(a&d) The maximum volume of the cylinder is 4 litres, and the compression ratio (max:min volume) is 18:1, so the (minimum) volume at the end of the compression stroke is V2 = 4/18=2/9=0.2222 litres.
Cylinder volume at start of expansion = volume at end of compression = 0.2222 litres. (Seems too easy for 5 marks each!)
(b&c) Compression is adiabatic, so the gas obeys the law p1V1^γ = p2V2^γ. p1=150kPa and V1/V2=18, therefore the pressure at the end of the compression stroke is
p2 = p1 x (V1/V2)^γ = 150 kPa x 18^1.4 = 8579.7 kPa.
The temperature at this point is T2 where p1V1/T1=p2V2/T2. p2/p1=18^1.4, V1/V2=18, and T1=273+47=320K, so
T2 = T1 x (p2/p1) x (V2/V1) = 320 x 18^1.4 x 18^-1 = 320 x 18^0.4 = 1016.86 K.
H=6 kJ of heat energy is now added instantaneously and raises the temperature at constant volume. The increase in temperature depends on the mass of gas, which is not given - neither is there enough information to calculate it, eg the density of air at some temperature and pressure. THE QUESTION IS FAULTY.
From data tables, the density of dry air at STP (0°C and 760mmHg = 101.3 kPa) is 1.293 kg/m^3 = 1.293x10^-3 kg/litre. The air in the cylinder occupies 4 litres at 47°C and 150 kPa. Using the ideal gas law p1V1/T1= p2V2/T2, the volume which this gas would occupy at STP is
V2 = V1 x (p1/p2) x (T2/T1) = 4 x (150/101.3) x (273/320) = 5.053 litres.
The mass of gas in the cylinder is m = 1.293x10^-3 kg/litre x 5.053 litres = 6.534x10^-3 kg.
The temperature rise of the gas ΔT is related to heat injected H by H = mCvΔT. Therefore
ΔT = H/mCv = 6 kJ/(6.534x10^-3 kg x 0.718 kJ/kgK) = 1278.93 K.
The temperature after heat injection is therefore 1016.86+1278.93 = 2295.79 K = 2020 °C (after rounding).
Using p1/T1=p2/T2, the pressure after heat injection is
p2 = p1 x (T2/T1) = 8579.7 kPa x ( 2295.79 K/ 1016.86 K) = 19,370 kPa = 19.4 MPa.
(NOTE: parts b&c carry the same marks as parts a&d but require a lot more work. Coupled with the missing density of air, this is a badly designed question.)