I think I understand the question, but let me repeat the question a
little differently to make sure.
If we have pure H2O at 273.16 K and 611.73 kPa we can have all three
phases present in equilibrium. If we change the temperature or
pressure, we will lose at least one phase. ***Question:*** Now instead
of changing the temperature or water-vapor pressure, suppose we
introduce an impurity, such as nitrogen gas (N2), such that the overall
pressure exceeds 611.73 kPa, but we keep it at 273.16 K. Are we still
at the triple point of water such that solid, liquid, and vapor water
are all present in stable equilibrium with the partial pressure of the
water at 611.73 kPa?
No, because not only is the gas now a mixture, the solid and liquid also
become mixtures as the N2 dissolves into the liquid and solid. We now
have a completely different situation. A mixture is very different than
a pure substance. For example, I would presume the melting point of the
mixture H2O-N2 is different from 273.16 K.
I suppose there could be a temperature and overall pressure for the
H2O-N2 mixture in which there would be a triple point for the mixture
such that we had all three phases present, but this would not be the
water triple point, and the percent composition of the three phases
would not be the same. I'm not sure if you would call this a "triple
point" or not. That is, does the use of the words "triple point" (1)
require a pure substance, or (2) can it be used if we have three phases
of the same composition, or (3) are we allowed to use those words to
describe any substance that is at a temperature and pressure were we see
three phases in equilibrium.
Personally, if the three phases do not have the same composition, I
would not be inclined to call it a triple point. But I do not know if
there is any official ruling on that, and I admit to being too lazy to
check it out right now.
I think the bottom-line answer to the original question is that if
another gas is present in the vapor phase besides H2O, then that other
gas is also present in solution with the liquid phase and the solid
phase, and if you should be able to find a T and P that you see three
phases present in apparent phase equilibrium, that would not be 273.16 K
nor 611.73 kPa partial pressure of H2O.
There was a period of time when I was trying to dissolve some gases
(such as 3He) into cryogenic liquids (such as liquid argon) and I
observed some pretty interesting temperature-pressure phenomena that I
would not have predicted ahead of time. Solutions, as physical chemists
would know, often behave non-ideally. That means our customary
idealized formulas and "rule-of-thumb" don't always work (seldom
work?)with mistures.
Michael D. Edmiston, Ph.D.
Professor of Chemistry and Physics
Bluffton University
Bluffton, OH 45817
(419)-358-3270
edmiston@bluffton.edu