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Topic: Vapor pressure x temperature graph  (Read 1883 times)

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Offline INeedSerotonin

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Vapor pressure x temperature graph
« on: October 31, 2019, 01:38:35 PM »


Hello

I have to tell which of these (among A, B, C) are liquids and which are gases at 80ºC e 600 mmHg. I tried converting it to 760 mmHg, because I already have that line on the graph, and that would be easier for me to understand. After all, the liquids only become gas when their vapor pressure is equal to the atmospheric pressure, i.e., 760 mm Hg.

Using Gay-Lussac's Law:

P / T = P' / T'

600/80 = 760/x
600.x = 80.760
x = 80.760 / 600 = 101.33 °C

So I think that, at 80°C and 600 mmHg (which is the same as 101.33°C and 760 mmHg), all of them--A, B, and C-- are gases.

The thing is, the answer is "only A and B are gases; C is still liquid." I don't understand. Is my book wrong?

Thank you

Offline mjc123

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Re: Vapor pressure x temperature graph
« Reply #1 on: October 31, 2019, 02:00:13 PM »
You can't use G-L law. It describes how the pressure of a gas varies with temperature. It does not describe how the vapour pressure of a liquid varies with temperature. (And if it did, you would have to use absolute temperature, not °C.)
80°C and 600 mm Hg is not equivalent to 101.33°C and 760 mm Hg.
80°C and 600 mm Hg means that the "atmospheric" pressure applied to the liquids in this case is 600 mm Hg, so they will boil at this pressure, not at 760 mm Hg.
All you have to do is draw a line on the graph at p = 600 mm Hg. Is all this rigmarole because you were too lazy to do that?

Offline INeedSerotonin

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Re: Vapor pressure x temperature graph
« Reply #2 on: October 31, 2019, 02:12:53 PM »
Thank you!

Yeah, I tend to overcomplicate things. Why make things easier when we can make them more difficult, right? Right?

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