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EK Chemistry: In-Class Exam for Lecture 2, #26, #29

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ikjadoon

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I have a few questions about the EK Chemistry In-Class Exam for Lecture 2, if anyone understands what I was supposed to be thinking in #26 and #29.

On question #26,

Q: "Based on the information in the passage, under which of the following conditions does ammonia behave most ideally?"

A. Low temperatures and low pressures
B. Low temperatures and high pressures
C. High temperatures and low pressures [correct]
D. High temperatures and high pressures

The explanation in the back says "26. C is correct. You can figure this out from the passage, but it's a lot easier to fall back on your previous knowledge: gases behave most ideally at high temperatures and low pressures."

From, the passage, however, it looks as if LOW temperature behaves ideally (making me think A was the correct answer). I know high temperatures show the most ideal behavior, but the questions specifically asks the answer to be "based on the information in the passage." Why do I think low temperature looks to behave most ideally? This is from the passage, showing the "Z" factor curve for ammonia:
55f21b7dc91648f29d3c25d.png


It's a little crude because I don't have a scanner at home. :( So, right, Z = PV/nRT. For ideal gases, PV = nRT, so Z =1 for ideal gases. 300K seems the most ideal. 200K, what I considered "low temperature", seems next best and 500K, what I considered "high temperature", seems to be the least ideal.

The passage also states that a and b for ammonia are 4.3 and 0.037, respectively, in the Van der Waals equation:

images


How can I figure out from the passage that ammonia acts ideally at high temperatures? I understand the "background knowledge", but the question specifically asked from the information in the passage. Or do I need to ignore that and keep my background knowledge available for times like these?

#29

Q: "Why must absolute temperature be used in the Van der Waals equation?

A. Becaues the Van der Waals equation is a nonrelativistic equation.
B. Because it is impossible to have a negative absolute temperature.
C. Because ratios of temperatures on other scales, such as the Celsius scale, are meaningless. [correct]
D. Because international convention requires it.

I put B. I understand why both B and C are correct, but I would think that B is the more correct option. If I have a "negative" temperature, that implies I could have negative pressure or negative volume, both which are impossible (right?) and thus would totally mess up any numbers from the Van der Waals equation. But, I guess "C" is more right because the Van der Waals equation is used most often to compare (thus create ratios) between ideal and non-ideal states of gases?
 

MD9

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Sorry for the late response, and it might be useless now, but I wanted to comment because these two questions are also the two I got wrong when taking that practice exam. You also used the same "incorrect" answers as I did. My reasoning about why the answers we chose (i.e. #26-A, and #29-B) are more correct than the "correct" answers they provide is as follows:

#26

When the words "behaves most ideally" are used in the question, in reference to the ideal behavior of gases quantified by the ideal gas law, it implies the same thing as "has a compression factor Z=PV/(nRT) that is most closely approximated by Z=1, as per 'ideal' behavior given by the ideal gas law." When we talk about the "best approximations" concerning a data series being approximated by an equation, we can do least-squares analysis to find which data series is best. I don't mean that we need to do least-squares analysis on the MCAT or anything like that, but I am saying that "best approximation" means, fundamentally, the data series which deviates the least from the given mathematical function (which, in this case, is Z=1).

As shown in the image below, where I have drawn in blue the line symbolizing the compression factor of an ideal gas (Z = 1), it is clear that the data series for Z at T=200K and at T=300K fit the "ideal" behavior of the horizontal line Z=1 much better than the line at T=500K (i.e. if we sum the squares of differences between Z=1 and the lines made by T=200, T=300, and T=500, the sum would be much lower for the T=200 and T=300 lines than for the T=500 line, which deviates from Z=1 much more than the other two).

OdgnmVY.jpg


Yet, the "correct" answer simply expects us to ignore the graph of compression factors of ammonia, even though the question at hand asks specifically about the information in the graph--information that shows a result contrary to the general statement that gases behave most ideally at low pressures and high temperatures. I thought the MCAT was about applying general knowledge to situations in which exceptions might occur, and not instead ignore specific cases by blanketing them with general trends. It is obvious here that the best approximation occurs at low pressures (where the data series all converge to Z=1) and at the data series shown for T=200 and T=300, which are the lower temperatures and not the high temperatures that the "correct" answer specifies.

The solution/answer paragraph for this question is also extremely vague, stating that "You can figure [Choice C] out from the passage, but it's a lot easier to fall back on your previous knowledge: gases behave most ideally at high temperatures and low pressures." How exactly can we figure this out from the passage when the data displayed in the chart to which the question refers directly contradicts the answer given to the problem?

The justification for answer choice C, which they gave, would be to say that, "Given a low-pressure state, the high-temperature data series deviates the least from Z=1 in the low-pressure states occurring at the leftmost range of the plotted data," which would be true, since yes, the T=500K line does deviate the least from the line Z=1 at very low pressures (i.e. low molarities on the graph). Only when pressure increases does the 500K line deviate the most, so then the line that best fits Z=1 at low pressures is T=500K. If this is the case, then this should have been the justification in their solution explanation.

As for #29:
The question asks "Why must absolute temperature be used in the Van der Waals equation?"
Since answers A, B, and C are all true, we must choose the best answer among these three.

We can say off the bat that A is likely not the right answer because the reason absolute temperature is used has nothing to do with the equation's non-involvement with the principles of general relativity.

Next, we must consider answers B, "Because it is impossible to have negative absolute temperature," and C, "Because the ratios of temperatures on other scales, such as the Celsius scale, are meaningless."

We first notice that the Van der Waals equation here does not involve ratios in temperatures. It is not, for example, a derived, simplified form of the equation that may be used for comparison between two states like P1/P2=T1/T2. Instead, it simply requires that a value of temperature be "plugged in" to T when solving for some other unknown variable.

On the other hand, if a temperature scale in which it was possible to have a negative temperature were used, using such a negative temperature would yield a result in which either P is negative or V<nb, both of which are impossible statements.

As a result, using any temperature scale in which any temperature above absolute zero is given a negative value would completely invalidate the equation, whereas using ANY other linear temperature scale for which absolute zero was given the baseline T=0 would only require an adjusted value of R. For example, if we modified the Fahrenheit scale such that T=0 at absolute zero (which is defined as the Rankine temperature scale), we could use this modified-Fahrenheit scale without any problems in the Van der Waals equation as long as R was adjusted to use units of degrees Fahrenheit rather than Kelvin/Celsius (to take into account that given some temperature change, values on the Rankine and Fahrenheit scales increase more than the corresponding values on the Kelvin or Celsius scales).

Thus the only factor limiting the use of any temperature scale in the Van der Waals equation is that the scale be "zeroed" such that it reads 0 at absolute zero. From there the temperature reading on the temperature scale used could grow at any linear rate with respect to the temperature, and it would still work in the equation (as long as the units of R were modified accordingly).

Therefore, the best answer as to why "other" temperature scales cannot be used is because it is impossible to have a negative temperature in the context of the Van der Waals equation, which is defined from the context in which zero temperature corresponds to zero energy, and thus negative temperatures could not be used.

Another way to look at this is to consider a range of temperature scales. Let's consider the Celsius, Fahrenheit, Kelvin, and Rankine temperature scales (recalling that the Rankine temp. scale grows at the same rate as the Fahrenheit scale, except it is "zeroed" at absolute zero, much like the Kelvin scale grows at the same rate as the Celsius scale, except it is "zeroed" at absolute zero). Let's say we have the ratio between the temperatures 200 C and 5 C. The ratio 200C/5C equals 40. Converting these temperatures to Fahrenheit, we would get 392F and 41F, which gives us the ratio 392F/41F = 9.56. Clearly, 40 is not equal to 9.56, so ratio comparison between Celsius and Kelvin would yield invalid results. However, let's do the same comparison between the Kelvin and Rankine scales. In Kelvin, the corresponding temperatures would be 473.15K and 278.15K, giving us a ratio of 473.15K/278.15K = 1.70. On the Rankine scale, the corresponding temperatures would be 851.67R and 500.67R, giving us a ratio of 851.76R/500.67R = 1.70. We see that as long it is impossible to have negative temperature on a temperature scale, i.e. the temperature scale is "zeroed" to give a temperature of 0 at absolute zero, then we can use any such temperature scale in calculating ratios between temperatures and still have valid results.

This means that the answer choice C, "Because ratios of temperatures on other scales, such as Celsius, are meaningless," is only true for temperature scales that possess negative values for real temperatures, and thus answer choice C would only be valid if answer choice B were not valid, because only then would you run into problems with ratios of temperatures being meaningless. Thus B is the "best" answer, because it not only is the most relevant to the form of the Van der Waals equation to which the question refers, but also because the validity of answer C is dependent on whether answer B is true for whatever "other scales" answer C is considering.

Yet, in spite of this, the "correct" answer is listed in the book as being C ("because ratios..."), despite the fact that even at first glance, the unmodified Van der Waals equation itself possesses no ratios of temperatures to which answer C could apply.

Can anyone chime in on this? Is the real MCAT less likely or more likely to have "correct" answers like these, to which logic does not necessarily apply?
 
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1TB4RKSB4CK

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I think is mostly choosing the best and most pertinent answer that related directly to the question. BTWW, bumping since this guy did a pretty good job in explaining and I got that last question wrong. #26 was basically knowing your ideal gases rules. no intermolecular interactions so a high temperature in which molecules are flying all over the place, and low pressure in which molecules aren't succumbing to any forces that try to bring it closer together was my quick reasoning.
 

Schenker

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MD9 offers a thoughtful response, and this explanation

The justification for answer choice C, which they gave, would be to say that, "Given a low-pressure state, the high-temperature data series deviates the least from Z=1 in the low-pressure states occurring at the leftmost range of the plotted data," which would be true, since yes, the T=500K line does deviate the least from the line Z=1 at very low pressures (i.e. low molarities on the graph). Only when pressure increases does the 500K line deviate the most, so then the line that best fits Z=1 at low pressures is T=500K. If this is the case, then this should have been the justification in their solution explanation

for #26 is correct and much better than the out-of-the-book explanation that the OP quotes.

However, I disagree with the rest of MD9's response.

#26

it is clear that the data series for Z at T=200K and at T=300K fit the "ideal" behavior of the horizontal line Z=1 much better than the line at T=500K (i.e. if we sum the squares of differences between Z=1 and the lines made by T=200, T=300, and T=500, the sum would be much lower for the T=200 and T=300 lines than for the T=500 line, which deviates from Z=1 much more than the other two).

This is an unsafe assumption to make, because
1. there is no indication that the x-axis (Molarity, or pressure) is linear, and
2. there is no indication that the right edge of the graph represents an upper bound of molarity.
In fact, each T=200,300,500K line will continue on far past the right edge of the graph depicted in the question. The trend of concavity of the depicted portions of each graph (500K < 300K < 200K) suggests that the T=200K graph will surpass the T=300K graph on the y-axis and that the T=300K graph will surpass the T=500K graph. This implies ammonia acts more ideally at higher temperatures and is consistent with answer choice C.

Furthermore, it's unnecessary to make such assumptions about the differences between the T=200K graph, the T=300K graph, and the T=500K graph since from the answer choices it is clear that the question asks you to evaluate ideal behavior in terms of two variables (temperature and molarity). It is possible to unambiguously determine that answer choice C is the correct answer from the passage alone, without any background knowledge, and the first of my MD9 quotes is a good explanation of that.



#29

The correct answer is C. The question "Why must absolute temperature be used in the Van der Waals equation?" requires an answer that discusses differences between temperature scales, not an answer that confounds properties of temperature scales with properties of temperature.

Answer choice B makes a confounding error in that it describes a property of temperature, not a property of the absolute temperature scale. If we discovered that negative absolute temperature were possible, we wouldn't suddenly stop using the absolute temperature scale in the Van der Waals equation. In fact, it would be reasonable to assume that a gas with negative absolute temperature would exhibit a negative pressure. Thus, B cannot be the correct answer.

On the other hand, the fact that ratios on the Kelvin scale are meaningful is both necessary and sufficient to use the Kelvin scale in the Van der Waals equation. In fact, any temperature scale can be used in the Van der Waals equation as long as its ratios are meaningful. (Of course, one would have to be mindful of using an appropriate R.) Temperature scales that satisfy that condition are called absolute temperature scales.

This may be a little confusing, so here's an analogy:

Imagine that we are discovering physics from the ground up. We discover that objects can have positive electrical charge, and as far as we know, negative charge is undiscovered. We carefully derive Coulomb's Law, and we run experiments with positive charges to confirm. It checks out. Every time we bring two positive charges together, they repel each other with a force predicted by Coulomb's Law. In this world, a medical applicant sits for the MCAT and comes across the question "Why do we use Coulombs as a unit in Coulomb's Law?" and is presented with analogous answer choices to our temperature question.

The correct answer choice would be that we use Coulombs because ratios are meaningful; in fact, you can use any unit of charge in Coulomb's Law as long as ratios are meaningful, and it will work, just as you can use any temperature scale in the Van der Waals equation as long as ratios are meaningful. Of course, you would have to use appropriate units for k.

It's easy to see in this analogy why a test taker might choose the answer choice "because it is impossible to have negative Coulombs." As far as this test taker is concerned, negative charges are unknown; furthermore, the test taker is under the impression that it would be nonsensical for Coulomb's Law to predict a force with a negative sign, as that would imply that forces between charges can be attractive, which is unheard of to this test taker.
Because we know that charges can be negative, it's easy to see why in the hypothetical analogy the answer choice "because it is impossible to have negative Coulombs" is wrong.

I hope that helps.
 

phattestlewt

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I think is mostly choosing the best and most pertinent answer that related directly to the question. BTWW, bumping since this guy did a pretty good job in explaining and I got that last question wrong. #26 was basically knowing your ideal gases rules. no intermolecular interactions so a high temperature in which molecules are flying all over the place, and low pressure in which molecules aren't succumbing to any forces that try to bring it closer together was my quick reasoning.

There's a reasoning even quicker than that that has to do with phase changes.

Look at this beautiful work of art
upload_2013-12-7_18-57-42.png

It basically says that gases exist most ideally in high temp low pressure conditions. Solids are Low temp, high pressure and liquids are somewhere in the middle.

I hardly think you needed paragraphs of reasoning for this question.

That #29 though is a poor question as both B and C were equally valid imo.

Thanks for bringing this back from the dead....twice...
Poor guy had a first response 7 months after he posted it lol.
 
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