Question

A pure substance M has lesser density in solid state than in liquid state. The $\Delta S_\text{fusion}$ of M is +25 J K$^{-1}$ mol$^{-1}$. The CORRECT representative Pressure–Temperature diagram for the fusion of M is 

• A. A
• B. B
• C. C
• D. D

Hint:

When a solid is less dense than its liquid (like ice), increasing pressure lowers its melting point because $\Delta V_\text{fusion}$ is negative, giving a negative $\frac{dP}{dT}$ slope.

Answer 1

The Correct Option is B

Step 1: Understanding the problem.
The question gives that solid M is less dense than liquid M. That means, upon melting, the substance contracts (volume decreases). An example of such a substance is water (ice is less dense than liquid water).
Step 2: Using the Clausius–Clapeyron equation.
The slope of the solid–liquid equilibrium line is given by \[ \frac{dP}{dT} = \frac{\Delta S_\text{fusion}}{\Delta V_\text{fusion}} \] where $\Delta S_\text{fusion}$ is the entropy change upon fusion and $\Delta V_\text{fusion}$ is the volume change.
Step 3: Sign of the terms.
Given: $\Delta S_\text{fusion}$ = +25 J K$^{-1}$ mol$^{-1}$ (positive, since melting increases entropy).
Since the solid is less dense than the liquid, $\Delta V_\text{fusion}$ = $V_\text{liquid} - V_\text{solid}$<0.
Therefore, \[ \frac{dP}{dT} = \frac{(+)}{(-)} = \text{negative.} \] Step 4: Interpretation.
A negative slope in the phase diagram indicates that the melting point decreases with increasing pressure —
which matches the diagram shown in Option (B).
Step 5: Conclusion.
Hence, the correct pressure–temperature diagram is Option (B).