What does a positive or negative liquid-solid slope indicate in a phase diagram?

Question

9322 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-27T19:51:06

When you already know the $DeltabarH$ for the phase transition, you will know whether the solid is more dense or less dense than the liquid.

Consider the phase diagram of water.

![https://s3-us-west-2.amazonaws.com/](useruploads.socratic.org)

The liquid-solid coexistence curve is $bar(AD)$ , and the slope of a any two-phase equilibrium curve is given by the Clapeyron equation:

$(dP)/(dT) = (DeltabarH)/(TDeltabarV)$

Consider $bar(AD)$ .

We know that the enthalpy of fusion for melting water is $DeltabarH_"fus" = "6.02 kJ/mol"$ , or $"60.20 L"cdot"bar/mol"$ , a positive quantity.
We also know that the temperature $T$ in $"K"$ must be positive.

Since the slope of $bar(AD)$ is negative, i.e. $(dP)/(dT) < 0$ , we can say that:

$(-) = ((+))/((+)(?))$

Thus, the change in molar volume $DeltabarV$ for melting ice is negative.

In other words, ice contracts when it melts and water expands when it freezes.