Question

The enthalpy of vaporization of water at 373 K and 1 bar is 40.7 kJ/mol and the molar heat capacities are 75.3 J/(mol K) for liquid and 33.6 J/(mol K) for gaseous water. Find the enthalpy of vaporization of water at 273 K and 1 bar. Express result in kJ/mol and type in value only.

Answer 1

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