Average Separation between molecules in hydrogen at STP (Alonso Finn Problem)

[agnimusayoti]

Homework Statement

Using relative density data, estimate the average separation between molecules in hydrogen at STP (gas).

Relevant Equations

$$N (molecules) = n (mol) * N_A (molecules/mol)$$

The relative density data:
$${\rho_{H_{2},H_{2}O}}=\frac{\rho_{H_{2}}}{\rho_{H_{2}O}}=8.988\times10^{-5}$$
With avogadro number, thus I can obtain number of molecules per 1 $m^3$ of Hydrogen gas, that is:
$$N = \frac{{{\rho_{H_{2},H_{2}O}}}\times{\rho_{H_{2}O}}}{M_{r}}\times{N_{A}} $$
thus, I get
$$N = 2.685 \times 10^{25} \frac{molecules}{m^{3}}$$

The problem is I can't relate this result with the question which ask the average separation or average distance between the molecules.

Oh, btw, the solution mention 2 different geometries, that is ball and cubic.

Could you help me? Thanks guys!

 

[BvU]

How about converting to e.g. $\ m^3/\text{molecule} \$ ?

$\$

 

[haruspex]

the average separation between molecules

How do you think that is defined? Taken literally, you would have to sum the distances between all pairs of molecules in the volume. So it must be more in the sense of nearest neighbours. But it seems it's not what that would really be either.
From what I read, you have to pretend a particular model for how the molecules are arranged. The simplest is as a cubic lattice. In that, if the distance to nearest neighbours is D then each molecule effectively occupies a volume $D^3$.
The "ball" model cannot represent any actual structure since spheres don't pack without gaps, so I assume this means merely that you take each molecule as occupying an octant of volume $\pi D^3/6$ (but that's just my guess).

Out of interest, the real average distance to the nearest neighbour in a random arrangement I calculate to be $(\frac 3{4\pi\rho})^{\frac 13}\Gamma(\frac 43)\approx 0.55\rho^{-\frac 13}$.

 

[agnimusayoti]

What I think about average separation is the average distance to the nearest neighbour of the molecules. But, I can't relate the number of molecules in $1 m^3$ space that is $2.685\times{10^{25}}$ with the distance.
Then, I try to think like you said:

How about converting to e.g. $\ m^3/\text{molecule} \$ ?

$\$

and get one molecules occupies $3.724\times{10^{-26}} m^{3}$.

but, what is the relationship between this result and cubic lattice and either the spherical models.

For example, if there is 2 molecule along 1 m line , then I know that the nearest neighbour separated by 1 m. Now, I use these lines to make a cubic with 1 m side. Therefore, in my 1 $m^{3}$ cubic, I have 8 molecules.

Now, I try to reverse the process of thinking:
suppose 8 molecules occupy $1 m^{3}$ cubic. Therefore, each of the molecule occupies $0,125 m^3$ or 1/8 part of the cubic. If I take cubic root, then I come with 0.5 m. Then, what is that mean?

 

[agnimusayoti]

For example, if there is 2 molecule along 1 m line , then I know that the nearest neighbour separated by 1 m. Now, I use these lines to make a cubic with 1 m side. Therefore, in my 1 $m^{3}$ cubic, I have 8 molecules.

Now, I try to reverse the process of thinking:
suppose 8 molecules occupy $1 m^{3}$ cubic. Therefore, each of the molecule occupies $0,125 m^3$ or 1/8 part of the cubic. If I take cubic root, then I come with 0.5 m. Then, what is that mean?

I think my process of thinking relate with your explanation but still not clear yet.

From what I read, you have to pretend a particular model for how the molecules are arranged. The simplest is as a cubic lattice. In that, if the distance to nearest neighbours is D then each molecule effectively occupies a volume $D^3$.

 

[BvU]

but, what is the relationship between this result and cubic lattice and either the spherical models

So in a cubic lattice the distance between molecules is this to the power $\ \frac 1 3$ (check with @haruspex post #3 !). For stacked spheres there is a little more distance.

Now, the exercise asks to provide an estimate. Would the average distance be very different if the molecules move like crazy instead of sitting still ?

$\$

 

[haruspex]

average separation is the average distance to the nearest neighbour

As I tried to explain, that is only meaningful for a given model for the spatial arrangement. Regular arrangements, like a cubic lattice, lead to several molecules at about the same distance from a given molecule. A cubic lattice gives you six. In a real gas, there will often be one quite a bit closer than the rest, leading to a rather smaller average distance.
In the end, it depends what you intend to do with the value when you have found it.

 

[Charles Link]

@agnimusayoti When having molecules on the corners of a cube, you need to be careful how you count them. You stack these like blocks on each other=at each corner of the cube, you have only 1/8 of a molecule assigned to the cube. The adjacent cubes make up the other 7/8. Thereby each cube has just one molecule=8 corners x 1/8.

Edit: The alternative is to put the molecules at the center of the cubes and stack them. Once again there is one molecule per block, and you will find the molecules (the centers of adjacent cubes) are one meter apart.

 

[agnimusayoti]

@agnimusayoti When having molecules on the corners of a cube, you need to be careful how you count them. You stack these like blocks on each other=at each corner of the cube, you have only 1/8 of a molecule assigned to the cube. The adjacent cubes make up the other 7/8. Thereby each cube has just one molecule=8 corners x 1/8.

Edit: The alternative is to put the molecules at the center of the cubes and stack them. Once again there is one molecule per block, and you will find the molecules (the centers of adjacent cubes) are one meter apart.

Hmm, could you refer a link or website that supply the image or drawing of this explanation?

 

[agnimusayoti]

So in a cubic lattice the distance between molecules is this to the power $\ \frac 1 3$ (check with @haruspex post #3 !). For stacked spheres there is a little more distance.

Now, the exercise asks to provide an estimate. Would the average distance be very different if the molecules move like crazy instead of sitting still ?

$\$

Yeah, but I came with the different answer (see my #4 post) that is $(1/8)^{\frac{1}{3}} = 0.5$. Actually, I begin with 1 m apart of molecules.

 

[haruspex]

Yeah, but I came with the different answer (see my #4 post) that is $(1/8)^{\frac{1}{3}} = 0.5$. Actually, I begin with 1 m apart of molecules.

Here's another way to look at it.
Each molecule is at a corner of 8 1m cubes; each 1m cube has 8 molecules at its corners; so over a large region, there must be an average of 1 molecule per 1m cube.

 

[Charles Link]

Hmm, could you refer a link or website that supply the image or drawing of this explanation?

The chemistry book by Mahan has this type of diagram if I'm not mistaken. I think I've also seen it in the Solid State Physics book by Kittel. The designation "primitive cell" is used in the description.

 

[Steve4Physics]

For example, if there is 2 molecule along 1 m line , then I know that the nearest neighbour separated by 1 m. Now, I use these lines to make a cubic with 1 m side. Therefore, in my 1 $m^{3}$ cubic, I have 8 molecules.

Now, I try to reverse the process of thinking:
suppose 8 molecules occupy $1 m^{3}$ cubic. Therefore, each of the molecule occupies $0,125 m^3$ or 1/8 part of the cubic. If I take cubic root, then I come with 0.5 m. Then, what is that mean?

Your analysis is wrong because you have have placed the molecules at the corners of a cube and ignored each molecule’s ‘territory’ outside the cube.

Try thinking of it this way.

Take a 1m³ cubic box.

Imagine the box partitioned into 2x2x2 = 8 smaller cubes; volume of each is 1m³/8 = 0.125m³.

Put a molecule at the centre of each of the smaller cubes.

You now have 8 molecules in 1m³.

The volume per molecule is V = 0.125m³.

The side-length of each smaller cube is ∛V = ∛(0.125m³). = 0.5m.

The centre-to-centre distance between adjacent smaller cubes is 0.5m. We can therefore take the separation between, molecules as 0.5m.

 

[agnimusayoti]

Well, it is more confusing because there are 2 different answer depend on how I placed the molecules in the smaller cube.

Which one is true regarding the separation of 8 molecules in 1 $m^{3}$ cube? Is it 0.5 m or 1 m? Or both is true??

From the post it seems 1 m is true the molecules placed in the corner of the smaller cubes. But, if the molecules is in the center of the smaller cubes, the average separation is 0.5 m...

 

[BvU]

it seems 1 m is true the molecules placed in the corner of the smaller cubes

But the size of the cubes is only 0.5 m !

$\$

 

[Steve4Physics]

Well, it is more confusing because there are 2 different answer depend on how I placed the molecules in the smaller cube.

One of the placement methods is wrong!

Which one is true regarding the separation of 8 molecules in 1 $m^{3}$ cube? Is it 0.5 m or 1 m? Or both is true??

You need to decide for yourself! See if this helps...

Extend the volume by arranging eight of the 1m-sided cubes into a 2m x 2m x 2m cube.
You now have a 2m-sided cube (volume = 8m³) containing eight 1m-sided cubes.
And each 1m-sided cube contains eight 0.5m -sided cubes.

You have 64 molecules.

Suppose you use your (incorrect) Post #4 method and put one molecule at each corner of every 1m-sided cube.

This will produce some points where there are 8 molecules at exactly the same point (every point where eight of the small cubes meet).

Corection: This will result in 8 molecules at the centre point of the 2m-sided cube. But other points will have different numbers of molecules. For example each corner of the 2m-sided cube will only have one molecule.

You do not have an evenly-spaced arrangement of the molecules - so you can’t use it to find average spacing.

But if you put a molecule at the centre of each 0.5m-sided cube, the 64 molecules are evenly spaced which is what is needed to find the average separation.

You are told there are 64 molecules in a volume of 8m³ and you want the average separation. You would do this:
Volume per molecule = 8m³/64 = 0.125m³
Spacing = ∛V = ∛(0.125m³) = 0.5m

Sorry - I corrected some mistaskes. Hope this didn't cause confusion.

 

[Passers_by]

I don't know how do I get the average separation in the spherical model.

 

[Steve4Physics]

I don't know how do I get the average separation in the spherical model.

This thread has been inactive for 7+ months.

Static spheres can be packed in different ways. Gas particles have random motions. The average separation depends on what system you are considering.

It might help you if you start a new thread - with a complete and clear description of what you are trying to determine.