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Internal energy and state functions

  1. Dec 27, 2011 #1
    Hello all,
    I have following questions dangling in my mind, with a positive attitude of seeking help i want put them infront you geneus people, in my earlier post i get a good result and now i want to more insight of the very matter. kindly help me.
    Using the concept of state of system, i understand that the state represent the condition of the system which it has. Further some set minimum values of state functions are required to specify the full discription abouy the system. Am I right?
    I now want to know following for crystal clear understanding. THE FOLLOWING IS TRUE FOR ALL IDEAL GASES.
    We know
    two similar systems having same volume, may have different temperature, and(PV= nRT)
    two similar systems having same pressure, may have different temperature,
    1.CAN TWO SIMILLAR SYSTEMS HAVING SAME INTERNAL ENERGY, MAY HAVE DIFFERENT TEMPERATURE?
    2.IS IT POSSIBLE FOR TWO SIMILAR SYSTEMS AT SAME STATE ( THEY SURELY WERE AT SAME TEMPERATURE, ISN'T) TO HAVE DIFFERENT INTERNAL ENERGY?
    3.IS IT POSSIBLE FOE ANY PROCESS (TO OCCUR) WHERE THE INITIAL AND FINAL STATES ARE DIFFERENT THOUGH THE CHANGE IN INTERNAL INERGY IS ZERO?
    I expect the answer of last question in affermative just as two at two different states a system may have same temperature it may also have same internal energy, well i want response dor all my questions so that things may get fully clearified,
    THANKS A BUNCH:) AND REGARDS!!
     
  2. jcsd
  3. Dec 27, 2011 #2
    1. By definition of an ideal gas, the internal energy is independent of volume and dependent on Temperature. This is usually visualized as there is no interaction between the particles until the collide with each other. The internal energy is sum of internal KE and internal PE. The internal KE, by the definition of Temperature, is just another way of calling what Temperature is! Hence, the answer is NO if you don't consider any potential field being applied on the gas. The answer could turn out to be YES if there is an external field such that it exactly balances the change of Temperature from one state to the other.

    2. Internal Energy is a state function, i.e. for a given state there is exactly one (no more and no less) value for Internal Energy for a particular state. Hence, answer is NO.

    3. Yes, it is perfectly possible as mentioned in 1.
     
  4. Dec 27, 2011 #3
    THANKS! Oh this comes out to be what I was started assuming last night afrer getting the concept of state from my earlier post through simon bridge.. What I now understood is that at a PARTICULAR STATE all the state functions have a UNIQUE VALUE(for each) that gives the discription of the condition the system is in.This does however not mean that two system at DIFFERENT STATE can't have some of its state functions THE SAME.
    The relation berween the state and its properties(state functions) is like many-one with the condition that two different states can't have all the state functions the same (otherwise they will be said to be at same state). Is the interpretation true?
     
  5. Dec 30, 2011 #4
    I disagree on numbers one and two. 1. Kinetic gas theory considers gas temperatures to be a function solely of the kinetic energy of translation. Internal energies play no role in gas temperatures.

    2. Monatomic gases possess both translational KE and temperature. They have zero internal KE at all temperatures. Other gases have internal energies in keeping with their quantum states, and these states vary from one gas to another at the same temperature (keep in mind that gas temperatures are averages of individual molecular characteristics). Furthermore, a diatomic gas and a polyatomic gas will have different internal energies at the same gas temperature.

    Keep in mind that classical physics does not always apply on the molecular level. That is why the Physics Forum places both molecular and quantum matters in a different category.
     
  6. Dec 30, 2011 #5
    Addendum to my post #4: At temperature T, Helium will have an internal KE of 0, Oxygen will have an internal KE of 1kT (two degrees of rotation, no vibration is possible), and H2S will have an internal KE of roughly 3/2 kT (three degrees of rotation plus whatever vibration might occur). These values assume that all available modes are excited, and this is not always the case. For diatomic and polyatomic molecules, theory and observation do not always agree.
     
    Last edited: Dec 30, 2011
  7. Dec 30, 2011 #6

    Philip Wood

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    Kilmatos is using the term 'internal energy' in a non-standard way. He is using it to mean non-translational molecular energy.

    When most writers (including, I think, previous posters on this thread) use the term, they mean the sum of all forms of molecular KE and intermolecular PE for the system; the latter being zero for an ideal gas, but the former including translational KE. Thus the internal energy of an ideal monatomic gas of N molecules of mass m and rms speed crms is U = (3/2) N m crms2.
     
  8. Dec 30, 2011 #7
    To me, internal energy includes all sorts of of energy contained in a system (including potential energy). In the kinetic theory of gasses temperature is defined using ideal gasses, which have no internal energy other than (kinetic) translational energy. So in this case, the answer to the first question is no. However, if you consider molecules to be complex quantum mechanical systems, then there's certainly some potential energy that composes the internal energy. In that case, if you take two different gasses, the average kinetic energy (temperature) contained in in each of them might be different, even if the have the same internal energy. So the answer might be yes (does this make sense)?
     
  9. Dec 31, 2011 #8
    1) True, and I apologize for not stating this non-standard use in advance. However, as PW noted, it is obvious from my posting. Referring to KE of translation as external energy and KE of rotation and vibration/libration as internal energy is not that uncommon. The first can easily be measured (gas temperature), the second can only be calculated.

    2) True only for monatomic gases. As soon as you start to deal with other configurations, the internal energies (conventional usage) will not necessarily be the same at the same gas temperature.

    My physics work is almost entirely in atmospheric physics. The mixture of gases that make up the atmosphere (some 104 of them at the ppb range and higher, including isotopes but not including ions) have a very wide range of internal energies (conventional usage) at the same atmospheric temperature. This is evidenced by the wide range of individual specific heats.
     
  10. Dec 31, 2011 #9
    How so? A real gas' temperature will also depend on what you are referring to as internal KE.
     
  11. Dec 31, 2011 #10
    No, it does not. If it were true, gases with different specific heats (combined translational, rotational, and vibrational/librational kinetic energies per degree) could not have the same temperature. Yet they do.

    The same amount of heat will raise the temperature of one mole of hydrogen far more than one mole of oxygen. Conversely, one mole of oxygen at temperature T will have far more internal energy (your definition) than one mole of hydrogen.

    Temperature is usually defined by the equation T=2u/3k. Here u is the mean kinetic energy of translation along the true paths of the molecules and k is Boltzmann's Constant. Note that neither rotational nor vibrational energies play a role.

    For support, I refer you to almost any good work on the nature of specific heats.
     
  12. Jan 1, 2012 #11
    You're right, I'd completely forgoten about that. Gasses with more DOFs than simple translation will have higher heat capacity.
     
  13. Jan 1, 2012 #12

    Philip Wood

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    With respect, Klimatos, the internal energy of hydrogen and oxygen at 'ordinary' densities and temperatures are both given quite accurately by U = (5/2) n R T, in which n is the number of moles, R is the molar gas constant and T is the kelvin temperature. This formula neglects PE due to intermolecular forces, and allows for 2 rotational degrees of freedom and 3 translational.
     
  14. Jan 1, 2012 #13

    Philip Wood

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    With respect, the internal energy of hydrogen and oxygen at 'ordinary' densities and temperatures are both given quite accurately by U = (5/2) n R T, in which n is the number of moles, R is the molar gas constant and T is the kelvin temperature. This formula neglects PE due to intermolecular forces, and allows for 2 rotational degrees of freedom and 3 translational.
     
  15. Jan 1, 2012 #14
    Does anyone have a good definition of temperature? Because although klimatos' argument was good, I have a hard time believing that T is related only to translational KE and not to every kind of KE in a system.
     
  16. Jan 1, 2012 #15
    Will you take NASA's word for it?

    "The temperature of a gas is a measure of the average translational kinetic energy of the molecules."

    http://www.grc.nasa.gov/WWW/k-12/airplane/temptr.html

    Addendum: Please note that I am referring to gas temperatures only.
     
    Last edited: Jan 1, 2012
  17. Jan 2, 2012 #16

    Rap

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    The exact definition of temperature is not simple because of quantum effects. All degrees of freedom are quantized - the energies are discrete. If the energies of the particles are very great compared to the energy difference between the levels, then you can pretend the energy is continuous, and that degree of freedom is practically "unfrozen". The average contribution to a particle's energy for such a degree of freedom will be kT/2 , where T is the temperature.

    However, if the particles are only occupying the lowest few energy levels, then you cannot pretend that the energy is continuous, and that degree of freedom is starting to become "frozen". The average contribution to a particle's energy for this degree of freedom will be less than kT/2. If practically all of the particles are in their lowest possible energy state, then that degree of freedom is practically "frozen". The average energy contributed per particle is much less than kT/2, practically zero. If you raise the temperature a little, this degree of freedom will gain practically no energy, it is as if it were not even there, thermodynamically speaking.

    If you have a molecule that can vibrate, it has degrees of freedom in addition to translational. For an O2 molecule at room temperature, it has 3 rotational and 2 rotational degrees of freedom and all five are quite unfrozen. It also has vibrational degrees of freedom, but these are frozen. The average energy per particle is 5 times kT/2, and 3k/2 of that is translational. You could say that T=2U/5k where U is the average energy per particle or you could say that T=2u/3k where u is the average translational energy per particle

    As the temperature goes down, the rotational degrees of freedom freeze out, leaving only the translational degrees of freedom. The internal energy is 3 times kT/2, all of which is translational. Again, the temperature is 2u/3k, but it is no longer 5U/2k.

    The translational degrees of freedom also freeze out eventually, but this only happens at VERY low temperatures, near absolute zero. That's why people say that the temperature is given by 2u/3k where u is the average translational energy per particle. All such expressions eventually fail but the translational expression is the very last to go.
     
    Last edited: Jan 2, 2012
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