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Energy of particles in three phases of matter

  1. Oct 17, 2015 #1
    This is a restatement of a question I have been pondering whilst helping my daughter with GCSE physics. My previous post didn't get me the closure I was after so am trying again.

    I have three different GCSE text books.

    All three have a simple section on the kinetic theory of particles in solids vs liquids vs gases describing how the the particles move differently in all three - and that increase in temp increases their movement/kinetic energy.

    All three get across that temperature is a measure of the average kinetic energy of the particles in the solid or liquid or gas (as distinguished from the total kinetic/thermal energy in any given "body" of matter).

    All three include the "key" statement to the effect that in general the particles in a liquid are more energetic than the particles in solid and the particles in a gas are the most energetic. The form of energy is not stated as such, but the statements appear in the section on kinetic theory of matter so perhaps it is fair to assume it is kinetic energy only that is being referred to.

    Now, I am trying to understand what this last statement is intended to mean. Do you you think:

    1. it must be referring only to three states of the same matter - even though not explicitly stated ? It is easy to see that this must be true as you progress from ice to water to steam for instance. Or,

    2. do you think this statement is intended to hold true more generally as seems to be the meaning on face value, ie suggesting in general that the particles of any gas will be more energetic than the particles of any liquid (even of a different substance) which will in turn be more energetic than the partcles of any solid (even of another differance substance). Interpreted in this way the statement would suggest for example that the particles in the air in my room are more energetic than the particles in my room temperature glass of water, which are in turn more energetic than the particles in my room temperature paperweight.

    What bothers me is if only the first more strict interpretation is intended it would have been very easy for this to have been stated explicitly - but it isn't in any of the three books. If this omission implies the broader second interpretation then how does this sit with the definition of temperature - which seems clearly to say that if the air, water and paperweight in my room are all at the same temperature then their particles (on average) are all equally energetic.

    I have spent literally hours "Googling" this but cannot get clarity - i have found articles that include the same unqualified statement, and a smaller number that do seem restricted to the narrower interpretation but non which clearly discounts the broader interpretation (accepting that I have not ploughed through the more complex papers which might address this in a way not apparent to me).

    Hope someone can help.
  2. jcsd
  3. Oct 18, 2015 #2


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    Staff: Mentor

    I believe it's case number 1. I have a hard time believing that gaseous helium at 5 kelvin has more average kinetic energy than liquid water at 300 k.
  4. Oct 18, 2015 #3
    That does seem a good example. Though I have to say when I have been pondering the issue I focused mainly on solids vs liquids because I need to refresh my understanding of how gases behave because there is something here which puzzles me and probably shouldn't. Namely if I allow a gas to expand its temperature will drop yet it doesn't seem that the partcles will loose kinetic energy in the expansion. But this would then seem to contradict the definition of temperature as average kinetic energy of the partcles so there is clearly something important I am missing here.
  5. Oct 18, 2015 #4
    The temperature will drop if the gas does some work on the environment. For example on a moving piston.
    This will account for the transfer of kinetic energy from gas molecules to the outer medium.

    Of course, the temperature may decrease too if there is a heat transfer simultaneous with the expansion.
  6. Oct 18, 2015 #5


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    Staff: Mentor

    Indeed. Let's look at the combined gas law: P1V1/T1 = P2V2/T2
    Where P is pressure, V is volume, and T is temperature. The numbers are for our before and after states.

    Let's say we have a gas at a pressure of 760 torr, that occupies a volume of 1 liter, and is at a temperature of 300 kelvin. If we then double the volume without allowing the gas to perform work, then the new volume is 2 liters, the temperature is still 300 k, and the pressure falls to 380 torr.
  7. Oct 20, 2015 #6
    Thanks guys, very helpful and much appreciated.
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