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What does it mean when we say that average KE is equal to temperature?

  1. Aug 13, 2014 #1
    I know that delta h(Enthalpy) is heat and it is also the total kinetic energy of the system.

    Temperature is the average kinetic energy of the system from enthalpy.

    For example, a bathtub may have the same temperature as the ocean, but the total enthalpy will be greater for the ocean.

    How do we measure the average kinetic energy of heat to get temperature?

    Is it every atom's movement(rotational bonds, friction) of kinetic energy divided by the number of each individual movements, or is it divided by time?
     
  2. jcsd
  3. Aug 14, 2014 #2

    Andrew Mason

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    Not quite. The total energy of the system is U which consists of internal potential energy as well as internal kinetic energy.

    Since H = U + PV:

    ΔH = ΔU + ∫PdV + ∫VdP

    If there is no change in pressure (e.g. a process taking place at atmospheric pressure) then ΔH = ΔU + ∫PdV = Q. So, for constant pressure processes, ΔH = Q = the heat or heat flow during the process.

    I am not sure what this means. But except for extremely low temperatures, temperature is the average translational kinetic energy of the particles in the system as measured from the frame of reference of the centre of mass of the particle system.

    Temperature is usually measured indirectly (eg. a thermometer) rather than computing the average translational kinetic energy of the molecules in a substance.

    AM
     
  4. Aug 14, 2014 #3

    td21

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    Sorry. I am quite confused by this equality. So temperature in Kelvin = the K.E. of the particles in J?
     
  5. Aug 14, 2014 #4

    Andrew Mason

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    The units of temperature are related to the units of energy by:

    Average KE per particle of mass m = <KE> = <mv2/2> = 3kT/2 where k is the Boltzmann constant 1.3806503 × 10-23 J/K

    AM
     
  6. Aug 14, 2014 #5
    Hello Andrew

    This has been bugging me for a while . Why do we relate temperature to only average translational kinetic energy of the molecules , and not to the internal energy of the molecules ? In case of monoatomic molecules ,it makes sense as the internal energy comprises only translational KE .

    But what if we consider diatomic molecules. There we have rotational energy as well .The RKE also depends on the temperature .

    So why do we say that temperature is related to only translational KE ?
     
  7. Aug 15, 2014 #6

    Andrew Mason

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    The thermodynamic concept of temperature is tied to entropy: does heat flow from one body to another? If so, they have different temperatures ie. the body from which heat flows has a higher temperature than the one to which the heat flows.

    Experiment shows that bodies with more internal kinetic energy per molecule (due to molecules having more degrees of freedom) but the same translational kinetic energy as another, do not produce heat flow when put in thermal contact.

    Experiment shows that given two bodies of particles, A and B, with the same number of particles, if A's particles have 5 degrees of freedom and B's particles have 3, more heat will flow from an infinite reservoir at temperature T to A than to B (the flow continues so long as the temperature of A or B < T).

    This is because of the equipartition of energy. The relationship between heat flow and temperature change (heat capacity) of a body depends on the number of degrees of freedom of the constituent molecules of the body.

    AM
     
    Last edited: Aug 15, 2014
  8. Aug 16, 2014 #7
    Hi Andrew,

    Thank you for all your help! I'm getting a better idea of the concept.

    Lastly, how does temperature relate indirectly to average translational kinetic energy. Did we just set a standard and say that if the chemical in the thermometer rises this much, it is due to this much vibrating molecules in the system?

    And thus, we can measure the ocean temperature in a particular region and that amount of vibration can equal a temperature in a bathtub filled with water.
     
  9. Aug 16, 2014 #8

    td21

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    Dear Andrew,

    Sorry I am quite confused. What does internal kinetic energy in this context mean?
    Thank you very much.
     
  10. Aug 16, 2014 #9

    Andrew Mason

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    The concept of temperature has to do with heat flow. If heat flows from body A to body B, A has a higher temperature. If no heat flows, then A and B are in thermal equilibrium (bodies both at the same temperature).

    Thermal equilibrium means that the distribution of energies of the particles follows a Maxwell-Boltzmann distribution.

    Kinetic theory says that the average kinetic energy per particle per degree of freedom for a system of particles in thermal equilibrium at temperature T is kT/2. The Equipartition theorem, which is part of the Kinetic Theory, holds that there is, on average, an equal distribution of energy among all modes (each mode relating to a single degree of freedom). Kinetic theory agrees with experiment most of the time (the exception being where a mode is partially frozen out due to quantum effects).

    In summary, the concept of temperature has to do with whether heat flows between two systems of particles. No one defined temperature to be the average translational kinetic energy of the particles. Rather it is Kinetic theory that explains how the traditional concept of temperature is related to the average translational kinetic energy of the particles. Kinetic theory shows that if you have two systems of particles each with the same average translational kinetic energy per particle, the average kinetic energies per particle will not change (no heat flow) when they are put in thermal contact.

    AM
     
    Last edited: Aug 16, 2014
  11. Aug 16, 2014 #10

    Andrew Mason

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    Internal kinetic energy includes kinetic energies associated with all degrees of freedom, (3 translational degrees + rotational degrees - ie. one degree per axis of rotation + vibrational degrees - ie. one degree for each independent mode of vibration).

    AM
     
  12. Aug 24, 2014 #11
    Thanks .

    Sorry . I didn't understand .

    Isn't it contrary to what you explained before ? A diatomic gas molecule having more DOF has same average translational kinetic energy per molecule at the same temperature , so no heat should flow .

    Please excuse me if I may have misunderstood you.
     
    Last edited: Aug 24, 2014
  13. Aug 24, 2014 #12

    Andrew Mason

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    Heat capacity = Cv = ΔQ/ΔT. So the higher the heat capacity, the more heat flow required to change temperature by a given ΔT. A system, A, of n particles, each of which have 5 degrees of freedom, has a higher heat capacity than a system, B, of n particles each of which have 3 degrees of freedom. So more heat is required to flow to system A than to system B in order to increase each of their temperatures by ΔT.

    All I said is this: For an infinite heat reservoir at temperature T and with systems A and B (above) initially at temperature T-ΔT, more heat will flow from the reservoir to A than to B as equilibrium is reached (i.e. as the temperatures of A and B reach T).

    AM
     
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