Amount of work supplied to a closed system = 150 kJ
Initial volume = 0.6 m3
Pressure-volume relationship, p = 8 – 4V
The work done during the process is given by
But this work is equal to – 150 × 103 J as this work is supplied to the system.
∴ – 150 × 103 = 105[8V2 – 2V22 – 4.08]
or 2V22 – 8V2 + 2.58 = 0
Positive sign is incompatible with the present problem, therefore it is not considered
Final volume, V2 = 0.354 m3.
and, final pressure, p2 = 8 – 4V = 8 – 4 × 0.354
= 6.584 bar = 6.584 × 105 N/m2 or Pa.