Entropy of mixing - Ideal gas. What is x?

Click For Summary

Homework Help Overview

The problem involves calculating the total entropy change when two ideal gases, argon and xenon, mix in a connected system. The scenario describes two bottles of equal volume, each containing one mole of gas, at the same temperature and pressure, and explores the entropy changes during the mixing process and after the gases have mixed.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the entropy change using a reversible isothermal process and questions the need for two separate calculations of entropy change. They also inquire about the meaning of the variable x in their textbook's equation.
  • Some participants discuss the reasoning behind separate entropy changes for each gas during expansion and suggest that x represents a fraction of the total volume occupied by one gas.
  • Others express their understanding of the entropy change at the end of the mixing process, speculating that it may be zero due to the system reaching equilibrium.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the entropy changes involved in the mixing process. Some guidance has been offered regarding the meaning of x and the nature of entropy changes during mixing and after equilibrium is reached.

Contextual Notes

Participants are considering the implications of the gases being ideal and the conditions under which the entropy changes are calculated. There is a focus on the concepts of disorder and randomness as they relate to entropy.

navm1
Messages
44
Reaction score
0

Homework Statement


A bottle with volume v containing 1 mole of argon is next to a bottle of volume v with 1 mole of xenon. both are connected with a pipe and tap and are same temp and pressure. the tap is opened and they are allowed to mix. What is the total entropy change of the system? Once the gases have fully mixed, the tap is shut and the gases are no longer free. what is the entropy change with this process?

Homework Equations


ds= integral (nR/v) dv

The Attempt at a Solution



imagining it as a reversible isotherm. I used
deltaS_mix = deltaS_1+deltaS_2

deltaS_1 = n_1*R*ln(V_1+V_2)/V_1

deltaS_2 = n_2*R*ln(V_1+V_2)/V_2

then adding them together and cancelling down from 2V/V etc i ended up with
R(ln2+ln2) = Rln4.

My Questions are:

Why does it ask for two separate entropy change calculations in the question?
In my textbook it uses xV and (x-1)V for the respective volumes and it ends up as

deltaS = -NK_b(x*lnx+(1-x)ln(1-x))

What does x represent here?

Thanks
 
Physics news on Phys.org
navm1 said:
Why does it ask for two separate entropy change calculations in the question?

as the two gases have expanded there will be entropy change for each one of them i.e. under expansion from a state of order to more disorder.
 
navm1 said:
In my textbook it uses xV and (x-1)V for the respective volumes and it ends up as

well it might be taking V as the total volume of the system. and if x fraction of V is being occupied by one then 1-x times V must be the volume of the other one .
 
got it. so i used x as 1/2 and got Rln(2). makes sense to me and seems to match my course material.

I still don't quite understand why it asks for an entropy change of the system when they mix and when theyve mixed and the tap is shut so they can't mix any more. I am guessing the entropy change at the end is 0 because its just a homogenous mix now and theyve reached equilibrium now entropy is maximum
 
navm1 said:
I still don't quite understand why it asks for an entropy change of the system when they mix and when theyve mixed and the tap is shut so they can't mix any more. I am guessing the entropy change at the end is 0 because its just a homogenous mix now and theyve reached equilibrium now entropy is maximum

Entropy is a property reflected in the ways in which a system of N particles can get described.

the more ordered a system is- it gets to less entropy- and the opposite is also true.
as one opens the tap-
the two gases are free to diffuse throughout the volume of two containers. For an ideal gas, the energy is not a function of volume,

and, for each gas, there is no change in temperature. The entropy change of each gas is affected as for a reversible isothermal expansion from the initial volume to a final volume
In terms of the overall spatial distribution of the molecules of the two gases , one can say that
final state was more random, more mixed, than the initial state in which the two types of gas molecules were confined to specific regionsof space in the bottles..

Another way to say this is in terms of ``disorder;'' there is more disorder in the final state than in the initial state.
the perspective/background of entropy is thus that increases in entropy are connected with increases in randomness or disorder.no doubt in the final state they can not take any path of more randomness or disorder.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
846
Replies
16
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
10
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
4
Views
3K
Replies
4
Views
2K