#water vaporization

Specific Heat Capacity and Latent Heat

How much heat is required to convert 1 g of water from liquid at 100^circ C to vapor at 100^circ C at 1 atm ? (Latent he

**Latent heat** energy needed for phase change without temperature change, overcomes intermolecular forces, e.g., heating ice at 0°C to water at 0°C requires 334 kJ/kg, then heating water to 100°C requires c ΔT, then vaporization 2260 kJ/kg, illustrating two types of heat. Heat: Δ Q = m L . m = 1 g , L = 2256 J/g . Δ Q = 1 × 2256 = 2256 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η = 1 - T_c/T_h,

Ref: NCERT > Physics Book > Thermodynamics > Specific Heat Capacity and Latent Heat

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Third Law of Thermodynamics and Applications

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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First Law of Thermodynamics and Enthalpy and Internal Energy

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

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Enthalpy Changes - Reaction Formation Combustion and Hess's Law

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

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