# Thermal Physics - Change of Entropy

1. Apr 2, 2010

### nahanksh

Consider a very simple model of a computer memory, in which molecules are either found to reside in the left half of their memory cell (encoding a "0"), or in the right half (encoding a "1"). Imagine that we have a 10-bit register. Initially each cell is in the "0" state (i.e., all particles are in the left side of their respective cells); after the computation, they are in either half of the cell (depending on the specific computation). This doesn't necessarily require any work, e.g., if one simply pulls out (transverse to the axis of the memory) the dividing wall between the "0" and "1" side, the particles can by free expansion move from the "0" state into the "1" state.

Your task is to determine the energy cost to reset the 10-bit register to its initial state, where every particle is again in the "0" side of its cell. This can be done by using a piston to push the particles (~compressing the gas) so that they can only occupy the left side of the cell.

1. What is the change in (dimensionless) entropy in this process?

For this question, i figured that N=10 and final Volume should be half of Initial Volume) because it says it has to occupy only the left side of the cell which is half
and hence the change in entropy would be 10*ln(1/2)..

Am i doing correctly?

2. What energy is required to carry out the process?

I am totally stuck with this question.
I am guessing that the energy required would be neglect-ably small...?

Could someone help me out with these two questions ?

Thanks !

2. Apr 3, 2010

### nickjer

Unfortunately you are using the ideal gas law for this circumstance. I don't believe it is wise to use it for a liquid with molecules.

I believe you are supposed to make use of:
$$S = k_B ln(\Omega)$$

where $$\Omega$$ is the number of possible states that are equally likely.

So you will need to compute the number of states initially when the molecules can be in 0 or 1 for all 10 registers. And do the same for the final state.

You do end up with a very similar answer to what you originally came up with. I just don't think you would be able to justify your first answer.

3. Apr 3, 2010

### chrisk

You wrote:

"Consider a very simple model of a computer memory, in which molecules are either found to reside in the left half of their memory cell (encoding a "0"), or in the right half (encoding a "1")."

If a gas occupies the left half (0) and the wall is removed to allow free expansion then the final condition has molecules throughout the entire cell, not just the right half. The initial condition is all cells are in the 0 state, and the claimed final state has molecules in either the left or right half of the cell. The portion of the problem:

"This doesn't necessarily require any work, e.g., if one simply pulls out (transverse to the axis of the memory) the dividing wall between the "0" and "1" side, the particles can by free expansion move from the "0" state into the "1" state."

This cannot occur with a free expansion. Work must be done to move all the molecules to the right side of the cell.