Entropy of mixing - Ideal gas. What is x?

In summary, the two bottles of gas, one with 1 mole of argon and the other with 1 mole of xenon, are connected by a pipe and tap at the same temperature and pressure. When the tap is opened, the gases mix and the system experiences a total entropy change due to the expansion of the gases. After the gases have fully mixed, the tap is shut and the system can no longer mix. The entropy change in this process is due to the increase in randomness or disorder in the final state compared to the initial state.
  • #1
navm1
44
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
 
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  • #2
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.
 
  • #3
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 .
 
  • #4
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
 
  • #5
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.
 

Related to Entropy of mixing - Ideal gas. What is x?

What is the concept of entropy of mixing?

The entropy of mixing is a thermodynamic measure of the disorder or randomness in a system that occurs when two or more substances are combined together. It is a result of the increased number of possible arrangements of particles when they are mixed.

How is the entropy of mixing calculated for an ideal gas?

For an ideal gas, the entropy of mixing is calculated using the formula S_mix = -nR(xlnx + (1-x)ln(1-x)), where n is the number of moles of each gas, R is the gas constant, and x is the mole fraction of one of the gases.

Why is the entropy of mixing of an ideal gas always positive?

The entropy of mixing for an ideal gas is always positive because the mixing of gases leads to an increase in the number of possible microstates, resulting in an increase in entropy. This is in accordance with the Second Law of Thermodynamics, which states that the entropy of a closed system will always increase over time.

What is the physical significance of the ideal gas mixing entropy?

The mixing entropy of an ideal gas is a measure of the degree of disorder in the system. It reflects the randomness of the distribution of particles and the amount of energy dispersed in the system due to the mixing process. It is an important concept in understanding the behavior of gases and predicting the changes in entropy in various chemical reactions.

How does the entropy of mixing change with temperature and pressure?

The entropy of mixing increases with an increase in temperature and/or pressure. This is because higher temperatures and pressures lead to a greater degree of disorder and randomness in the system, resulting in an increase in entropy. Conversely, a decrease in temperature and/or pressure will result in a decrease in the entropy of mixing.

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