#vaporization

Ideal Gas Laws and Thermal Properties

How much heat is required to vaporize 0.25kg of ethanol at 78∘C? (Latent heat of vaporization of ethanol = 8.5×105J kg−1

Given: m = 0.25kg, Lv = 8.5×105J kg−1. Q = mLv = 0.25×8.5×105 = 212500J = 212.5kJ. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 212.5 kJ. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.

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Ideal Gas Laws and Thermal Properties

How much heat is required to raise 0.6kg of water from 25∘C to 75∘C and then convert 0.4kg of it to steam at 100∘C in a

Q1 = (0.6×4186+0.1×386)×(75−25) = (2511.6+38.6)×50 = 2550.2×50 = 127510J (to 75°C). Q2 = (0.6×4186+0.1×386)×(100−75) = 2550.2×25 = 63755J (to 100°C). Q3 = 0.4×2.256×106 = 902400J (vaporization). Total: Q = 127510+63755+902400 = 1093665J = 1093.67kJ.

Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.

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Temperature Scales and Triple Point

How much heat is required to raise 0.4kg of water from 15∘C to 85∘C and then convert 0.25kg to steam at 100∘C in a 0.2kg

Q1 = (0.4×4186+0.2×236)×(85−15) = (1674.4+47.2)×70 = 1721.6×70 = 120512J (to 85°C). Q2 = (0.4×4186+0.2×236)×(100−85) = 1721.6×15 = 25824J (to 100°C). Q3 = 0.25×2.256×106 = 564000J (vaporization). Total: Q = 120512+25824+564000 = 710336J = 710.34kJ.

Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.

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Temperature Scales and Triple Point

During vaporization, what does the supplied heat primarily do to the liquid?

During vaporization, the supplied heat (latent heat of vaporization) breaks intermolecular bonds to convert the liquid into vapor, without changing its temperature. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Converts it to vapor without temperature change. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.

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Gibbs Energy and Gibbs Energy Change and Equilibrium

For the vaporization of 72 g of water at 373 K ( Δ Hvap = 40.79 kJ/mol , molar mass = 18 g/mol), what is Δ Ssys ?

Moles = 72 / 18 = 4 mol . For vaporization, Δ Ssys = n × Δ Hvap / T = 4 × 40790 / 373 ≈ 437.48 J/K ≈ 437.5 J/K .

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Gibbs Energy and Gibbs Energy Change and Equilibrium

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Bond Enthalpy and Entropy and Spontaneity

What is the entropy change (Δ S) for the vaporization of 18 g of water at 373 K if Δ Hvap = 40.79 kJ/mol? (Molar mass of

Moles = 18 / 18 = 1 mol, Δ S = Δ Hvap / T = 40.79 × 10³ / 373 = 109.36 J/K·mol.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Bond Enthalpy and Entropy and Spontaneity

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Bond Enthalpy and Entropy and Spontaneity

Calculate Δ Ssys when 1 mol of water vaporizes at 373 K. ( Δ Hvap = 40.79 kJ/mol )

For H₂O(l) → H₂O(g) at 373 K, Δ Ssys = (Δ Hvap/T) = (40.79 × 10³/373) = 109.4 J/K , representing the entropy increase due to phase change from liquid to gas.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Bond Enthalpy and Entropy and Spontaneity

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