1
$\begingroup$

Given that these are single atoms, why should the specific heat difference be so great? And more specifically, why does it take 36 times as much energy to raise the temperature of a given mass of Helium compared to Xenon?

$\endgroup$
  • 2
    $\begingroup$ How does it compare on a per-mole basis? $\endgroup$ – Jon Custer Mar 25 '15 at 13:44
1
$\begingroup$

The key question is "Per mole or per gram?" Because the both values can be found tabulated as "specific heat" in various sources. Perhaps it would be useful to distinguish "molar specific heat" from "specific heat per unit mass".

You seem to be using the intuition for the molar quantity, so if the table is by mass, the answer is simply that you need more moles of helium.

$\endgroup$
  • $\begingroup$ By mass. But why should the number of molecules matter? $\endgroup$ – user56903 Mar 25 '15 at 15:58
  • $\begingroup$ Because temperature is related to the average energy per mode, and when there are more particles there are more modes. $\endgroup$ – dmckee --- ex-moderator kitten Mar 25 '15 at 16:35
  • 1
    $\begingroup$ Err ... "Intuitively, temperature equals velocity" Is wrong two ways. First, temperature is proportional to the energy not the velocity, and secondly it is the average energy per mode, not the total (which should be very clear because total energy is extensive and temperature is intensive). So for a change in energy $\Delta E$ the temperature change is proportional to $\Delta E/N$ where $N$ is the number of modes (which reduces to the number of atoms for an ideal gas). $\endgroup$ – dmckee --- ex-moderator kitten Mar 25 '15 at 16:44
  • 1
    $\begingroup$ In a kinetic gas it is the kinetic energy, but that is proportional to velocity squared. For systems that are not non-interacting there are also potential terms. $\endgroup$ – dmckee --- ex-moderator kitten Mar 25 '15 at 16:48
  • 1
    $\begingroup$ In an ideal gas atoms have three degrees of freedom, the $x$, $y$ and $z$ directions, and each DOF gets about $\tfrac{1}{2}kT$ of energy. So the total internal energy is $\tfrac{3}{2}kT$ times the number of atoms. $\endgroup$ – John Rennie Mar 25 '15 at 16:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy