Behavior of atom's wave packets in a gas It is my understanding that the wave packet of a free localized particle spreads with time. My question is what is the best description of the particles in a gas inside a closed container:
Do they look like localized packets? if so, what prevents them to spread? 
Or is each particle in a gas better described as a delocalized entity spread across most of the container? 
 A: First, the geometric extent of the (quantum mechanical) wave packet doesn't mean that the particle (atom in your case) has become diluted all over the volume like if it were fog. 
Instead, the right interpretation of the wave packet is in terms of probabilities. For example, assume that there is 1 atom in a box. The initial wave packet is nearly localized but the packet spreads as a function of time. 
It doesn't mean that each cubic inch of the box contains a part of the atom. Instead, only one place of the box contains the atom and the rest is exactly empty – and the spread wave packet only means that we don't know where the atom is. It has some probability density to be here or to be there – but not both at the same moment.
If one shines light into the box, we may observe where the reflected light comes from. In that way, we may determine where the atom was located. Once the location is observed, e.g. with a light bulb, the wave packet shrinks or "collapses" to one of the places. The wave function for the atom becomes localized again.
Also, it's important to realize that in general, $N$ different atoms are not described by $N$ functions $\psi(x,y,z)$ in 3D space. Instead, they are described by 1 function in the $3N$-dimensional space, $\psi(\vec r_1, \vec r_2,\dots, \vec r_N)$. For any arrangement of the position vectors $\vec r_1,\vec r_2,\dots ,\vec r_N$ of the $N$ atoms, there exists a probability amplitude $\psi$ – and a probability density $|\psi|^2$ – that the particles' positions are near these values.
When you allow the atoms of an ideal gas to move freely for a long enough time, the probability densities for positions and velocities to be $\vec r_i,\vec v_i$ approach the so-called "Maxwell-Boltzmann distribution", basically
$$ \rho \sim \exp\left[\frac{ -\sum_j \frac{m_j|v_j|^2}{2m} - U(\vec r_j) }{kT} \right] $$
where $k$ is the Boltzmann's constant, $T$ is the absolute temperature in kelvins, $U(\vec r_j)$ are potential energies for the $j$-th particle, $m_j$ is the mass, the terms in the exponential are the kinetic and potential energy, and so on.
Note that this particular formula for $\rho$ can be written as a product over $j$ - product of probability densities for individual particles. This is equivalent to the "independent properties" of the particles (probabilities of independent conditions are products of their probabilities). In general, the probability density doesn't have to be a product because correlations between particles may arise. The same is true in quantum mechanics when we write the probability density $\psi$ instead of $\rho$. The independent particles' wave function $\psi$ is a product of wave functions that only depend on the $\vec r_j$ separated variables. But whenever the particles interact and get correlated in some way, the independent breaks down and the wave function has a more general form than a simple product.
Note that this probability distribution is nonzero "everywhere" in the box and for all finite values of the velocities. But again, it doesn't mean that the particles have physically "spread" over the whole box like butter. It only means that our knowledge about their precise positions has become uncertain. In the case of quantum mechanics, much of the ignorance is unavoidable – no observed and not even "God" can know better. But once the position or velocity of a particle is measured, the wave packet always collapses and the measured quantity becomes well-defined immediately after the measurement.
