Answer:
a) the entropy of vaporization is ΔS = 0.08779 kJ/(mol*K)
b) the entropy of the surroundins is ΔS surr ≥ -0.08779 kJ/(mol*K )
Explanation:
from the second law of thermodynamics
ΔS ≥ ∫dQ /T
where ΔS= change in entropy of trichloromethane , Q= heat exchanged with the surroundings, T= absolute temperature
since the boiling of trichloromethane is at constant temperature if it is in a pure state ( with no other substances), then
ΔS ≥ ∫dQ /T = 1/T ∫dQ = ΔH/T
ΔS ≥ ΔH/T
since the process is also reversible
ΔS = ΔH/T = 29.4 kJ mol−1/ 334.88 K = 0.08779 kJ/(mol*K)
for the surroundings
ΔS universe = ΔS + ΔS surr ≥ 0 (the entropy of the universe always increases)
ΔS surr ≥ -ΔS = -0.08779 kJ/(mol*K)
ΔS surr ≥ -0.08779 kJ/(mol*K )
