3}=\frac{P}{2}{\left(\frac{1}{8}\right)}^{5/3}=\frac{P}{2}{\left(\frac{1}{2^3}\right)}^{5/3}=\frac{P}{64}$

(4)

[IMAGE 116]

In a cyclic process work done is equal to the area under the cycle and is positive if the cycle is clockwise and negative if the cycle is anticlockwise.

As it is clear from figure,

$W_{AEDA}=+\text{area of }∆AED=+\frac{1}{2}{P}_0{V}_0$

$W_{BCEB}=-\text{area of }∆BCE=-\frac{1}{2}{P}_0{V}_0$

The net work done by the system is

          $W_{\text{net}}=W_{AEDA}+W_{BCEB}=+\frac{1}{2}{P}_0{V}_0-\frac{1}{2}{P}_0{V}_0=\text{zero}$