Determine how many microstates and macrostates (Thermodynamics)

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Homework Help Overview

The discussion revolves around determining the number of microstates and macrostates in the context of a two-state paramagnet with 40 magnetic dipoles. Participants explore the definitions and calculations related to microstates and macrostates, questioning the assumptions and interpretations involved.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of microstates, initially suggesting it might be ##40^2##, but later correcting it to ##2^{40}##. There is uncertainty about how to define and count macrostates without explicit outcomes. Questions arise regarding the total number of macrostates and the criteria for defining them, particularly in relation to magnetic dipole moments.

Discussion Status

There is an ongoing exploration of the definitions of microstates and macrostates, with some participants providing alternative interpretations. The discussion includes attempts to clarify the relationship between the number of heads and tails in a coin analogy and the corresponding macrostates. Some guidance has been offered regarding the total number of macrostates being 41, though this is still under examination.

Contextual Notes

Participants note the importance of understanding the definitions of microstates and macrostates in the context of thermodynamics, particularly as it relates to the properties of the system being analyzed. There is an acknowledgment of the need for explicit outcomes to accurately determine the number of macrostates.

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Homework Statement
A two-state paramagnet has 40 magnetic dipoles. Determine the amount of microstates and macrostates?
Relevant Equations
##\Omega(N,n)=\frac{N!}{n!(N-n)!}## or ##\omega(N_{\uparrow})=\frac{N!}{N_{\uparrow}!N_{\downarrow}!}##
Since this is a two-state paramagnet where N = 40, therefore the microstate is ##40^2##? But I am not sure how to proceed to count the number of macrostates? Because from what I understand of what a macrostate is, shouldn't there a specific outcome to be stated in order to determine how many macrostate there are? For instance suppose we have 40 coins then the number of microstate (all different states of heads and tails) will be ##40^2## but in order to compute macrostate then you will have to explicitly provide the state it is in, for example determine the amount of macrostate of the 40 coins that only have two heads then you can use $$\Omega(N,n)=\frac{N!}{n!(N-n)!}$$ $$\therefore\Omega(40,2)=\frac{40!}{2!(40-2)!}=780?$$ And since the question didn't explicitly provide the macrostate then will it just be the general case of $$\Omega(40,n)=\frac{40!}{(40-n)!}?$$

Is my reasoning correct?
 
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learningastronomy said:
Is my reasoning correct?
I don't think so. Not my area but this might help.

You have 3 coins in a line. How many head-tail patterns are possible? (It's not 3².) What about 40 coins (it's not 40²)?

As for macrostates, I would assume the macrostates are simply (continuing the 40-coin example):
0 heads, 40 tails
1 head, 39 tails
2 heads, 38 tails
etc.
 
there are 41 macro states total. A macro state is different from each other by its macroscopic properties in this case it would be the total magnetic dipole moment.
learningastronomy said:
Homework Statement:: A two-state paramagnet has 40 magnetic dipoles. Determine the amount of microstates and macrostates?
Relevant Equations:: ##\Omega(N,n)=\frac{N!}{n!(N-n)!}## or ##\omega(N_{\uparrow})=\frac{N!}{N_{\uparrow}!N_{\downarrow}!}##

Since this is a two-state paramagnet where N = 40, therefore the microstate is ##40^2##? But I am not sure how to proceed to count the number of macrostates? Because from what I understand of what a macrostate is, shouldn't there a specific outcome to be stated in order to determine how many macrostate there are? For instance suppose we have 40 coins then the number of microstate (all different states of heads and tails) will be ##40^2## but in order to compute macrostate then you will have to explicitly provide the state it is in, for example determine the amount of macrostate of the 40 coins that only have two heads then you can use $$\Omega(N,n)=\frac{N!}{n!(N-n)!}$$ $$\therefore\Omega(40,2)=\frac{40!}{2!(40-2)!}=780?$$ And since the question didn't explicitly provide the macrostate then will it just be the general case of $$\Omega(40,n)=\frac{40!}{(40-n)!}?$$

Is my reasoning correct?
 
Steve4Physics said:
I don't think so. Not my area but this might help.

You have 3 coins in a line. How many head-tail patterns are possible? (It's not 3².) What about 40 coins (it's not 40²)?

As for macrostates, I would assume the macrostates are simply (continuing the 40-coin example):
0 heads, 40 tails
1 head, 39 tails
2 heads, 38 tails
etc.

Ops, I meant to say ##2^{40}## not ##40^2##, thanks for catching that.
 
guv said:
there are 41 macro states total. A macro state is different from each other by its macroscopic properties in this case it would be the total magnetic dipole moment.
Hmm can you please elaborate why it is 41 macro states?
 
learningastronomy said:
Hmm can you please elaborate why it is 41 macro states?
@guv is saying that a macro state is defined by its total dipole moment. How many different possibilities are there for that total?
 
haruspex said:
@guv is saying that a macro state is defined by its total dipole moment. How many different possibilities are there for that total?
Oh I see, I may need to revisit the definition of macrostate then because I was thinking of a different interpretation of it. Also, the total different possibilities will be 40+1 therefore 41.
 
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learningastronomy said:
Oh I see, I may need to revisit the definition of macrostate then because I was thinking of a different interpretation of it. Also, the total different possibilities will be 40+1 therefore 41.
Macrostates are whatever matters to you in a particular context. For the result of a tennis match, we might only care who won, or we might care what the individual set scores were, etc.
 
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