Van der Waals Equation: How Surface is Considered

[FranzS]

TL;DR Summary

How is surface taken into account in van der Waals equation?

How is surface taken into account in van der Waals equation?

( P + a n²/V² ) ( V - nb ) = nRT

The term a n²/V² makes for a reduction of the expected pressure, due to surface effects (gas molecules near the surface sense a net attraction force towards the "inside of the volume").
But how is the amount of surface area taken into account? The container may be a ball, a cube, or a corrugated shape, hence different surface areas should make for a different amount of surface effect (pressure reduction).

 

[BvU]

Hello @FranzS , !

due to surface effects

And here's me thinking this term originates from intermolecular forces. From what reference comes your claim ?

Did you know this equation of state is useful for gases as well ?

Check the links below

 

[FranzS]

Thanks for your reply!
Here's what Wikipedia says. Is it inaccurate?

Next, we introduce a (not necessarily pairwise) attractive force between the particles. Van der Waals assumed that, notwithstanding the existence of this force, the density of the fluid is homogeneous; furthermore, he assumed that the range of the attractive force is so small that the great majority of the particles do not feel that the container is of finite size.[citation needed] Given the homogeneity of the fluid, the bulk of the particles do not experience a net force pulling them to the right or to the left. This is different for the particles in surface layers directly adjacent to the walls. They feel a net force from the bulk particles pulling them into the container, because this force is not compensated by particles on the side where the wall is (another assumption here is that there is no interaction between walls and particles, which is not true, as can be seen from the phenomenon of droplet formation; most types of liquid show adhesion). This net force decreases the force exerted onto the wall by the particles in the surface layer. The net force on a surface particle, pulling it into the container, is proportional to the number density. On considering one mole of gas, the number of particles will be NA

.
The number of particles in the surface layers is, again by assuming homogeneity, also proportional to the density. In total, the force on the walls is decreased by a factor proportional to the square of the density, and the pressure (force per unit surface) is decreased by

,
so that

Upon writing n for the number of moles and nVm = V, the equation obtains the second form given above,

It is of some historical interest to point out that van der Waals, in his Nobel prize lecture, gave credit to Laplace for the argument that pressure is reduced proportional to the square of the density.[citation needed]

 

[FranzS]

In addition, I'm working with pressure vessels which are charged at a certain pressure (~200 bar) and then compressed (via a piston). Two vessels with the same volume but with different internal layout (surface area), charged at the same pressure (~200 bar) and then compressed to half the initial volume, came out with a different final pressure. I remember I came up with some coefficient (sort of surface to volume ratios) which I used to normalize the experimental data, and in fact the normalized final pressures then matched.

 

[BvU]

No problem with #3. Note that in that context surface effects does not include interaction between the surface and the container. It's a pure state equation. So for

Two vessels with the same volume but with different internal layout (surface area), charged at the same pressure (~200 bar) and then compressed to half the initial volume, came out with a different final pressure.

I see only two possibilities:
1) The experiment wasn't carried out isothermally
2) The interaction with the container walls can't be ignored

Pinging @Chestermiller : do you agree ?

 

[TeethWhitener]

Pressure is force per unit surface area, so there won’t be any dependence of pressure on the surface area.

Van der Waals’s goal here was to give an intuitive argument for why the attractive intermolecular force should enter the equation of state as the inverse square of the molar volume. Basically, in the bulk, assuming a homogeneous fluid, there won’t be any net force on average on a single molecule from the surrounding molecules. However, at a (non-interacting) surface, there will be a net force: away from the surface, assuming that the intermolecular forces are attractive.

Now, for an individual molecule at the surface, it will be pulled back toward the bulk with a force in proportion with the (number) density of molecules. Basically, the more molecules there are to tug on an individual molecule at the surface, the larger the force on that individual molecule. In addition, the total number of molecules at the surface will also be proportional to the (number) density. So the total force, which is the sum of all the individual molecular forces, will have to be proportional to the square of the density. Since this logic applies regardless of the size of the box, the correct term that enters into the equation of state is the molar volume, and it does so (inverse) quadratically.

 

[Chestermiller]

No problem with #3. Note that in that context surface effects does not include interaction between the surface and the container. It's a pure state equation. So for

I see only two possibilities:
1) The experiment wasn't carried out isothermally
2) The interaction with the container walls can't be ignored

Pinging @Chestermiller : do you agree ?

The Van der Waals equation is a relationship independent of surface effects. It applies in the interior of a fluid. Provided the surface to volume ratio of the fluid is small, it applies throughout the fluid.

 

[FranzS]

Thank you all for your replies.
So, how about an accurate equation for "weakly-interacting gases" (such as N₂) at T ~ room temperature (293.15 K) and for relatively high pressures (~150 to 500 bar)?
As I wrote above, I'm working with high pressure dynamic vessels (think piston compression). This is experimental result: initial pressure 200 bar, volume is compressed to half its initial value, measured final pressure is about 500 bar (= ideal gas result of 400 bar x 1.25). Van der Waals predict final pressure of 720 bar (= ideal gas result x 1.8). Gas is nitrogen N₂.
Is it better to use compressibility factor charts for room temperature and 150-500 bar pressure ranges?
Thanks a lot.

 

[Chestermiller]

What value of z do the z-factor charts give?

 

[FranzS]

No problem with #3. Note that in that context surface effects does not include interaction between the surface and the container. It's a pure state equation. So for

I see only two possibilities:
1) The experiment wasn't carried out isothermally
2) The interaction with the container walls can't be ignored

Pinging @Chestermiller : do you agree ?

1) The compression from initial to final volume (V to 0.5V) was not isothermal, but I waited some minutes for everything to settle back to room temperature before measuring pressure.
2) Could it be quite the opposite? If the molecules of the container's inside surface behave more or less like the gas molecules, the net force experienced by the gas molecules near the container surface will be zero, such as for the gas molecules "deep inside the volume". If there is no interaction between the container's inside surface and the gas, the gas molecules near such surface should sense an "inward" net force. In this last case, the amount of surface area wouldn't be negligible.
What do you think? Thanks.

 

[FranzS]

What value of z do the z-factor charts give?

I have to check. I just found some Z charts on the internet, but no source Excel data sheets whatsoever, in order to rebuild the chart myself. Maybe you know where to find them? Thanks.

 

[Chestermiller]

1) The compression from initial to final volume (V to 0.5V) was not isothermal, but I waited some minutes for everything to settle back to room temperature before measuring pressure.
2) Could it be quite the opposite? If the molecules of the container's inside surface behave more or less like the gas molecules, the net force experienced by the gas molecules near the container surface will be zero, such as for the gas molecules "deep inside the volume". If there is no interaction between the container's inside surface and the gas, the gas molecules near such surface should sense an "inward" net force. In this last case, the amount of surface area wouldn't be negligible.
What do you think? Thanks.

I think that the surface has nothing to do with this.

 

[FranzS]

I think that the surface has nothing to do with this.

Provided the surface to volume ratio of the fluid is small, it applies throughout the fluid.

Thanks again. Could you please elaborate on this second statement of yours?

 

[Chestermiller]

I confirm your results using the vdw equation, I’ve also tried the z-factor method, based on plots on the literature at higher reduced pressures (the 500 bar case is Pr of 14.5), and the z-factor at the case of compressed volume was only about 1.3, consistent with the 500 bar pressure. So the compressibility factor plots are more accurate.

 

[FranzS]

I confirm your results using the vdw equation, I’ve also tried the z-factor method, based on plots on the literature at higher reduced pressures (the 500 bar case is Pr of 14.5), and the z-factor at the case of compressed volume was only about 1.3, consistent with the 500 bar pressure. So the compressibility factor plots are more accurate.

Thanks so much for taking your time to check the data. Could you please suggest a reliable compressibility factor plot (for nitrogen)? Thanks again!

 

[Chestermiller]

Thanks so much for taking your time to check the data. Could you please suggest a reliable compressibility factor plot (for nitrogen)? Thanks again!

Google for it at high pressures (at least up to reduced pressures of 15). Google “generalized compressibility factor plots.”

 

[Chestermiller]

 

[vanhees71]

It's indeed not a surface effect but an effect of two-particle interactions. You can formally derive the van der Waals equation from the socalled "virial expansion" of classical mechanics which holds for dilute gases. A very good treatment of this not easy subject can be found in Landau&Lifshitz vol. 5.

 

[hutchphd]

It is my understanding that the effect of the long range attraction produces a "surface tension" when the gas is restricted. This is complicated because the gas is compressible and the the density therefore nonuniform at the these edges. This may be relevant to the OP's original issues but I have no feel for the relative size of this.
Is this not a surface effect ?

I took a look at Landau & Lifshitz vol.5 and believe any edge effects are very small because of the falloff of the Van der Waals attraction and the energy of the well. I very much recommend their treatment.

 

[Swamp Thing]

You can formally derive the van der Waals equation from the socalled "virial expansion" of classical mechanics which holds for dilute gases.

So the surface analysis, where we think of molecules being pulled back into the gas (leaving less momentum to create pressure on the wall) is just one of those nice heuristic pictures that happen to give useful results? In some sense similar to the popular "waves between two ships" picture of the Casimir force...

Another question: what do we actually mean by pressure of gas in the middle of a container? Is it the momentum per unit time per unit area due to collisions on a hypothetical test surface? In that case, the bulk picture can be linked with the surface picture by introducing a hypothetical test surface which doesn't interact with the molecules, thus leading to a situation like the actual container walls. Maybe... or maybe there is a definition of internal bulk pressure that doesn't involve a special surface?

 

[FranzS]

So, how about an accurate equation for "weakly-interacting gases" (such as N₂) at T ~ room temperature (293.15 K) and for relatively high pressures (~150 to 500 bar)?

I reply to my own question, in case anyone's interested. I found a great article which provides an analytic function (equation 14 on the article - the article also provides all the required constants/parameters) for the compressibility factor Z, without need to always rely on charts. Great for calculations in Excel.
This model is said to be really accurate in reproducing the experimental data for the following ranges of temperatures and pressures:
1.15 < T pseudo-reduced < 3
0.2 < P pseudo-reduced < 15
which works great for my needs (nitrogen gas, room temperature, pressures ranging from 10 to 500 bar).

Full PDF article downloadable from: https://link.springer.com/article/10.1007/s13202-015-0209-3

Thanks again everyone.

 

[Chestermiller]

I reply to my own question, in case anyone's interested. I found a great article which provides an analytic function (equation 14 on the article - the article also provides all the required constants/parameters) for the compressibility factor Z, without need to always rely on charts. Great for calculations in Excel.
This model is said to be really accurate in reproducing the experimental data for the following ranges of temperatures and pressures:
1.15 < T pseudo-reduced < 3
0.2 < P pseudo-reduced < 15
which works great for my needs (nitrogen gas, room temperature, pressures ranging from 10 to 500 bar).

Full PDF article downloadable from: https://link.springer.com/article/10.1007/s13202-015-0209-3

Thanks again everyone.

My comment was that the Z factor analysis is more accurate than the VDW equation for the range of values that interests you.

What value of the volume and z-factor did your article predict at 293K and 200 bars, and what value of pressure did it predict when the volume was reduced by a factor of 2?

That reference you suggested sounds great. I will download it. Thanks.

 

[FranzS]

My comment was that the Z factor analysis is more accurate than the VDW equation for the range of values that interests you.

I saw it, thanks!

What value of the volume and z-factor did your article predict at 293K and 200 bars, and what value of pressure did it predict when the volume was reduced by a factor of 2?

That reference you suggested sounds great. I will download it. Thanks.

Just rebuilt the Z-factor graph in Excel. I'll throw in some values and let you know.

 

[FranzS]

What value of the volume and z-factor did your article predict at 293K and 200 bars, and what value of pressure did it predict when the volume was reduced by a factor of 2?

Well, it wasn't as easy as I thought. Maybe you can help, thanks.
Known parameters are:

• initial volume (real) = 1 L
• initial pressure (real) = 200 bar
• final volume (real) = 0,5 L

Goal is to find final pressure (bar).

As said, I rebuilt the Z-factor v. Pressure chart in Excel using equation 14 in the article. Seems to work great!
Problem is that any Z-factor formula or chart needs you to know the pressure and gives you the "correction factor" (Z-factor) for the volume, whereas I'd like to have the opposite.
What I did in Excel was:

• column A = pressure (bar) [starting from 50 bar and up to 600 bar, with 0.1 bar increments]
• column B = Z-factor formula (from equation 14 in the article)
• calculating n (number of moles) from P_initial and V_initial (known values) and from Z_initial (easily calculated plugging P_initial in Z-factor equation)
• column C (starting from the row corresponding to P_initial = 200 bar): I plugged in the formula to calculate the Volume (i.e. V = (Z*n*R*T)/P ) for all following rows up to (actually, down to - if we talk Excel-wise!) 600 bar, for each of the in-between 0.1 bar increments)
• finally, I searched column C for the value of V closest to 0,5 L and looked at the corresponding pressure in column A

Is this the only way to "calculate" final pressure knowing the final volume? Or am I missing something trivial here? Is there an explicit formula for P_final? If I write down the EOSs, I always end up with two variables (P_final and Z_final) and can't solve them.

By the way, with the method described above, I got a final pressure P_final = 547.7 bar (for a corresponding Z-factor of 1.401).
For nitrogen, 547.7 bar is around 16.15*P_critical, so it's just out of the "super-accuracy range" (the extended chart, up to 600 bar, seems nice and smooth though).

Sorry for the long post! Thanks for your patience.

 

[Chestermiller]

You are trying to solve the implicit equation $$P=\frac{RT}{v}z(P)$$where v is the specific volume. This can be solved iteratively. Start with the ideal gas value $$z=z_0=1.0$$ and then calculate an initial estimate of P: $$P_0=z_0\frac{RT}{v}$$ Then apply the iterative equation $$P_n=\frac{RT}{v}z(P_{n-1})$$starting with n=1. This should converge pretty quickly for both P and z, so keep iterating until neither changes significantly.

 

[Swamp Thing]

Is this the only way to "calculate" final pressure knowing the final volume?

I haven't looked at your equations in detail, but I think you want to solve an equation that cannot be explicitly arranged with the unknown variable on one side.

Have you looked at the Solver feature in Excel? In case you haven't used it before, what you do is create an error-square cell somewhere in the sheet, that calculates the total of the squares of errors between the LHS and RHS of your equation(s). Then you ask the solver to vary the unknown variable cells until the error is minimized. I have found this very useful over the years.

It helps if your initial values for the unknowns are reasonably good, but it's often not very important.

 

[FranzS]

Thanks for the suggestions.
I managed to solve it in Excel by calculating the volume as $\frac{ZnrT}{P}$ for each single pressure and corresponding z-factor values (numbler of moles n is calculated from initial pressure, initial volume, and initial z-factor - all known values) and by automatically searching for the smallest difference between actual final volume (known value) and calculated volumes, and finally returning the pressure value corresponding to that volume (final pressure, which was the goal).

 

[Chestermiller]

Thanks for the suggestions.
I managed to solve it in Excel by calculating the volume as $\frac{ZnrT}{P}$ for each single pressure and corresponding z-factor values (numbler of moles n is calculated from initial pressure, initial volume, and initial z-factor - all known values) and by automatically searching for the smallest difference between actual final volume (known value) and calculated volumes, and finally returning the pressure value corresponding to that volume (final pressure, which was the goal).

To finish things off, why don't you just linearly interpolate at the very end.

 

[FranzS]

To finish things off, why don't you just linearly interpolate at the very end.

That would be a nice final touch, but I'm already using pressure increments of $0.1$ bar which are already way more precise than I actually need.
Afterthought: it's quite odd to think that someone came out with such a comprehensive formula for mimicking experimental z-factor values with good accuracy over a wide pressure range just a few years ago (I'm talking about the article I came across).

 

[Swamp Thing]

If I may, I'd like to bump up a question from my earlier post on this thread: How do we define the pressure in the middle of a volume of non-ideal (van der Waals) gas? Is it the pressure that would exist on an imaginary surface at that location? It there a definition that doesn't refer to such a surface?

 

[Chestermiller]

If I may, I'd like to bump up a question from my earlier post on this thread: How do we define the pressure in the middle of a volume of non-ideal (van der Waals) gas? Is it the pressure that would exist on an imaginary surface at that location? It there a definition that doesn't refer to such a surface?

That is the way I envision it. The normal force per unit area on the imaginary surface. Just like the tension in the middle of a string.

 

[Swamp Thing]

In that case one would have to assume a surface that screens nearby molecules against forces from molecules on the other side of the surface. Then the molecules hitting the surface would have reduced momentum due to attraction of molecules further away from the surface -- hence reduced pressure a la van der Waals.

In that sense the virial expansion picture (bulk effect) would be equivalent to the surface effect picture.

 

[Chestermiller]

In that case one would have to assume a surface that screens nearby molecules against forces from molecules on the other side of the surface. Then the molecules hitting the surface would have reduced momentum due to attraction of molecules further away from the surface -- hence reduced pressure a la van der Waals.

In that sense the virial expansion picture (bulk effect) would be equivalent to the surface effect picture.

It's an imaginary surface, not a real one. So it would do nothing. The pressure is the force per unit area exerted on the molecules on one side of a massless entity by the molecules on the other side of the massless entity. Are you not able to visualize a surface (massless) in space?

 

[Swamp Thing]

Are you not able to visualize a surface (massless) in space?

When I imagine a non-shielding surface, it seems to me that such a model might not capture the van der Waals deviation from ideal gas behavior. But let me have another think about this..

 

[Chestermiller]

When I imagine a non-shielding surface, it seems to me that such a model might not capture the van der Waals deviation from ideal gas behavior. But let me have another think about this..

In addition to the repulsive forces exerted by molecules from one side of the conceptual plane surface on those situated on the other side of the surface, in the case of a gas, there is actually flux of molecules (and accompanying momentum) from one side of the conceptual surface to the other other side. For an ideal gas, the latter is the only contributor to what we call the pressure.

The flux of molecules from one side of the conceptual surface to the other, in conjunction with the flux of molecules from the other side across in the opposite direction is equivalent to molecules at an actual rigid surface colliding with the rigid surface and bouncing back (i.e., exerting a force on the rigid surface).