Answer
(a) $W = -9.0 \times 10^4 J$
(b) $\Delta U = - 2.30 \times 10^5J $
(c) No. It doesn't matter
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) $
$W = (1.80×10^5 Pa)(1.20 m^3 −1.70 m^3)$
$W = -9.0 \times 10^4 J$
$(b) $ The absolute value of total heat added is
$\Delta U = \Delta Q - W$
$\Delta Q = \Delta U + W$
$\Delta Q = (-1.40 \times 10^5 J)+ (-9.0 \times 10^4 J)$
$\Delta U = - 2.30 \times 10^5J $
The negative sign indicates that heat is flowing out of the system.
$(c)$ 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.
