4
$\begingroup$

we know that internal energy of the system is defined in terms of temperature as $(3/2)kT$. so if temperature is zero so internal energy is zero. and that means that particle will not have much kinetic energy. so is the entropy zero, system may not move to different microstates?

$\endgroup$
  • 1
    $\begingroup$ My intuition says yes. Since entropy is defined by boltzman constant times the logarithm of number of different configurations the system can exist in, i.e. $S=k_b\ln(\Omega)$. Hence, if the temperature is zero, then it means that every particle is in its lowest energy state, and that it therefore only exist one available configuration. But it should be noted that if we have an ideal gas enclosed in a finite volume $V$, then the volume adds to the entropy since we can place the particles at different positions in the volume. $\endgroup$ – Turbotanten Apr 17 '18 at 19:16
  • $\begingroup$ @Turbotanten This is incorrect. The only systems which have zero entropy at zero temperature are those with a non-degenerate ground state (i.e. "perfect crystals," as per the Third Law of Thermodynamics). If there exists a degenerate ground state (as is true of any system of fermions, for example), then there is a nonzero entropy associated with the multiple possible ground states. So the question we have to ask is: what is the ground state of this ideal gas? This is not typically well-defined. $\endgroup$ – probably_someone Apr 17 '18 at 19:33
  • $\begingroup$ Doesn't the ideal gas have zero volume at 0K? $\endgroup$ – PM 2Ring Apr 17 '18 at 22:52
  • $\begingroup$ @PM2Ring If you cool it down at constant pressure, then yes. But now you've got a very tricky situation, in which you have a bunch of particles of zero radius that occupy the same point, all of which are stationary, but because of the nature of the limit, these stationary, infinitely-close zero-radius particles exert a finite pressure on whatever their container is. Even in this case, there is nonzero entropy at absolute zero, because taking the limit requires acknowledging that there are many different infinitesimal momentum-space configurations that give the same pressure. $\endgroup$ – probably_someone Apr 18 '18 at 2:57
  • $\begingroup$ @PM2Ring But you can see why I stuck to the constant-volume case, since the above is profoundly unintuitive. $\endgroup$ – probably_someone Apr 18 '18 at 2:58
2
$\begingroup$

At zero temperature, a system must be in its ground state. By the Third Law of Thermodynamics, if there is only one possible non-degenerate ground state (i.e. the object is a "perfect crystal"), then the entropy is zero at zero temperature, because there is only one possible configuration for the system to adopt.

This is manifestly not true in an ideal gas. One of the assumptions of an ideal gas is that there are no interactions whatsoever between individual particles, except for collisions. Therefore, any possible spatial arrangement of particles with zero velocity would have the same internal energy, since there are no interactions to favor one arrangement over another. As such, there are a multitude of possible ground states for the ideal gas, which means that its entropy is nonzero at zero temperature.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ well its true that kinetic energy of particles be cannot be completely zero at T=0 but if we say that it is zero then even if there are different positions possible, how can ideal gas particles change its position. (since is there is no movement,no "randomness").As collisions in case of ideal gas are necessary for the particles to change positions between different states., and acquire different microstates. $\endgroup$ – Kritika Apr 18 '18 at 1:49
  • 2
    $\begingroup$ @tiffany The argument above assumed that the kinetic energy of all the gas particles is zero. This is the lowest-energy state for an ideal gas. The fact that ideal gas particles don't change their position at zero temperature is irrelevant, because that's not how entropy is defined. Entropy is the number of possible configurations of a system; at zero temperature, this is equal to the number of lowest-energy states of the system. By the above, there are many such possible states (any given sample of gas adopts only one at a time), so the entropy is nonzero. $\endgroup$ – probably_someone Apr 18 '18 at 1:56
  • $\begingroup$ so just to be clear, entropy is not randomness.? will ideal gas be able to switch among different states at T=0 $\endgroup$ – Kritika Apr 18 '18 at 2:24
  • $\begingroup$ @tiffany It is not. It can sometimes, in a qualitative, imprecise, and colloquial sense, be thought of as randomness or disorder, but that won't get you very far in extreme conditions. In this case, you can sort of cast this result in terms of randomness in the following way: if a system randomly adopts a single ground-state configuration upon cooling to absolute zero, then the entropy at absolute zero is the randomness you would get by cooling a bunch of separate, originally-identical systems down to absolute zero. $\endgroup$ – probably_someone Apr 18 '18 at 2:28
  • $\begingroup$ @tiffany The ideal gas will not be able to switch between different ground states at absolute zero. It will adopt one of the possible ground states, and it will stay in that ground state. $\endgroup$ – probably_someone Apr 18 '18 at 2:29

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.