Answer
a) $m = 10.027\ kg$
b) $Q = 2705.78\ kJ$
Work Step by Step
At the initial state:
$P_1 = 160\ kPa,\ x=0.40,\ saturated$
From the R-134a tables:
$T_1 = -15.60 °C,\ v_{1,L} = 0.0007435\ m³/kg,\ v_{1,G} = 0.12355\ m³/kg $
$u_{1,L} = 31.06\ kJ/kg,\ u_{1,G} = 221.37\ kJ/kg $
Therefore:
$v_1 = v_{1,L}+x(v_{1,G}-v_{1,L}) = 0.049866\ m³/kg $
$u_1 = u_{1,L}+x(u_{1,G}-u_{1,L}) = 107.18\ kJ/kg$
a) Since the initial volume is 0.5 m³:
$m = 10.027\ kg$
b)
For the final state:
$P_2 = 700\ kPa,\ v_2=v_1$
Extrapolating from the table
$u_2 = 377.03\ kJ/kg,\ T_2 = 169.83°C$
Since the process occurs at a constant volume closed system without changes in potential or kinetic energies:
$Q=\Delta U$
$Q = m.\Delta u = 2705.78\ kJ$
The P-V diagram of the process would be a straight vertical line starting below the saturation curve and finishing to the right of it.
