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).

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?