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NEET CHEMISTRYThermodynamicsMedium

Question

The entropy change involved in the conversion of 1 mole of liquid water at 373 K to vapour at the same temperature will be [ΔHvap=2.257 kJ/g\Delta H_{vap} = 2.257\text{ kJ/g}] [MP PET 2002]

A

0.119 kJ/K

B

0.109 kJ/K

C

0.129 kJ/K

D

0.120 kJ/K

Step-by-Step Solution

To calculate the entropy change (ΔS\Delta S) for a phase transformation occurring at equilibrium (like boiling at 373 K), we use the relationship between enthalpy and temperature: ΔS=ΔHvap/T\Delta S = \Delta H_{vap} / T .

  1. Identify the molar mass of water (H2OH_2O): The molar mass is the sum of the atomic masses of hydrogen and oxygen, which is approximately 18.02 g/mol18.02\text{ g/mol} .
  2. Convert enthalpy of vaporization to molar enthalpy: The given value is 2.257 kJ/g2.257\text{ kJ/g}. To find the molar enthalpy (ΔHvap\Delta H_{vap} per mole), multiply by the molar mass: ΔHvap=2.257 kJ/g×18.02 g/mol40.67 kJ/mol\Delta H_{vap} = 2.257\text{ kJ/g} \times 18.02\text{ g/mol} \approx 40.67\text{ kJ/mol} .
  3. Calculate the entropy change: Using the boiling point temperature (T=373 KT = 373\text{ K}): ΔS=40.67 kJ/mol/373 K0.10903 kJ/(mol\cdotK)\Delta S = 40.67\text{ kJ/mol} / 373\text{ K} \approx 0.10903\text{ kJ/(mol\cdot K)}.

For one mole of water, the change in entropy is approximately 0.109 kJ/K0.109\text{ kJ/K}, which matches Option B .

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Thermodynamics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYThermodynamicsentropychangeinvolvedconversionliquid

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