Negative Excess Molar Volume

Factual Questions
1

This is not homework. I got a free college algebra book at a library book sale some time ago. The other day I was looking at some of the problems and I came across this problem:

“An automobile radiator contains 16 L of antifreeze and water. This mixture is 30 % antifreeze. How much of this mixture should be drained and replaced with pure antifreeze so that the final mixture will be 50 % antifreeze.”

This problem might lead students to think that volume is a conserved property when dealing with mixtures of liquids. This is not always the case. For example, at Normal Temperature and Pressure, one liter of ethanol mixed with one liter of water will produce approximately 1.92 liters of ethanol/water mixture. This may seem counterintuitive (or, as we say back in Arkansas, f***ed up) at first but think about what happens when you add one liter of fine sand to one liter of coarse gravel. This phenomenon of reduced total volume of mixtures is not uncommon when dealing with polar liquids and is sometimes referred to as negative excess molar volume. Excess molar volume of a particular binary system can be described by a form of the Redlich-Kister equation, however some of the coefficients used in the equation must be empirically determined.

To analyze this problem further, it will be necessary to make some assumptions. The first assumption is that “antifreeze” means pure ethylene glycol. Admittedly, most commercially available automotive antifreezes contain minor components other than ethylene glycol and some may be based on something other than ethylene glycol. However, with the vast multitude of products available on the market, we will never get anywhere if we have to consider each individual case. Because excess molar volume is temperature and pressure dependent, the second assumption is that all measurements will be made at Normal Temperature and Pressure (68 degrees F. and an absolute pressure of 1 atmosphere). The third assumption concerns the definition of concentration. There are several different ways of defining concentration, including, but not limited to, for example, mass/mass concentration (mass of a constituent divided by the mass of the mixture), mass/volume concentration (mass of a constituent divided by the volume of the mixture), molar fraction (the number of moles of a constituent divided by the number of moles of all components of a mixture), and volume concentration (the volume (prior to addition) of a constituent divided by the final volume of the mixture). If the problem is referring to molar fractions then further analysis of the problem would be quite simple. However, in the context of this algebra book and with the mention of the volume of the radiator, I am quite certain that the author had volume concentration in mind and this is the assumption I will make.

Ethylene/glycol water mixtures have been well studied. It is known that ethylene glycol/water mixtures have a negative excess molar volume. I am unable to find literature which includes the molar fraction of ethylene glycol in ethylene glycol/water mixtures described as 30 % ethylene glycol by volume or 50 % ethylene glycol by volume. However, if I had precise and accurate data for the densities of 30 % ethylene glycol by volume and 50 % ethylene glycol by volume mixtures, then I believe I would have enough information to go forward. Does anyone disagree that this would give me adequate information to solve the problem?

Does anyone know where I can find the densities (or similar properties) of ethylene glycol/water mixtures when the concentration is expressed in percentage of ethylene glycol by volume?

There will certainly be some critics who say that it is silly to worry about negative molar volume when adjusting the antifreeze concentration in your personal automobile. They would be correct. Nonetheless, I am pretty sure that the answer to the problem will include a negative excess molar volume that is easily measurable by readily available measuring devices (but maybe not readily available at your house). What if this were part of some critical and expensive research project concerning automotive cooling systems and precise and accurate information was essential? What if we were talking about 10 million radiators?

2

Does this page contain numbers you can use? It has “Specific Gravity” (density relative to pure water) of Ethylene Glycol based Water Solutions by % volume. And “Densities” of Ethylene Glycol based Water Solutions by mass fraction.

3

It does not have the specific gravity for 68 degrees F. but I could interpolate between 40 degrees F. and 80 degrees F. and get an approximate specific gravity. I can convert from specific gravity to density easily enough. (The density of water is not exactly 1 g/mL like a lot of people think and it does vary with temperature. Water’s density at various temperatures is well documented. With only four significant digits in the ethylene glycol data, it probably doesn’t make any difference anyway.)

Thank you Frankenstein Monster.

Does anyone know of more precise data at 68 degrees F.?

4

Here’s another: PDF link

Figure 17 (page 28) has graphs and a formula that should let you interpolate more accurately.

Table 6 (page 11) has a conversion table between weight % and volume % at 68degF

5

Are you assuming that the percentages are by volume? And if so, are you assuming that they sum to 100% (i.e., the volume of one pure fluid divided by the sums of volumes of pure fluids), or that the percentage is volume of pure fluid divided by volume of mixture (in which case, a solution could be 52.1% ethanol, 52.1% water)?

6

Your 2nd assumption, 68ºF, seems to be a bit unrealistic in actual practice.

The time when one checks the antifreeze in vehicles, especially if increasing the percentage of antifreeze, is late fall. Probably quite a bit colder than 68º at that time.

P.S. I think you’re doing what is called ‘overthinking the question’. Consider what the designer of the test expected as a correct answer. Given that it’s an algebra textbook, not a science one, probably all these assumptions & considerations were never considered.

My first guess would be as follows:
16L of 30% = 4.8L of antifreeze + 11.2L of water.
desired result of 16L of 50% = 8L each of antifreeze & water.
Therefore add 3.2L of 100% antifreeze.

But I don’t think that is correct.

7

That won’t work, because if you drain out 3.2L of the original mixture, you’d also lose some antifreeze, and so you’d come up short.

But yes, I think the OP knows full well that he’s overthinking this, and that the whole point of this is just for fun.

8

I am using the conventional definition of concentration by volume which is the volume of the constituent of concern (before being added to the mixture) divided by the final volume of the mixture. Concentration by volume is a dimensionless quantity.

9

It is true that I am doing this just for fun. However, I was also trying to make a point with some non-SDMB people that sometimes textbooks can mis-educate (is that a word?) students. In this example, a student who does do not study the natural sciences might remember this problem and think for the rest of his or her life that volumes are conserved when mixing two or more different liquids and this almost never the case.

I have a long meeting at work tomorrow morning. I can work on this problem and they will think that I am enthusiastically paying attention and taking notes. Would anyone at the SDMB be interested in my results?

10

I only chose 20 deg. C (68 deg. F) because that is often used when defining “normal” conditions. Perhaps we can find other solutions to the problem at different temperatures using the reference provided by Frankenstein Monster.

11

I just want to point out that concentration by weight is also a dimensionless quantity and is what chemists typically use to characterize the chemical activity of different chemical species in a solution. It also conveniently sidesteps the whole issue that volumes are not conserved when mixing solutions. In fact, in many solutions, such as concentration of fluoride in your drinking water, or the carbon content of a tool steel, the concept of concentration by volume is rather meaningless.

And, yes, I know that alcoholic beverages are typically listed as alcohol by volume (ABV). I suspect this is because alcohol is often taxed based on concentration, and for the same concentration, the percentage by volume will be greater than the concentration by weight, so basing the taxes on ABV allows the government to collect more taxes.

12

That’s only if you’re assuming they’d be using the same numbers in the tax tables either way. They could just as easily have made it by weight and increased the numbers, to get the same taxes. More likely, it’s just a historical accident that alcohol concentration is traditionally measured by volume, and so when the laws were written, they used the same standard everyone else was.

13

Jasmine’s solution:

A gentleman drives into my garage and makes the above request. After some thought and after applying some “new math”, I come up with this solution:

“Sir, in checking your system, we noticed a lot of particles in the coolant indicating that flushing the entire system is definitely in order. We can do that for you and give you fresh coolant at a 50/50 ratio within the hour!” :slight_smile:

(As Jasmine counts her extra money) “Hah! Algebra, schmalgebra!”

14

I think the issue is that you’re looking at a math (Calculus?) book, not a chemistry book. While what you’re discussing, I assume, comes up in chemistry, mixing problems are common Calc/Diff Eq work. Plenty of things are ignored/assumed. For example, it’s assumed that the two fluids are, and will be perfectly mixed, nothing clings to the walls, nothing evaporates etc.
It’s homework to help you understand how to work the equations, the physics and chemistry would get added in later as you work towards a specific major/grad degree.

As a math student, I wouldn’t be lead to believe that the sum of the two volumes equals the new volume, it’s what I would have already believed. Had I pursued chemistry, I assume I would have been taught that. But math classes are primarily there to teach you how to work with equations and doesn’t tend to deal with fringe issues like this.

15

I would be interested in seeing your results!

I worked out the answer to the original problem. You pour out 32/7ths of a liter (= approx 4.57 liters) of mix, which gets rid of 3.2 liters water leaving 8. Then add 32/7ths of a liter ethylene glycol to get fifty-fifty.

There was originally 33.6/7ths of a liter of ethylene glycol, and since you poured out 9.6/7ths and add back 32/7ths you finish with 56/7ths or 8 liters of both.

16

That’s what I got working the problem the way that the textbook author intended. I had just got started on working the problem using the second cite you gave me but then my boss found a spider in the women’s restroom and now I am sitting at my desk with a spider in a beaker and googling New Mexico spider species. I may be able to waste the rest of the day doing this. My preliminary guess is that it is a Western Parson Spider.

17

OK, I want Ynnad’s job.

18

nm

19

Correct. As an earlier poster pointed out, this is a problem from an ALGEBRA textbook, not a CHEMISTRY textbook; the point of the exercise is to develop a mastery volume/concentration calculations, not to explore how volume/concentration calculations are interdependent with the arcane properties of chemistry and chemical processes.

20

Negative Excess Molar Volume and Western Parson Spider are both great band names.