# How are the conditions of pressure and temperature shown on a [phase] diagram at which two phases coexist in equilibrium?

Jul 7, 2017

Well, luckily for us, phase diagrams are made to be straightforward to read. The curves are literally the lines of two-phase coexistence.

They indicate one degree of freedom (i.e. parallel to the curve at each instantaneous displacement), in accordance with the Gibbs' phase rule.

Consider the phase diagram of water:

If you are unsure how many phases there are in a given region on a phase diagram, refer to the Gibbs' phase rule to check:

$\boldsymbol{f = c - p + 2}$

• $f$ is the number of degrees of freedom, i.e. the number of coordinate directions you are allowed to move in a phase diagram without moving out of the bounds of consideration.
• $c$ is the number of components in the substance (could be a solution full of electrolytes... could be interacting).
• $p$ is the number of phases.

Below are example calculations for cases we should already be able to verify physically.

TEST CHECK 1: TRIPLE POINT

For example, consider the triple point of water at ${0.01}^{\circ} \text{C}$ and $\text{0.0060 atm}$.

• $f = 0$, because you can't move around in a triple point in any way, unless you are no longer at the triple point.
• $c = 1$ for pure substances, because there is only one of itself in itself.

This indicates that the number of phases at a triple point is:

$\boldsymbol{p} = c + 2 - f = 1 + 2 - 0$

$= \boldsymbol{3}$ phases in equilibrium at the triple point.

And in fact, that's why it's called a triple point, because three phases coexist at a triple point.

TEST CHECK 2: COEXISTENCE CURVE

Or, consider the liquid-vapor coexistence curve, $\overline{A E}$. We should expect TWO coexisting phases: liquid and vapor. Thus, we expect $p = 2$. We have:

• $f = 1$, because if you move parallel to the curve for each instantaneous displacement, you won't move off the curve.
• $c = 1$, because our water sample is assumed to be a pure substance.

And we get:

$\boldsymbol{p} = c + 2 - f = 1 + 2 - 1$

$= \boldsymbol{2}$ phases in equilibrium with each other, as expected.

TEST CHECK 3: SINGLE-PHASE REGION

Or, consider just being in a single-phase region, like at ${50}^{\circ} \text{C}$ and $\text{1.00 atm}$. We should just have one phase: the liquid. We have:

• $f = 2$, because you can move in spirals if you wanted, which requires hybridizing two particular coordinate directions, while still staying in the same two-dimensional region.
• $c = 1$, and we know why at this point.

So,

$\boldsymbol{p} = c + 2 - f = 1 + 2 - 2$

$= \boldsymbol{1}$ phase exists by itself, which we already said... it's the liquid phase.