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Let's say $\ce{CO2}$ and $\ce{C2H6}$ are both in gas form, not vapour. Now which one will be easier to liquify? My take is that the molecule of $\ce{C2H6}$ should be much larger than $\ce{CO2}$ and hence have greater van der Waals forces. And also the critical temperature of $\ce{C2H6}$ is higher than that of $\ce{CO2}$, so it should be relatively easy to make $\ce{C2H6}$ vapour. So I think the answer should be $\ce{C2H6}$.

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    $\begingroup$ Note that in context of chemistry, vapour is gaseous phase of substances considered liquid or solid at normal conditions. So gas vapour does not make sense for CO2 or ethane. $\endgroup$ – Poutnik Feb 22 at 10:37
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You can liquify ethane at standard pressure (1 bar) simply by lowering the temperature (cooling), as indicated by its phase diagram (see for instance here). On the other hand, if you look at the phase diagram of $\ce{CO2}$ you find that starting at standard pressure (1 bar) and room temperature, it cannot be cooled into the liquid state. Rather it will solidify upon encountering the transition temperature (-78.464 °C, see e.g. the wikipedia) and become dry ice. In order to liquify carbon dioxide you need to compress the gas above its triple point pressure (518 kPa) before cooling. This makes the process of liquifying $\ce{CO2}$ somewhat more complicated.

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