Respuesta :
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
In the case when heat is removed from the system and transferred to the environment so this kind of process is not possible.
Second law of thermodynamics:
Since
ΔS≥ ∫dQ/T
Now
for a constant temperature process ( condensation)
ΔS≥ 1/T ∫dQ
Now from first law of thermodynamics
ΔH = Q - ∫VdP , but P=constant → dP=0 → ∫VdP=0
Q=ΔH
So,
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
So,
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
Also,
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
Hence, the process is not possible
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