Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: themedicalgardener on December 28, 2021, 05:00:45 PM
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I'm studying from the Kaplan book and there is an excerpt that states "[...] as the temperature of a gas is reduced towards its condensation point (which is the same as its boiling point), intermolecular attractions cause the gas to have a smaller volume that that which would be predicted by the ideal gas law."
I already understand why the volume would be low according to the ideal gas law. V=nRT/P. A decrease in T (or even n) would cause a decrease in V.
BUT, I was watching a Khan Academy video: https://www.youtube.com/watch?v=UABFOI1sb7A where he basically said assuming a real gas scenario where the temperature has been reduced to the condensation point but the volume remains large, we would expect a lower pressure. He then goes to say that the ideal gas law would anticipate a low pressure already, but considering the increase in intermolecular forces between the gas particles, they are less likely to hit the walls of the container (and even if they did, it would be done with less vigor) and thus pressure would be even lower than predicted from the ideal gas law.
So now I'm confused (or I've confused myself). I thought low temperatures led to a smaller volume (given intermolecular forces) which meant higher pressure, since volume and pressure are inversely related? Is the Khan Academy example an exception since temperature is low BUT volume is still large????
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I thought low temperatures led to a smaller volume (given intermolecular forces) which meant higher pressure, since volume and pressure are inversely related?
That is not how it works. Lowering the temperature makes the volume smaller at a constant pressure.