Answer:
The process is not possible
Explanation:
if we want to determine if the process is possible , we can check with the second law of thermodynamics
ΔS≥ ∫dQ/T
for a constant temperature process ( condensation)
ΔS≥ 1/T ∫dQ
and from the first law of thermodynamics
ΔH = Q - ∫VdP , but P=constant → dP=0 → ∫VdP=0
Q=ΔH
then
ΔS≥ΔH/T
from steam tables
at P= constant = 200 Kpa → T= 120°C = 393 K
at P= constant → H vapor = 2201.5 kJ/kg , H liquid = 1.5302 kJ/kg
, S vapor= 7.1269 kJ/kg , S liquid 1.7022 kJ/kg
therefore
ΔH = H vapor - H liquid = 2201.5 kJ/kg - 1.5302 kJ/kg = 2199.9698 kJ/kg
ΔS = S vapor - S liquid = 7.1269 kJ/kg - 1.7022 kJ/kg = 5.4247 kJ/kg
therefore since
ΔS required = ΔH/T = 2199.9698 kJ/kg/(393 K)= 5.597 kJ/kg K
and
ΔS= 5.4247 kJ/kg ≤ ΔS required=5.597 kJ/kg K
the process is not possible
