#### Answer

a) 60.3 Joules
b) 1.516 L

#### Work Step by Step

a) We first find the pressure after the initial process is complete. It is adiabatic, so we find:
$(100)(4)^{1.4}=P(2)^{1.4} \\ P = 263.9 \ kPa$
Next, we find the work done over this process:
$W = \frac{(263.9)(4)-(100)(2)}{.4}=319.5 \ J$
We now need to consider the next process, which is isobaric, for it occurs at a constant pressure. Thus, we find the new volume to be:
$V = .3789 \times 4 = 1.516 \ L$
We now use the equation for work in isobaric processes:
$ W = -(263.9)(1.516-2)=127.8 \ J$
Finally, we consider the last process, which is isothermal:
$W = -400ln(\frac{4}{1.52})=-387 \ J$
Adding these up, we find that the net work is: $\fbox{60.3 J}$
b) Using the above work, we see that the minimum volume is 1.516 liters.