Entropy of mixing - Ideal gas. What is x?

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SUMMARY

The discussion focuses on calculating the total entropy change when mixing one mole of argon and one mole of xenon in two connected bottles of equal volume, v, at constant temperature and pressure. The entropy change during mixing is derived using the formula ds = integral (nR/v) dv, leading to a total entropy change of Rln4. The conversation also addresses the need for separate entropy change calculations for each gas and clarifies the use of variable x in the context of volume fractions, ultimately concluding that the final entropy change is zero once the gases reach equilibrium.

PREREQUISITES
  • Understanding of thermodynamics, specifically entropy and its calculations.
  • Familiarity with the ideal gas law and properties of ideal gases.
  • Knowledge of isothermal processes and reversible expansions.
  • Basic mathematical skills for logarithmic functions and integration.
NEXT STEPS
  • Study the derivation of entropy change for mixing ideal gases using the formula deltaS = -NK_b(x*lnx+(1-x)ln(1-x)).
  • Explore the concept of entropy in relation to disorder and randomness in thermodynamic systems.
  • Learn about the implications of entropy changes in real-world applications, such as chemical reactions and phase transitions.
  • Investigate the differences between reversible and irreversible processes in thermodynamics.
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Students of thermodynamics, physicists, and chemists interested in understanding entropy changes during gas mixing processes and their implications in physical chemistry.

navm1
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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
 
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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.
 

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