# Is entropy zero at temperature zero for ideal gas?

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?

• 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. – Turbotanten Apr 17 '18 at 19:16
• @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. – probably_someone Apr 17 '18 at 19:33
• Doesn't the ideal gas have zero volume at 0K? – PM 2Ring Apr 17 '18 at 22:52
• @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. – probably_someone Apr 18 '18 at 2:57
• @PM2Ring But you can see why I stuck to the constant-volume case, since the above is profoundly unintuitive. – probably_someone Apr 18 '18 at 2:58