Work done and total heat transfer for piston-cylinder device

Baartzy89

Hi everybody, this question is on an old exam paper given to us for exam prep purposes. No answers have been supplied however. Am I on the right track?

Question:
A mass of 0.2kg of saturated refrigerant-134a is contained in a piston-cylinder device at 200kPa. Initially 75% of the mass is in the liquid phase. Now heat is transferred to the refrigerant at constant pressure until the cylinder contains only vapor. Determine:
a) the work done, and
b) the total heat transfer

Assumptions:
Piston-cylinder device is frictionless, constant pressure, quasi-equilibrium, no heat lost to surroundings

a) the work done
Formula:
W = P (V2-V1) where,

V2 = Volume at state 2
V1 = Volume at state 1
P = Pressure
W = Boundary work

Properties:
P = 200 kPa
V2 = 0.2 x vg where vg = 0.099867 (gas table) = 0.0199734 m3/kg
V1 = 0.2 x (0.75 x vf + 0.25 x vg) = 0.005106345 m3/kg

Analysis:
W = P(V2-V1)
= 200 x (0.0199734 - 0.005106345)
= 3.89 kJ boundary work

b) total heat transfer
Formula:
Q = h2 - h1 where,

Q = total heat transferred
h2 = Enthalpy state 2
h1 = Enthalpy state 1

Properties:
h2 = hg = 244.46 kJ/kg (from gas tables)
h1 = .75 x hf + .25 x hg = .75 x 38.43 + .25 x 244.46 = 89.9375 kJ/kg

Analysis:
Q = h2 - h1
= 244.46 - 89.9375
= 154.52 kJ heat

rock.freak667

I think that is all correct.