Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: G O D I V A on November 12, 2009, 09:07:44 AM
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Why does ΔU = 0 for an isothermal process? Is it still equal to 0 is it is isothermal expansion?
I understand why ΔH = 0 for isothermal, but I can't work out why it is for U by using the same reasoning.
If ΔU = q + w then surely for expansion, since a perfect gas's pressure and volume are not taken into account w = 0. Therefore ΔU = q, and isothermal means that ΔT = 0, not q = 0, so shouldn't ΔU = q, (or why does q = 0)
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U represents the internal energy of a gas. The contributions to U are the kinetic energy of the gas molecules and the energy from interactions between particles. Because ideal gases, by definition, do not have any intermolecular interactions, U depends solely on the kinetic energy of the gas. Since temperature is a measure of the kinetic energy of the gas, we say that U depends solely on the temperature of the gas. If the temperature of the gas does not change, it is neither gaining nor losing any kinetic energy, so it's internal energy must remain the same.
Qualitatively, we can express the relationship between internal energy and temperature as:
dU = CvdT
where
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.forkosh.dreamhost.com%2Fmimetex.cgi%3F%7B+C_v+%3D+%5Cleft%28%5Cfrac%7B%5Cpartial+U%7D%7B%5Cpartial+T%7D%5Cright%29+_v+%7D&hash=ceb90ee256ec1a8ab30361ec78da45e9a07759db)
Also, if you understand that ΔH = 0 for an isothermal process, what will Δ(PV) be for an ideal gas undergoing an isothermal process? Therefore, what must ΔU be?
Finally,If ΔU = q + w then surely for expansion, since a perfect gas's pressure and volume are not taken into account w = 0.
This statement is not true. Ideal gases have pressure and volume (perhaps you are confused by the assumption that the gas molecules of an ideal gas are assumed to have no volume. This does not mean that the gas itself has no volume). Unless the gas is expanding into a vacuum, w < 0 for any expansion.
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U is only dependent on temperature, regardless of the ideal gas. There is a fairly easy, but long derivation you can do to get it using basic physics with an integration thrown in, which I actually started typing out, but it would take way too long to finish, especially in this format. If you play around with the ideal gas equation, using Ygg's suggestion, you should get it straight. But if you're really interested, I could upload some notes explaining it.
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:delta:U = q + w =0 . So q = -w. So does it mean that all the heat is converted into work ???
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What about a reaction?
for example:
H2O(l) :rarrow: H2O(g)
this process is isothermal, but :delta: U isn't zero.
Is there any reaction with :delta: U = 0 ?