Answer
(a) $2.6MJ$
(b) $1100K$
(c) $5MJ$
Work Step by Step
(a) We can find the work done form A to C as
$W_{AC}=[200(10-2)+\frac{1}{2}(6-4)(600-200)+\frac{1}{2}(8-6)(600-200)+(400-200)]\times 10^3J=2.6MJ$
(b) We can find the required temperature as follows:
$T_f=\frac{V_fT_i}{V_i}$
We plug in the known values to obtain:
$T_f=\frac{10m^3}{2m^3}(220K)=1100K$
(c) We know that
$\Delta U=\frac{3}{2}(\frac{P_iV_i}{RT_i})R\Delta T$
We plug in the known values to obtain:
$\Delta U=\frac{3}{2}(\frac{(200KPa)(2m^3)}{220K})(1100K-220K)$
$\Delta U=2.4MJ$
Now $Q=\Delta +W$
We plug in the known values to obtain:
$Q=2.4MJ+2.6MJ=5MJ$
