Clausius–Clapeyron relation: For which types of phase-change processes is the Clausius–Clapeyron equation applicable?

Difficulty: Easy

Correct Answer: all (a), (b) & (c)

Explanation:


Introduction / Context:
The Clausius–Clapeyron equation connects the slope of a phase boundary in the pressure–temperature plane to the latent heat and volume change of the transition. It is fundamental for predicting how saturation pressure varies with temperature and applies to any two-phase equilibrium involving a first-order phase transition.


Given Data / Assumptions:

  • Two phases in equilibrium (e.g., solid–vapor, solid–liquid, liquid–vapor).
  • Latent heat and molar volume change defined at the coexistence condition.
  • Reversible, equilibrium phase change.


Concept / Approach:
The general form is dP/dT = ΔH_trans / (T ΔV_trans), where ΔH_trans is the molar latent heat for the specific transition and ΔV_trans is the change in molar volume between phases. This derivation makes no restriction on which pair of phases is considered, so it holds for sublimation (solid–vapor), melting/fusion (solid–liquid), and vaporisation (liquid–vapor). The “Clausius” simplification often refers to liquid–vapor with ΔV ≈ V_vapor and ideal vapor behavior, but the base Clapeyron equation itself is universal to first-order transitions.


Step-by-Step Solution:

Recognize that Clapeyron comes from equality of Gibbs free energy of two phases and its temperature/pressure derivatives.Identify that ΔH and ΔV are pathway-independent state changes at coexistence.Conclude the relation applies to any two-phase equilibrium line (solid–liquid, solid–vapor, liquid–vapor).Therefore the correct choice is “all (a), (b) & (c).”


Verification / Alternative check:
Textbook P–T phase diagrams show slope predictions (e.g., positive for most melting curves, negative for water’s solid–liquid line near 0 °C), all derived from the Clapeyron equation.


Why Other Options Are Wrong:

  • Choosing only one phase change ignores the general derivation that does not depend on which two phases are in equilibrium.
  • None of these is contradicted by the standard proof.


Common Pitfalls:
Confusing the general Clapeyron equation with the “Clausius” ideal-vapor approximation, which is a special case for vaporisation only.


Final Answer:
all (a), (b) & (c)

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