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Topic: Internal Energy and Temperature  (Read 2781 times)

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Offline galpinj

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Internal Energy and Temperature
« on: February 01, 2016, 10:52:51 PM »
Hey guys,

So I've been trying to understand why ΔU equals zero when ΔT is zero. I know that U = q + w, and that U =(3/2)PV or U = (3/2)nRT. I can't figure out how these equations would prove that a zero change it temperature automatically makes ΔU zero.

For example, if a system had a change in pressure and volume, this could represent a change in internal energy, no? In both U = q + w and U =(3/2)PV, it appears that internal energy would change regardless of temperature. I can't consolidate this difference.

Please help

Offline Borek

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Re: Internal Energy and Temperature
« Reply #1 on: February 02, 2016, 03:40:52 AM »
So I've been trying to understand why ΔU equals zero when ΔT is zero.

I can be wrong, but I fail to see why this should be true in general. There is a reason why the heat related to phase changes is called a "latent heat" - it is absorbed or released without a temperature change.
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Offline Burner

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Re: Internal Energy and Temperature
« Reply #2 on: February 02, 2016, 06:18:53 AM »
In general, Internal energy=Total potential energy+Total kinetic energy (Only studied internal energy in physics briefly)

Temperature affects kinetic energy of substances mostly. But potential energy can change without changing temperature.
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Offline Enthalpy

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Re: Internal Energy and Temperature
« Reply #3 on: February 02, 2016, 10:32:25 AM »
Pay great attention to when something applies. For instance, (3/2)nRT is for monoatomic perfect gases only. Take nitrogen, usually "perfect" but diatomic, it's already wrong. Take liquid xenon, it's wrong. Take any boiling compound, it's wrong.

Here you have a case in thermodynamics, but this is one of the most frequent misunderstandings in any area.

So, under which assumptions would the resulting ΔU have to be zero?

Offline galpinj

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Re: Internal Energy and Temperature
« Reply #4 on: February 02, 2016, 06:59:38 PM »
Pay great attention to when something applies. For instance, (3/2)nRT is for monoatomic perfect gases only. Take nitrogen, usually "perfect" but diatomic, it's already wrong. Take liquid xenon, it's wrong. Take any boiling compound, it's wrong.

Here you have a case in thermodynamics, but this is one of the most frequent misunderstandings in any area.

So, under which assumptions would the resulting ΔU have to be zero?

That's exactly right, and I apologize for not specifying earlier. This is when looking at a monoatomic ideal gas, which allows us to look at internal energy as essentially a measure of kinetic energy (PE can, for all intensive purposes, be ignored). Under these circumstances, can you help explain why a ΔT of zero will always make ΔU zero?

Offline Enthalpy

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Re: Internal Energy and Temperature
« Reply #5 on: February 03, 2016, 05:36:27 PM »
U = (3/2)nRT then.

More in depth:

"Perfect gas" means that the attraction, repulsion and so on between the molecules are negligible. Then, the density that makes interactions more frequent plays no role, so U, H and others depend only on T.

In the case of a perfect monoatomic gas, everything boils down to language in fact. Temperature being the energy per degree of freedom (measured in Kelvin instead of Joule for historical reasons only) and monoatomic perfect gases having just the three translations, U = (3/2)nRT only means "the temperature is the temperature".

Polyatomic perfect gases add only degrees of freedom, for instance two rotations but no vibration for nitrogen under usual conditions, so that U becomes (5/2)nRT instead of (3/2) for the three translations.

Translations are practically not quantized hence store (1/2)nRT each. Rotations are so finely quantized that usually the molecules rotate freely (cryogenic hydrogen is an exception) and store (1/2)nRT each too. Vibrations are less simple:
  • Each stores nRT if fully excited, not (1/2)nRT
  • They are more coarsely quantized, hence get progressively excited when the temperature suffices.
  • N2, H2 don't vibrate at room temperature, but heavier and less stiff Br2 and CO2 do to some extent, and the number of modes increases with the number of atoms.
  • So these vibrating molecules often show a heat capacity that increases with the temperature.
  • Metals are an extreme case, because 1023 atoms make the vibrations modes very finely quantized hence completely excited at room temperature. The crystal having 3 times as many vibration modes as atoms gives then a constant heat capacity of 3*nRT.
  • All this still gives U depending only on T, but for instance if the gas' density approaches the liquid's one, interactions between the molecules let the energy depend on the density too.
« Last Edit: February 03, 2016, 06:05:09 PM by Enthalpy »

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