Entropy and configurations of microstates

[lost captain]

I was watching a Khan Academy video on entropy called: Reconciling thermodynamic and state definitions of entropy.
So in the video it says:
Let's say I have a container. And in that container, I have gas particles and they're bouncing around like gas particles tend to do, creating some pressure on the container of a certain volume. And let's say I have n particles.
Now, each of these particles could be in x different states. Now, if each of them can be in x different states, how many total configurations are there for the system as a whole?
Then it is said that the number of configurations will be x^n . But i don't understand this, shouldn't it be x(x-1)(x-2)....(x-(n-1))?
Can 2 molecules have exactly the same microstates? I mean sure they can have the same velocity but also be in the same position?

 

[.Scott]

You are, of course, presuming that it's the same 'x' states for every particle.
But given the context, that assumption is acceptable.

Apparently the author is only interested in a rough estimate - and is presuming that 'x' will be much bigger than 'n'.

Regarding: "Can 2 molecules have exactly the same microstates?"
In this context "No".

Except: Can cryogenic Helium-4 atoms share the same state?

 

[lost captain]

You are, of course, presuming that it's the same 'x' states for every particle.
But given the context, that assumption is acceptable.

Apparently the author is only interested in a rough estimate - and is presuming that 'x' will be much bigger than 'n'.

Regarding: "Can 2 molecules have exactly the same microstates?"
In this context "No".

Except: Can cryogenic Helium-4 atoms share the same state?

Thank you for taking the time to reply,
I've been searching online for an answer, there's statistical mechanics and quantum mechanics involved...so i kinda gave up on understanding this.
A sentence that is being repeated though is that particles are indistinguishable and also mathematically we view them as points and points take up no volume...
Any opinions on the above? If don't get to understand this fully maybe i can get an intuition

 

[.Scott]

The particles are indistinguishable - but that doesn't help to explain the use of x^n - in fact, it gives another reason why x^n is not precise.
Even though the molecules are taken as "points", you still cannot conclude that there are an infinite number of states. For example, as best we can tell, electrons are "points", but you still cannot encode an infinite amount of information in a 1cc volume containing 1 electron. There are quantum limits.

But, although x will always be finite, it will still be huge - generally much bigger than "n". So, I would take the x^n as a reasonable approximation.

 

[Demystifier]

I was watching a Khan Academy video on entropy called: Reconciling thermodynamic and state definitions of entropy.
So in the video it says:
Let's say I have a container. And in that container, I have gas particles and they're bouncing around like gas particles tend to do, creating some pressure on the container of a certain volume. And let's say I have n particles.
Now, each of these particles could be in x different states. Now, if each of them can be in x different states, how many total configurations are there for the system as a whole?
Then it is said that the number of configurations will be x^n . But i don't understand this, shouldn't it be x(x-1)(x-2)....(x-(n-1))?
Can 2 molecules have exactly the same microstates? I mean sure they can have the same velocity but also be in the same position?

Here it is tacitly assumed that $x\gg n$. Under this approximation
$$x(x-1)(x-2)....(x-(n-1)) \approx x^n$$

 

[Chestermiller]

In statistical thermodynamics, the following approximation is frequently used: $$\ln{(n!)}=\ln(1)+\ln(2)...\ln(n-1)+\ln(n)\approx \int_0^n{\ln{n'}dn'}=n\ln(n)-n$$so $$n!\approx n^ne^{-n}$$