Answer
(a) $1200KJ$
(b) No
Work Step by Step
(a) We can find the required work done as follows:
$W=\Sigma P\Delta V$
We plug in the known values to obtain:
$W=[200(6-2)+\frac{1}{2}(6-4)(600-200)]\times 10^3J$
$W=1200KJ$
(b) We know that the work done is calculated by measuring the area under the pressure-volume curve and hence the work done by the fluid does not depend on whether the fluid is an ideal gas or not.
