For the process H₂O(l) → H₂O(g) at 373 K, Δ H = 40.79 kJ/mol, what is Δ Ssurr?
Δ Ssurr = -Δ H / T = -40.79 × 10³ / 373 = -109.4 J/K·mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications
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What is the change in internal energy ( Δ U ) for the vaporization of 1 mol of water at 373 K if Δ H = 40.79 kJ/mol ? (
For H₂O(l) → H₂O(g) , Δ ng = 1 . Using Δ H = Δ U + Δ ng RT , where RT = 8.314 × 373 × 10⁻³ = 3.1012 kJ , we get Δ U = Δ H - Δ ng RT = 40.79 - 3.1012 = 37.69 kJ/mol .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy
The enthalpy change for vaporization of 1 mol of water at 373 K is 40.79 kJ/mol. What is Δ U if water vapor behaves as a
For H₂O(l) → H₂O(g) , Δ ng = 1 . Using Δ H = Δ U + Δ ng RT , where RT = 8.314 × 373 × 10⁻³ = 3.1012 kJ , Δ U = 40.79 - 3.1012 = 37.69 kJ/mol .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Enthalpy Changes - Reaction Formation Combustion and Hess's Law