Answer
(a) $318KJ$
(b) $795KJ$
(c) $1113KJ$
Work Step by Step
(a) We can find the required work done as follows:
$W=A_{rectangle}A_{triangle}$
We plug in the known values to obtain:
$W=(3m^3-1m^3)(106KPa)+\frac{1}{2}(3m^3-1m^3)(212KPa-106KPa)$
$W=318KJ$
(b) The change in the internal energy of the gas can be determined as:
$\Delta U=\frac{3}{2}nR[T_f-T_i]$
$\Delta U=\frac{3}{2}nR[\frac{P_fV_f}{nR}-\frac{P_iV_i}{nR}]$
$\Delta U=\frac{3}{2}[P_fV_f-P_iV_i]$
We plug in the known values to obtain:
$\Delta U=\frac{3}{2}[(212\times 10^3Pa)(3m^3)-(106\times 10^3Pa)(1m^3)]$
$\Delta U=795KJ$
(c) We can find the required heat added to the system as follows:
$Q=\Delta U+W$
We plug in the known values to obtain:
$Q=795KJ+318KJ$
$Q=1113KJ$
