The triple point of Iodine occurs at 0.12 atm and at standard temperature it is a solid on the phase diagram.
I understand that therefore, when heating iodine at 1 atm, it (at least some of it) will pass through the liquid phase, and it is also obvious from observing heated Iodine that it yields liquid when heated gradually. Also, just as liquids can evaporate below the boiling point, I understand that some Iodine can escape from a sample in the vapor form when the vapor pressure is lower than the atmospheric pressure.
My question is 2 part:
1) Is it correct to call the change form solid to gas below equilibrium vapor pressure "sublimation". In other words, is sublimation equivalent to the term "boiling" or is it equivalent to the term evaporation in the change of phase from liquid to vapor, or both? (Is there a separate term, even if out of use, for the solid to vapor transition below equilibrium pressure?)
2) At 1 atm, when Iodine molecules in a solid sample escape into the vapor form, either at standard temperature, or as the sample is heated, do they all briefly pass through the liquid phase first, either at a vapor pressure below 1 atm or as vapor pressure is raised above 1 atm. .
Iodine's crystal vapor pressure will of course be <0.12 atm below the triple point pressure as can be seen on both graphs. A converse but equivalent question would be whether vaporous Iodine at 1 atm can crystalize and deposit without liquifying.