Question #344cb

Without comparing to a fully-labeled diagram, we know that:

• at low temperature and high pressure, a solid would exist, because the particles are slow and condensed.
• at medium temperature (higher than the melting point) and high pressure, a liquid would exist.
• at high temperature and low pressure, a gas would exist, because the particles are fast and not condensed.

And so, the solid should be to the upper-left, liquid to the upper-right, and gas to the lower-right. That immediately gives:

• #b# contains only liquid
• #c# contains only solid
• #d# contains only gas

#a# is interesting, because it lies on the liquid-gas coexistence curve. That means it is at a phase equilibrium point where both liquid and gas exist at the same time ("coexistence"). That is,

• #a# contains liquid and gas

Gibbs' phase rule, shown as

#f = 2 + overbrace(c)^("components") - overbrace(p)^"phases"#,

does allow us to determine degrees of freedom, but we could first do it intuitively.

• If you are in a one-phase region (#b,c,d#), you have one phase existing by itself, and #bb2# degrees of freedom; you can change #T# and #P# at the same time without breaking a steady state.
• If you are on a two-phase coexistence curve (#a#), you have two phases coexisting, and #bb1# degree of freedom; you can change either #T# or #P#, but not both at the same time, before you break the phase equilibrium.
• If you are on a triple point, there are three phases coexisting, and you have #bb0# degrees of freedom. You cannot change either #T# nor #P# without breaking the equilibrium.

To verify that,

#a#: #" "##f = 2 + 1 - 2 = bb1#, for a two-phase coexistence curve

#b#: #" "##f = 2 + 1 - 1 = bb2#, for a one-phase region

#c#: #" "##f = 2 + 1 - 1 = bb2#, for a one-phase region

#d#: #" "##f = 2 + 1 - 1 = bb2#, for a one-phase region

where there is only one component (#c = 1#) because it is a pure substance. Try verifying this for a triple point.