Question

How do I make a phase diagram for water?

Answer 1

By knowing where the normal boiling and freezing points are (at `"1 atm"`), critical point and triple point are, and the slope of the liquid-solid, liquid-vapor, and solid-vapor coexistence curves.

Note that the phase diagram is simply a pressure vs. temperature graph.

---

We know that `T_f = 0^@ "C"` at `"1 atm"` and `T_b = 100^@ "C"` at `"1 atm"` are the normal freezing and boiling points, respectively.

The critical point is when the liquid and vapor exist at the same time, at some `(P_c,T_c)` on a `P` vs. `T` graph. For water, `P_c = "218.3 atm"` and `T_c = 374.2^@ "C"`.

The triple point is when the solid, liquid, and vapor exist at the same time, at some `(P_"trpl", T_"trpl")`. For water, `T_"trpl" = 0.01^@ "C"`, and `P_"trpl" = "0.006032 atm"`.

The Clapeyron equation describes the slope of a coexistence curve, `(dP)/(dT) = (DeltabarH_"trs")/(TDeltabarV_("trs"))`. We omit the derivation, but:

• `DeltabarH_"trs"` is the molar enthalpy of the phase transition.
• `DeltabarV_"trs"` is the change in molar volume due to the phase transition.
• `T` is the temperature in `"K"`.

Using the Clapeyron equation:

• For the liquid-solid coexistence curve in the phase diagram of water, `(dP)/(dT) < 0`, since `DeltabarV_((l)->(s)) > 0`, and `DeltabarH_"frz" < 0`, which indicates that water expands when it freezes (this is unusual).
• For the liquid-vapor coexistence curve, `(dP)/(dT) > 0`, since `DeltabarV_((l)->(g)) > 0`, and `DeltabarH_"vap" > 0`, which indicates that water expands when it vaporizes (as usual).
• For the solid-vapor coexistence curve, `(dP)/(dT) > 0`, since `DeltabarV_((s)->(g)) > 0`, and `DeltabarH_"sub" > 0`, which indicates that water expands when it sublimates (clearly).

Put all that together into a phase diagram:

Simply note where each point is, and you should be able to at least sketch its general shape. Note that the phase diagram is simply a pressure vs. temperature graph.