On the other hand, if I know the resulting gas in volume $2V$, I can time reverse so that I know the initial configuration and vice versa.
If you know all positions and velocities of the particles, it is meaningless to talk about entropy in any time.
To see it, consider first that you have one particle inside a chamber of volume $V$. All that I will say for this particle applies for a gas of $N$ non-interacting particles and you could just change all quantities that I'm calculating (energy, volume, entropy) by (energy per particle, volume per particle,...). The advantage to consider a single particle is to understand the relation with mechanics, as you will see.
Initially, you are not exactly tracking its position and/or velocity, but you know for sure that it is somewhere in the volume $V$. Now, you open a wall, increasing the total volume by $2V$, giving the particle the access of a second chamber of volume $V$. You lost the information about where the particle is, if it is in the first or second chamber. But you know it is somewhere in the volume $2V$.
In this experiment, we could say that the variation of entropy when you opened the wall was $\Delta S = k_b\ln 2$. Your uncertainty about the particle position is two times bigger.
Now, back to the beginning of this experiment and instead of to remove the wall, you changed it by a moveable wall, that can continuously expand the chamber to the volume $2V$. The particle, travelling a free path inside the chamber, will collide many times against this moveable wall, transferring momentum and realizing work. Let's say that all the system is in contact with a thermal reservoir, keeping the temperature constant. It means that the particle will realize isotermic expansion.
The total work realized by the particle can be calculated using the relation $pV = k_bT$, which gives
$$
W = \int_V^{2V} pdV' = k_bT \ln 2
$$
The difference between the two processes is that in first, the particle expanded freely, and know it expanded realizing some work. The reservoir keeps the temperature constant, which means that the energy of the particle does not change. It means that the system received an amount of reversible heat
$$
dQ_{rev} = k_bT\ln (2)
$$
And its entropy increased, by definition, by an amount $k_b\ln(2)$.
Now, imagine that, as you said, you know the particle position after expansion to the volume $2V$. Imagine that in an instant $t$, it is in the left chamber of volume $V$. It will go to the right part at some moment later, because it is traveling freely in the volume $2V$. But before it go there, you insert a moveable wall between the two Chambers. The particle will expand realizing work again, receiving heat from the reservoir. But observe that this system is converting the total amount of heat in work, and each time you know the position of the particle and you insert the wall, it is back to the initial state. Analysing this way, it seems that this system is a perfect engine, converting the total amount of reversible heat to work in a cyclic process.
But in fact, it is not. The ace in the hole is that every time you learn the particle location, the entropy of the system decrease. The only way to solve this paradox is connecting the thermodynamic concept of entropy with the informational concept of entropy. This is the principle behind Szilard engine. So it is possible to convert information in energy and the only way to not violate the principle of conservation of energy is to consider the physical representation of this information.
So for example, in order to learn the particle location, you have to store this information somewhere, and to spent some energy to erase previous informations stored in the memory in each new round. A good source about this is the James Sethna's book chapter 5.
In conclusion, when you suppose that you know the particles positions and velocities, you are already assuming that the system has zero entropy, and you can use it to extract work from the system as you wish. One if the first interpretations of entropy, well before information theory to be a thing, as it was energy degradation of a system, and in this sense, it connects exactly to this lack of knowledge about the exact system state, in order to put it to realize work.
I hope it helps
