<p>In short, the issue is that we have no law of conservation of volume for mixing unequal components. The illustration is for macroscopic particles. It applies equally at the molecular to atomic scale.</p>
<p>Imagine a set of 100 equally sized solid spheres. Pack them in a cubic arrangement. Now pack them in a close-packed arrangement. The volumes will be different.</p>
<p>Imagine two sets of 100 spheres each, A and B. Pack them in different relative proportions, each always equaling 100 total spheres. By example 10 A + 90 B or 50 A + 50 B. The diagonal lines in the figure show the volumes in each set as you pick different numbers of spheres A and B. The sum is the topmost diagonal line. In the case of true conservation of volume, the total volume of the mixture will be the sum of the volumes of each set. The total volume will always follow the topmost diagonal line. As the particle mixture deviates from conservation of volume, the total volume of the mixture follows the individual lines rather than the total line.</p>
<p>With the above in mind, the figure is stating that mixtures with large + small spheres follows more closely to total volume being controlled by each individual sphere sets. This is the greatest deviation from conservation of volume. Fundamentally, the smaller spheres intercalate within the larger spheres. As you make mixtures with more equally-sized spheres, they behave more closely to conservation of volume.</p>
<p>Now, to appreciate the straight lines on the graph, consider that the discussion you quote talks about infinitely larger particles compared to the smaller particles. Infinitely larger is only possible when the size of the smallest particles is at some point allowed become zero volume. This is a <em>hypothetical</em> state, not a real state. Even with the medium sized particles, they seem still to treat the ratio as infinitely larger at the limit of 100% of the smallest particles. Also note that a system with 100% fine particles is not shown as having zero volume, it is shown with a volume just greater than 100% course particles.</p>
<p>So, start with the leftmost figure. As you add smaller particles you must remove larger particles <em>to keep some aspect of relative amounts (perhaps total number of particles) the same</em>. This latter restriction on relative amounts must be explained din the source article. Hence, as you add smaller particles, the total volume decreases along a linear path as larger particles are removing more than smaller particles are adding back. At some point, the smaller particles become effective at increasing volume even though you are still removing larger particles. As the smaller particles become larger in size (right most figure), the point where they affect the volume change shifts to lower relative amount (shifts to the left on the graph. Also, the effect is less abrupt (the transition curve is smoother and further displaced from the two straight lines).</p>

CC BY-SA 4.0
2023-09-24T21:19:41.723Z
Just for good measure, without any knowledge of the paper or topic... this is primarily talking about solids, yes???
JeopardyTempest
CC BY-SA 4.0
2023-09-24T21:47:21.727Z
@JeopardyTempest The same analysis applies to liquids. We use the term partial molar volume in liquid mixtures rather than molar volume, which is solely for pure liquids.
Jeffrey J Weimer
CC BY-SA 4.0
2023-09-30T04:50:58.417Z
@JeffreyJWeimer any idea why the line is being drawn to touch at bottom right corner which means 100% fine particles and apparent volume become zero which doesn't make sense. any hints on how to find the equation of this straight line?
Sagar Rawal
CC BY-SA 4.0
2023-10-02T02:58:11.300Z
I did as best I could with an addendum. I encourage that a closer reading of the journal article must help clear what is likely a confusion on exactly what is plotted versus what in the figure. Note for example that 100% fine particles are not being shown to have zero volume. They are being shown to have a volume just greater than 100% coarse particles.
Jeffrey J Weimer
CC BY-SA 4.0
2023-09-24T20:55:24.277Z
2023-10-02T03:00:06.477Z
2023-10-02T03:00:06.477Z
Jeffrey J Weimer
<p>This is meant as a supplement to Jeffrey Weimer's fine answer</p>
<p>The apparent volume is based on the largest particle if the diameter ratios are large enough. If the smaller particle is much smaller (the diameter ratios in the graphs show  50.5 and 6.3) then it can occupy the space in-between the larger particles. If all the small particles fit then the apparent volume doesn't change. It will be the volume of the space occupied by the larger particles which is not the sum of the volumes of the larger particles.</p>
<p>As a small thought experiment, fill a small box with tennis balls packed as efficiently as possible. The apparent volume is the volume of the box. The addition of sand won't alter the apparent volume (since it will settle in the empty space between tennis balls) until there is so much sand it overflows out of the box.</p>
<p>Now (after the box is fully packed with tennis balls and sand) remove all of the tennis balls. The apparent volume of the remaining sand will be much less than the volume of the box.</p>

CC BY-SA 4.0
2023-10-03T17:51:02.697Z
2023-10-03T18:28:47.113Z
2023-10-03T18:28:47.113Z
Stevan V. Saban
<p>This is from paper &quot;<a href="https://ceramics.onlinelibrary.wiley.com/doi/abs/10.1111/j.1151-2916.1930.tb16222.x" rel="nofollow noreferrer">The packing of particles</a>&quot; by A.E.R Westman and H.R. Hugill.</p>
<blockquote>
<p>Here they give result as, &quot;it is evident that if the diameter ratio is
unity, ie., the coarse and fine particles have the same diameter,
there will be no shrinkage on mixing and the Va curve will be the
straight line,  CF. On the other hand, if the coarse  particles have
a diameter which is infinitely  greater than that of the fine particles,
a different Vu curve will be obtained. On adding fine particles to a
quantity of coarse  particles, the fine will pack in the voids in the
coarse and the apparent volume of the mix will be equal to that of the
coarse particles. The Va curve corresponding to FIG. 4. -Packing in
two-size systems: Diameter ratio = 8.0. this state of affairs is a
straight line joining the point C in Fig. 2 with the right hand lower
corner of the diagram.  On adding coarse particles to a large quantity
of fines, the solid coarse particles will be immersed in the fines and
the Vu curve will be the straight line joining F, Fig. 2, to the point
1.00 at the left side of the diagram. These straight lines cross at R and thus CRF gives the Vu  curve for the case when the coarse
particles are infinitely larger than  the fine particles.&quot;</p>
</blockquote>
<p><a href="https://i.sstatic.net/rzKIz.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/rzKIz.png" alt="enter image description here" /></a></p>
<p>Can anyone explain what they mean and what they were trying to prove?</p>

<geology><geophysics>
How is the apparent volume affected by particle packing?
Sagar Rawal
# How is the apparent volume affected by particle packing?

(Asked by: username_0 on 2023-09-23)

This is from paper "[The packing of particles](https://ceramics.onlinelibrary.wiley.com/doi/abs/10.1111/j.1151-2916.1930.tb16222.x)" by A.E.R Westman and H.R. Hugill.

> Here they give result as, "it is evident that if the diameter ratio is
> unity, ie., the coarse and fine particles have the same diameter,
> there will be no shrinkage on mixing and the Va curve will be the
> straight line, CF. On the other hand, if the coarse particles have
> a diameter which is infinitely greater than that of the fine particles,
> a different Vu curve will be obtained. On adding fine particles to a
> quantity of coarse particles, the fine will pack in the voids in the
> coarse and the apparent volume of the mix will be equal to that of the
> coarse particles. The Va curve corresponding to FIG. 4. -Packing in
> two-size systems: Diameter ratio = 8.0. this state of affairs is a
> straight line joining the point C in Fig. 2 with the right hand lower
> corner of the diagram. On adding coarse particles to a large quantity
> of fines, the solid coarse particles will be immersed in the fines and
> the Vu curve will be the straight line joining F, Fig. 2, to the point
> 1.00 at the left side of the diagram. These straight lines cross at R and thus CRF gives the Vu curve for the case when the coarse
> particles are infinitely larger than the fine particles."

[![enter image description here](https://i.sstatic.net/rzKIz.png)](https://i.sstatic.net/rzKIz.png)

Can anyone explain what they mean and what they were trying to prove?

**Tags:** geology, geophysics


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## 2 answers

### Answer by username_1 on 2023-09-24. Score: 5

In short, the issue is that we have no law of conservation of volume for mixing unequal components. The illustration is for macroscopic particles. It applies equally at the molecular to atomic scale.

Imagine a set of 100 equally sized solid spheres. Pack them in a cubic arrangement. Now pack them in a close-packed arrangement. The volumes will be different.

Imagine two sets of 100 spheres each, A and B. Pack them in different relative proportions, each always equaling 100 total spheres. By example 10 A + 90 B or 50 A + 50 B. The diagonal lines in the figure show the volumes in each set as you pick different numbers of spheres A and B. The sum is the topmost diagonal line. In the case of true conservation of volume, the total volume of the mixture will be the sum of the volumes of each set. The total volume will always follow the topmost diagonal line. As the particle mixture deviates from conservation of volume, the total volume of the mixture follows the individual lines rather than the total line.

With the above in mind, the figure is stating that mixtures with large + small spheres follows more closely to total volume being controlled by each individual sphere sets. This is the greatest deviation from conservation of volume. Fundamentally, the smaller spheres intercalate within the larger spheres. As you make mixtures with more equally-sized spheres, they behave more closely to conservation of volume.

Now, to appreciate the straight lines on the graph, consider that the discussion you quote talks about infinitely larger particles compared to the smaller particles. Infinitely larger is only possible when the size of the smallest particles is at some point allowed become zero volume. This is a *hypothetical* state, not a real state. Even with the medium sized particles, they seem still to treat the ratio as infinitely larger at the limit of 100% of the smallest particles. Also note that a system with 100% fine particles is not shown as having zero volume, it is shown with a volume just greater than 100% course particles.

So, start with the leftmost figure. As you add smaller particles you must remove larger particles *to keep some aspect of relative amounts (perhaps total number of particles) the same*. This latter restriction on relative amounts must be explained din the source article. Hence, as you add smaller particles, the total volume decreases along a linear path as larger particles are removing more than smaller particles are adding back. At some point, the smaller particles become effective at increasing volume even though you are still removing larger particles. As the smaller particles become larger in size (right most figure), the point where they affect the volume change shifts to lower relative amount (shifts to the left on the graph. Also, the effect is less abrupt (the transition curve is smoother and further displaced from the two straight lines).

> **Comments:**
> **username_2**: Just for good measure, without any knowledge of the paper or topic... this is primarily talking about solids, yes???  
> **username_1**: @username_2 The same analysis applies to liquids. We use the term partial molar volume in liquid mixtures rather than molar volume, which is solely for pure liquids.  
> **username_0**: @username_1 any idea why the line is being drawn to touch at bottom right corner which means 100% fine particles and apparent volume become zero which doesn't make sense. any hints on how to find the equation of this straight line?  
> **username_1**: I did as best I could with an addendum. I encourage that a closer reading of the journal article must help clear what is likely a confusion on exactly what is plotted versus what in the figure. Note for example that 100% fine particles are not being shown to have zero volume. They are being shown to have a volume just greater than 100% coarse particles.  

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### Answer by username_3 on 2023-10-03. Score: 1

This is meant as a supplement to username_1 username_1's fine answer

The apparent volume is based on the largest particle if the diameter ratios are large enough. If the smaller particle is much smaller (the diameter ratios in the graphs show 50.5 and 6.3) then it can occupy the space in-between the larger particles. If all the small particles fit then the apparent volume doesn't change. It will be the volume of the space occupied by the larger particles which is not the sum of the volumes of the larger particles.

As a small thought experiment, fill a small box with tennis balls packed as efficiently as possible. The apparent volume is the volume of the box. The addition of sand won't alter the apparent volume (since it will settle in the empty space between tennis balls) until there is so much sand it overflows out of the box.

Now (after the box is fully packed with tennis balls and sand) remove all of the tennis balls. The apparent volume of the remaining sand will be much less than the volume of the box.


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