Answer
$T_0 = 433.0 K$
$T_f = 173.2 K$
$W = -540 kJ$
Work Step by Step
For ideal gases:
$PV = nRT$
With Helium' molar mass: $M = 4.0026 \frac{kg}{kmol}$, and mass $m = 1\ kg$
$n = \frac{m}{M} = 0.25 kmol$
$R = 8.314 \frac{kJ}{kmol.K}$
$T_0 = \frac{P_0V_0}{nR}$, $T_f = \frac{P_fV_f}{nR}$
With $P_0=P_f = 180kPa, V_0=5m³, V_f=2m³$
$T_0 = 433.0 K$
$T_f = 173.2 K$
For the isobaric work:
$W = P_0(V_f-V_0)$
$W = -540 kJ$
