Entropy and Energy exchange of systems.

[nahanksh]

Homework Statement

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

Homework Equations

The Attempt at a Solution

For the first 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?


For the 2nd question,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 !

 

[mark726]

I would like to first clarify some points in the given scenario. The model described is a simplified version of computer memory and it is not clear what type of particles are being used. Also, the process of pulling out the dividing wall may not be feasible in a real computer memory system.

Now, to answer the first question, the change in entropy can be calculated using the equation ∆S = Nkln(Vf/Vi), where N is the number of particles, k is the Boltzmann constant, Vf is the final volume and Vi is the initial volume. In this case, N=10 and Vf=Vi/2, so the change in entropy would be 10*ln(1/2). Your calculation is correct.

For the second question, the energy required to compress the gas can be calculated using the equation W = P∆V, where W is the work done, P is the pressure and ∆V is the change in volume. In this case, the work done would be the energy required to push the particles back to the left side of the cell. The pressure can be calculated using the ideal gas law, P = nRT/V, where n is the number of moles, R is the gas constant and T is the temperature. Since the temperature and number of moles remain constant, the energy required would depend on the change in volume. This can vary depending on the specific setup of the system and the type of particles being used.

In conclusion, the energy required to reset the 10-bit register to its initial state would depend on the specific conditions of the system and cannot be determined without further information. However, we can say that it would require some energy to compress the gas and push the particles back to the left side of the cell, and this energy would be dependent on the pressure and change in volume.