Question

What is the relation between critical temperature and boiling point or vapor pressure?

Answer 1

Higher `P_(vap) harr` lower `T_c`

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This is easier to see with a phase diagram. Consider the phase diagram of `"CO"_2`:

The critical temperature `T_c` is the temperature at which some critical pressure `P_c` gives the critical point, `(T_c, P_c)`, where the liquid and vapor phase are indistinguishable. i.e. it is the temperature above which the vapor cannot be liquiefied.

For `"CO"_2`, `T_c ~~ "300 K"`, and `P_c ~~ "60 bar"`.

When we consider the substance with the higher vapor pressure, it thus vaporizes more easily, having more vapor in equilibrium with the liquid phase, giving it a lower boiling point.

But going from left to right (i.e. heating), a liquid needs to boil to become a gas. Thus, if boiling is easier, reaching the critical temperature should be easier as well.

Thus (with exceptions in mind), `T_c` should tend to be lower for the substance with the higher vapor pressure.

For example... see the trend here. All data are derived from NIST phase change data.

`"CO"_2`:
`P_(vap) = "64.01 bar"` at `25^@ "C"`, `T_c ~~ "304 K"`
`T_b = -78.5^@ "C"`

`"CH"_3"Cl"`:
`P_(vap) ~~ "6.5 bar"` at `25^@ "C"`, `T_c ~~ "416 K"`
`T_b = -24.2^@ "C"`
(vapor pressure estimated, b/c missing data points)

`"CH"_3"CH"_2-"O"-"CH"_2"CH"_3`:
`P_(vap) ~~ "0.6688 bar"` at `25^@ "C"`, `T_c ~~ "467 K"`
`T_b = 34.6^@ "C"`

`"CH"_3-("C"="O")-"CH"_3`:
`P_(vap) ~~ "0.3078 bar"` at `25^@ "C"`, `T_c ~~ "508 K"`
`T_b = 56.2^@ "C"`

`"H"_2"O"`:
`P_(vap) ~~ "0.0317 bar"` at `25^@ "C"`, `T_c ~~ "647 K"`
`T_b = 100.0^@ "C"`

`"I"_2`:
`P_(vap)` `"<<"` `"0.0112 bar"` at `25^@ "C"`, `T_c ~~ "819 K"`
`T_b = 184.4^@ "C"`