I have a rudimentary question about phase diagrams, my professional training is as a mechanical engineer so I don't deal with them often.

I have an experiment where I need a mixture of air and CO$_2$ with a certain mass fraction of CO$_2$. I am accomplishing this by putting dry ice into a sealed container of air. I know that CO$_2$ is already a component of air, but suppose for simplification that it is a small enough fraction to be ignored. As part of this experiment, I need to correctly interpret the phase diagram for CO$_2$ (see below).

When using the phase diagram for CO$_2$ in a mixture of gases, do I use the partial pressure of CO$_2$ or the total pressure of the mixture?

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1 Answer 1


For gases, only the partial pressure matters. Each gas can be considered separately; use just partial pressure of $\ce{CO2}$, in your example.

For solids and liquids, though, the phase diagram is more complex.

Pb-Sn Phase Diagrma Phase Diagram of Pb-Sn Alloy

Tin-lead solder, for example, has a melting point minimum for a roughly 60/40% alloy (eutectic) that is below that of pure tin or pure lead.

  • $\begingroup$ Thank you for your answer - I don't understand the physical reason for using the partial pressure. It seems like the total mixture pressure would be relevant in determining whether the CO2 molecules sublimate or condense. For air and CO2 ice, is the partial pressure of the air irrelevant in determining the behavior of the CO2? $\endgroup$
    – nwsteg
    Commented Jul 10 at 1:24
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    $\begingroup$ Ah! The CO2-water mix does not behave as an ideal gas where there is liquid water. Consider the seltzer bottle... As mentioned above, it applies only to gases, and no gas is ideal, anyway. $\endgroup$ Commented Jul 10 at 1:30
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    $\begingroup$ @nwsteg Total pressure is mostly irrelevant (not 100%, but close enough for most purposes) to the gas/liquid or gas/solid equilibrium. However, it may affect the rate at which equilibrium is reached. This is more familiar with water. If a glass of water is placed in a 20 degree C room with dry air (low partial pressure of water vapor), that water will slowly evaporate until it's all in the vapor phase -- that is the equilibrium state. In a near-vacuum with the same low partial pressure of water vapor, the same water will boil and disappear much more quickly (if its temperature is kept at 20). $\endgroup$
    – anon
    Commented Jul 10 at 1:40

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