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Isothermal process

  1. Jul 23, 2010 #1
    I understand that an isothermal process occurs at constant Temperature. And for an ideal gas, the the internal energy is a function of temperature only. Therefore, when Temp = 0, then U = 0 also, meaning the internal energy of the gas doesn't change... and we have q = - w or basically q = PΔV. All of the heat added to the system is used to do work.

    However, I read that Temperature = internal energy/ volume. When the gas does work, the volume increases. But if the volume decreases, it would affect the above equation and thusly would affect the temperature, so then how is the temperature constant?

  2. jcsd
  3. Jul 24, 2010 #2
    Can you cite the source you found this: "Temperature = internal energy/ volume"? It's odd to me. Besides, the equation does not satisfy the balance of dimensions of both sides.
  4. Jul 24, 2010 #3
    exam krackers haha. they didn't write it out as an equation but they said temperature was internal energy per volume
  5. Jul 24, 2010 #4
    Then it's probably unreliable. You can see that if internal energy = temperature * volume, then temperature must have the same dimension as pressure, which is impossible.
  6. Jul 24, 2010 #5
    Forget that comment that "temperature is internal energy per volume", which is wrong. Think of temperature as an indicator of the AVERAGE internal energy of a particle, that is, the total internal energy for all N particles in the sample, divided by N.

    Also, it's CHANGES in internal energy that are really meaningful, not internal energy itself. In other words, if translational kinetic energy changes typically, and in some cases rotational kinetic energy and vibrational potential energy may also exist and may change, then they contribute to the changes in internal energy, But if some forms of internal energy never change at all, say, nuclear bond energy, then those forms of internal energy can be neglected as if they didn't even exist. So a value for U is an ambiguous idea, but delta U is a definite idea. There is limited meaning in saying that T represents U, but a lot of meaning in saying that delta T represents delta U.


    If you are given that volume is decreasing and temperature is constant, then heat is being exchanged. The equation requires that, if delta U is zero and delta W is nonzero, then delta Q is nonzero. Heat is crossing the system boundary.
  7. Jul 24, 2010 #6
    thanks guys it makes sense i think... i probably misread it in the book or got confused during my studying.
    What about this statement:
    "The same amount of the same substance can have the same amount of energy and be at different temperatures."

    Is it because of the potential energy that exists... some substances may have more potential energy than others? yet temperature is only kinetic energy right? so thats why they can have the same energy yet different temperatures?
  8. Jul 25, 2010 #7
    I guess the statement means that there are many kinds of energy, one of which is microscopic kinetic energy, which is represented by temperature. For example, gas in a moving chamber has both microscopic and macroscopic kinetic energies, not to mention potential energy if it is not ideal gas or placed in some external field.
  9. Jul 25, 2010 #8
    It that an exact quotation, and the entire statement provided, or did they say more about it? The meaning is not clear to me.

    A principle called equipartition says that the average atom or molecule will store internal energy in all of the ways that are possible for it to store internal energy, and it will distribute that stored energy equally among all of these possible ways. For a monoatomic gas such as He, translational kinetic energy is the only way for the atom to store internal energy. For a diatomic gas (two atoms in a molecule -- H, N, O, F, Cl, Br and I) it also has rotational kinetic energy. For polyatomic gases (three or more atoms in a molecule) it also has a form of electrical potential energy in atomic bond vibrations. The temperature has to indicate the total of all of these.
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