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B Energy of an ideal gas

  1. Dec 18, 2016 #1

    When calculating the energy of an ideal gas we neglect the potential energy and calculate the kinetic energy using:

    K.E = 3 /2 n R T

    My question is why do we not consider the electrostatic energy of the gas?

    If I am trying to work out the internal energy of 1 mol of Radon, why do I only require the kinetic energy and not the energy that was required to bring the electrons and protons together? Is this energy not held within the gas?
  2. jcsd
  3. Dec 18, 2016 #2


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    That's the meaning of ideal gas, a gas whose particles don't interact with each other. Its an approximation! Of course you can consider interactions and have a more precise model, but that means you should do harder calculations.
  4. Dec 18, 2016 #3


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    If your gas is so hot that electrons can be removed from atoms (=your gas is actual a plasma), you cannot approximate it as ideal gas any more.
  5. Dec 19, 2016 #4
    When snooker balls collide it is not necessary to consider the "electrostatic energy" of the snooker balls..molecules are considered to be 'snooker balls' in the simple kinetic theory of gases
  6. Dec 20, 2016 #5


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    Be aware that this formula is only valid for a monatomic ideal gas. It is not valid if the gas molecules have 2 or more atoms -- e.g. O2, N2, CO2, etc.

    I've personally known physics teachers who were unaware of this.
  7. Dec 21, 2016 #6
    oops....should perhaps have qualified 'translational kinetic energy'
    this is more or less implied among physicists when the model is based on snooker balls, as far as I know there are no diatomic snooker balls.
    Such a simple model has many limitations but is surprisingly informative.
    The model also breaks down at high pressures and low temperatures, I imagine most physics teachers would be aware of this or at least suspect it.
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