Is the wavefunction of particles inside a gas spread or localized?

In the context of solid-state physics, a closely related question has been an area of active research in the past few years. Most interacting systems do indeed thermalize (and thus delocalize) over long time scales. However, certain systems whose disorder is much stronger than their interactions experience "Many-Body Localization," in which the individual particles remain "stuck" indefinitely. This has many macroscopic consequences, like lack of electrical conduction (because the electrons aren't free to move) and entanglement entropy that only scales as an area law rather than a volume law for every eigenstate (not just the ground state). There are far too many papers on this topic to list, but the original paper that started it all is Basko, Aleiner, and Altschuler (2006).

A dense gas with relatively strong interactions (by gas standards) might roughly be thought of as a very strongly disordered, weakly interacting solid, so I suspect that just as in the solid-state case, your thought experiment might lead to either a delocalized or a localized state, depending on the details of the density of the gas and the strength of its interactions.

To solve your problem exactly, you would have to solve the Schrödinger equation

$$i frac{partial}{partial t} Psi (vec r_1 dots vec r_N,t)= H Psi(vec r_1 dots vec r_N,t)$$

where $Psi (vec r_1 dots vec r_N,t)$ is the wave function of the $N$ particles and

$$H=sum_i^N frac{p_i^2}{2 m} + sum_{i<j}^N u_{ij}+V_{text{ext}}$$

where $u_{ij}$ is some pair potential and $V_{text{ext}}$ is the box potential. Of course, you also need an initial condition

$$Psi (vec r_1 dots vec r_N,0) = tilde Psi (vec r_1 dots vec r_N)$$

The first thing that you must notice is that it is inappropriate to talk about "the wave function of each particle", because you have to consider the total wave function of the $N$ particles ($Psi$). If the particles are indistinguishable, this function must posses some symmetries, depending on what kind of particles you are considering (bosons or fermions).

It is really difficult to solve such a problem analytically or numerically. A good starting point to get a qualitative idea would be to solve the corresponding equations for two particles and see what happens.

This has been done numerically by the authors of this article using a gaussian and square potentials, with distinguishable and indistinguishable particles. in the latter case, symmetrized (bosonic) and antisymmetrized (fermionic) wave functions were considered:

$$Psi'(x_1,x_2) = frac 1 {sqrt{2}} [Psi(x_1,x_2)pm Psi(x_2,x_1)]$$

The wave function at $t=0$ is assumed to be the product of two gaussian wave packets:

$$Psi(x_1,x_2, t=0) = g(x_1,x_1^0,k_1,sigma) g(x_2,x_2^0,k_2,sigma)$$

Where

$$g(x,x^0,k) = e^{i k x} e^{-frac{(x-x^0)^2}{4 sigma^2}}$$

and $k_2 = - k_1$.

In the following plot you can for example see the result of the collision of two distinguishable particles with equal mass $m$ interacting via a square potential (the curves are the probability densities obtained by the wave functions):

The numbers in the top left corners indicate the time (in units of $100 times $ time step) and the edges of the frames correspond to the walls of the box.

You can see that there is indeed a "spreading" of the wave packets (to be more precise, thier modulus squared, that is, the probability densities) after the collision (cfr. 1 and 120).

Quoting from the article:

In Fig. 5 we show nine frames from the movie of a repulsive m–m collision in which the mean kinetic energy equals one quarter of the barrier height. The initial packets are seen to slow down as they approach each other, with their mutual repulsion narrowing and raising the packets up until the time (50) when they begin to bounce back. The wavepackets at still later times are seen to retain their shape, with a progressive broadening until they collide with the walls and break up.

I suppose you mean that the gas is contained in a magic box. Otherwise the walls become part of the system, exchanging momentum/energy with the 'particles'. I have no answer for you; I don't know. What I do know is that none of the particle-particle collisions can be characterized other than by using a probability distribution. Common sense demands that assuming you know a particles approximate position and momentum at time t=0, and assuming the distribution of other particles in the box is random and they are all at thermal equilibrium with the walls, the position or momentum will be more and more evenly distributed over the volume (phase space) at times t>0. So, you tell me, are you using the particle as the origin of the coordinates of said wave equation, or is the frame of reference fixed by the box? As far as I can see, your question isn't useful and lacks any predictive value. It seems to me your question is about the state of the wave equation in between observations. Which (maybe because I'm a fan of the Copenhagen Interpretation) seems profoundly misguided. You probably can get the particle localized, once (t=0) but then what? How do you observe it a 2nd time? Clearly that 2nd observation will not be at an equivalently localized place. (barring some trap).

Quantum effects appear if the concentration of particles satisfies, $$frac{N}{V} ge n_q$$ where $N$ is the number of particles, $V$ is the volume, and $n_q$ is the quantum concentration, for which the interparticle distance is equal to the thermal de Broglie wavelength, so that the wavefunctions of the particles are barely overlapping. As the quantum concentration depends on temperature, high temperatures will put most systems in the classical limit unless they have a very high density e.g. a White dwarf or the very early Universe.

The quantum nature of the particle manifests itself in that bosons obey Bose–Einstein statistics and fermions obey the Fermi–Dirac statistics. Both Fermi–Dirac and Bose–Einstein statistics are well approximated by the classical Maxwell–Boltzmann statistics at high temperature and low concentration where the quantum effects are negligible.

It really depends on the boundary conditions. For boundary conditions like a 3D box with reflecting walls, the initial quantum state $Psi$ will stay a quantum state with the unique wave function depending on variables of each particle: $$Psi({bf{r}}_1,...,{bf{r}}_n, t).$$ If the boundary conditions are such that allow exchange with the environment, then the density matrix approach may become more appropriate.

In both cases the positions of particles are statistically predicted, but in the latter case there may not be interference phenomenon (or it will be less pronounced).

Consider your case for a double slit experiment without and with particle position measurements, measurements before the screen.

The behaviour of the molecules in your gedanken experiment can be approached by using decoherence. But I do not believe you can get a definitive answer until somebody makes a full scale simulation (or until some expert's answer can make a formal proof of what really happens, but I am not skilled to do that). The decoherence effects can be argued heuristically, but taking that approach the answer I found is ambiguous.

On one hand, a single molecule is not isolated, it interacts with a macroscopic environment (gas, electromagnetic radiation, walls). In most cases this environment will act to reduce the entangled state of as single molecule and make it to apparently collapse into a preferred basis. The prefered basis is usually the position basis, but this is not always the case and depends on the details of the set up.

If the preferred basis is the position basis, then the wavefunction of individual molecules will not spread for too long, its interaction with the rest of the environment in which it is situated will make the probabilities to "collapse" to a smaller packet (as mentioned by P. Shor in the comment, it cannot be a pure eigenstate of the position operator). But if the preferred basis is different, for instance the momentum basis, then the end result will be a largely delocalized wave packet, This second option seems to be more consistent with the fact that once reaching thermodynamic equilibrium, the gas molecules satisfy the Maxwell Boltzmann statistics.