# What is the temperature at which the motion of particles theoretically ceases?

Jul 6, 2017

I think that you will have to tell us that one........

#### Explanation:

Google $\text{temperatura assoluta............}$ or $\text{the Kelvin Scale}$. You might already have come across this scale.

Jul 6, 2017

The temperature at which the motion of particles theoretically ceases is ${0}^{0} A$ or absolute zero, Absolute ${0}^{0}$ is the definition of ${0}^{0}$ on the Kelvin and Rankine temperature scales.

#### Explanation:

Absolute zero computes to $- {459.67}^{0} F$, which is also $- {273.15}^{0} C .$

At this temperature, both the enthalpy (heat content) and entropy (state of randomness or disorder) approach zero. Effectively, the molecules of the gas are slowing down towards being motionless.

Absolute zero also describes a gas reaching a temperature from which no more heat can be removed. Experiments have shown that molecules continue to vibrate at absolute zero.

There are more thoughts on this topic here:
https://www.thoughtco.com/what-is-absolute-zero-604287

Jul 6, 2017

$\text{0 K}$, kelvin. But that is not strictly true, as all molecules vibrate at $\text{0 K}$ at their zero-point energy, $\frac{1}{2} h \nu$.

ATOMIC MOTION AT 0 K

In regards to atoms, yes, the motion of particles will stop at $\text{0 K}$, or $- \text{273.15"^@ "C}$, because average atomic kinetic energy (which for atoms is entirely translational) depends on the temperature and any intermolecular forces present.

When we consider non-interacting atoms in the classical limit, the equipartition theorem gives for the average per-particle kinetic energy:

<< K_(avg,"trans") >> -= K_(avg,"trans")/N = 3/2 k_BT, in $\text{J/particle}$

where the $3$ comes from the three cartesian directions ($x , y , z$), $N$ is the number of particles, ${k}_{B}$ is the Boltzmann constant, and $T$ is the temperature in $\text{K}$.

MOLECULAR MOTION AT 0 K

Molecules, on the other hand, have chemical bonds, which obviously means they can stretch and/or bend. To a first-order approximation, neglecting rotation at $\text{0 K}$ (which is entirely valid), they can be modeled by the simple harmonic oscillator system, with energy

E_(upsilon) = hnu(upsilon + 1/2) = ℏ omega (upsilon + 1/2),

where $\upsilon$ is the vibrational quantum number, $h$ is Planck's constant, and $\nu$ is the fundamental vibrational frequency in ${\text{s}}^{- 1}$.

$\omega$ is the angular frequency, or $2 \pi \nu$, and ℏ = h//2pi is the reduced Planck's constant.

At $\text{0 K}$, everything is in its ground state, and the vibrational ground state has $\upsilon = 0$, so

${E}_{0} = h \nu \left(0 + \frac{1}{2}\right) = \frac{1}{2} h \nu$

and all molecules continue to vibrate in their ground state at $\text{0 K}$.