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🧪 Chemistry  ·  Thermodynamics  ·  NEET & JEE

The entropy change for melting of ice at 0°C (ΔH_fusion = 6 kJ/mol) is:

Answer: 21.98 J K⁻¹ mol⁻¹.

  • A 6 J/K in general practice
  • B 21.98 J K⁻¹ mol⁻¹
  • C 22 kJ/K as frequently described
  • D 0.022 J/K in most textbook accounts

Correct answer: B. 21.98 J K⁻¹ mol⁻¹

Explanation: ΔS = ΔH/T = 6000 J / 273 K = 21.98 J K⁻¹ mol⁻¹.

P-V Diagram: Isothermal vs Adiabatic ExpansionVPIsothermal (T constant)Adiabatic (q=0)Adiabatic curve is steeper: no heat enters to cushion the pressure drop

Isothermal expansion follows a gentler curve (heat flows in to keep T constant) while adiabatic expansion drops in pressure more steeply (no heat exchange, so internal energy and temperature fall as the gas does work).

Concept context

Study energy changes in chemical reactions. Understand enthalpy, entropy, Gibbs free energy, and the laws of thermodynamics that decide whether a reaction will occur spontaneously or not.

Read the full Thermodynamics notes →