Answer
$(a) W = 34 650 J $
$(b) \Delta U = 80350 J $
$(c) $ It doesn't matter if the gas is ideal or not
Work Step by Step
$(a)$ The process is a constant pressure process. The work done by the gas is
$W = P(V_f -V_i) $
$W = 1.65 \times 10^5 Pa(0.320 m^3- 0.110 m^3 ) $
$W = 34 650 J$
$(b) $ The total internal energy is
$\Delta U = \Delta Q - W$
$\Delta U = 1.15 \times 10^5 J- 34 650 J $
$\Delta U = 80350 J $
$(c) $ It doesn't matter if the gas is ideal or not because the energy of the system will remain conserved, obeying the First Law of Thermodynamics.
