Answer:
The enthalpy of vaporization of water at 273 K and 1 bar = 44.9 KJ/mol
Explanation:
Enthalpy of vaporization of water at 273 K, ΔHvap(T₂) is given as;
ΔHvap(T₂) = ΔHvap(T₁) + ΔCp * (T₂ - T₁)
where ΔCp = molar heat capacity of gas - molar heat capacity of liquid
Therefore, ΔCp = (33.6 - 75.3) = -41.70 J/(mol K) = 0.0417 kJ/(molK)
substituting ΔCp = 0.0417 kJ/(mol K) in the initial formula
;
ΔHvap(T) = ΔHvap(T1) + ΔCp * (T₂ - T₁)
ΔHvap(T₂)= 40.7 kJ/mol + {-0.0417 kJ/(mol K) * (272 - 373 K)}
ΔHvap(T₂) = 44.9 kJ/mol
Therefore, enthalpy of vaporization of water at 273 K and 1 bar = 44.9kJ/mol
Answer:
smooth and skeleton muscle
Explanation:
they are considered as musculoskeletal system.
Answer:
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