Given that :
m = 1 kg
P1 = 98 kPa = 98 × 103 Pa;
T1 =36°C = 36 + 273 = 309 K,
P2 = 35 bar = 35 × 105 Pa
V3 – V2 = 0.06VS
For air standard cycle P1V1 = mRT1
98 × 103 × V1 = 1 × 287 × 309
V1 = 1.10 m3; V1 = V2 + V3 = 1.10
As process 1-2 is adiabtic compression process,
= 1539580 × 0.415 Pa = 6389257 Pa = 6389.3 KPa