TBR Passages

[Majik]

I'm a little confused with these passages. I feel like I'm guessing on half the answers. I don't understand how we're expected to know this stuff, unless these questions are meant to be wrong. For example:

Passage 8 - Elevator Tension (Chapter 2)

Question 53 in the old book asks: "The tension in the cable connected to the passenger compartment is T1. The tension in the cable connected to the counterweight is T2. The mass of the passenger compartment is M, and the mass of the counterweight is m. If the pulley is locked by means of a brake, what can you conclude about the two tensions?"

I assumed that the answer was D. T1=T2 under any circumstances, but the correct answer was C. "T1 does not equal T2 unless Mg=mg" and the reasoning was because: "When the pulley is locked, the tensions in the left and right parts of the cable are not necessarily equal. The locked pulley grips each part of the cable and makes the tensions in the left and right parts independent, as though there were two separate cables rigidly attached to the locked pulley."

Is this a random fact we're expected to know or was I suppose to reason this out. Because honestly, I don't see how.

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[whiteshadodw]

I'm a little confused with these passages. I feel like I'm guessing on half the answers. I don't understand how we're expected to know this stuff, unless these questions are meant to be wrong. For example:

Passage 8 - Elevator Tension (Chapter 2)

Question 53 in the old book asks: "The tension in the cable connected to the passenger compartment is T1. The tension in the cable connected to the counterweight is T2. The mass of the passenger compartment is M, and the mass of the counterweight is m. If the pulley is locked by means of a brake, what can you conclude about the two tensions?"

I assumed that the answer was D. T1=T2 under any circumstances, but the correct answer was C. "T1 does not equal T2 unless Mg=mg" and the reasoning was because: "When the pulley is locked, the tensions in the left and right parts of the cable are not necessarily equal. The locked pulley grips each part of the cable and makes the tensions in the left and right parts independent, as though there were two separate cables rigidly attached to the locked pulley."

Is this a random fact we're expected to know or was I suppose to reason this out. Because honestly, I don't see how.

LOL, that is one of the TBR lol questions that just makes you LOL. its like, ok, find i buy your reasoning, but i doubt that'll be on the test.

so far i've taken AAMC 3, 4, 5, and 7, and that sort of a question comes up on NONE of them.

 

[PingPongPro]

When I did that passage/question I got it right and had the same line of thinking as the BR explanation. I don't think this specific question is really a "random fact". It could be explained better, but the lock makes the two cables two separate systems. Its kind of an obscure example, but if you encountered the set-up in real life, you probably would be able to reason it out pretty easily. I think that is the key for a lot of the physics passages. Try to think about it with real life application and use your physics knowledge to choose the best answer.

It is true that BR often has experiments or setups that aren't explained very clearly, but don't let it deter you too much.

 

[Majik]

Alright good to know. Thanks for the advice guys. Glad to see I'm not the only one who was totally thrown off by that, ha.

 

[oasis67]

For passage V of the Phases and Phase Change chapter, question 29, how do you know that the vapor pressure of II is greater than that of I. I did all the calculations just to make sure, and they both come out to a vapor pressure of .59. I'm confused as to how do do this without any math and also I don't understand the reasoning behind II being greater than I. Hopefully, someone has the book and can look up the passage. Any help will do! Thanks!

 

[salim271]

For passage V of the Phases and Phase Change chapter, question 29, how do you know that the vapor pressure of II is greater than that of I. I did all the calculations just to make sure, and they both come out to a vapor pressure of .59. I'm confused as to how do do this without any math and also I don't understand the reasoning behind II being greater than I. Hopefully, someone has the book and can look up the passage. Any help will do! Thanks!

Just look for the beaker that has the most methanol. I and III use grams and moles, and ethanol has the higher molecular mass so when adding the same amount of grams and moles, you get more ethanol in the solution than methanol so the vapor pressure is lower. In number II you're adding mLs, so you look at the densities to find out how many grams are added in solution per mL. Methanol has a slightly higher density than ethanol so more methanol is present in beaker II and it has the highest vapor pressure.

Doing it this way you dont even really have to even touch a pencil to paper or try to use Raoult's law in your head.

 

[oasis67]

Just look for the beaker that has the most methanol. I and III use grams and moles, and ethanol has the higher molecular mass so when adding the same amount of grams and moles, you get more ethanol in the solution than methanol so the vapor pressure is lower. In number II you're adding mLs, so you look at the densities to find out how many grams are added in solution per mL. Methanol has a slightly higher density than ethanol so more methanol is present in beaker II and it has the highest vapor pressure.

Doing it this way you dont even really have to even touch a pencil to paper or try to use Raoult's law in your head.

Thanks...that really helps. But I still don't understand how you know the vapor pressure of II is greater than I. How can you compare the two mole fractions without calculating them? Maybe I am not understanding the concept of density properly? The density they give means grams of methanol/milliliters of H2O right? I know that methanol vapor pressure is greater in all the beakers but I don't know how to compare beakers I and II.

 

[oasis67]

Also, how did they know the original mixture was initially 32 torr for 2-prop and 4 torr for acetophenone for question 43?

 

[salim271]

For your first question, you dont have to do the actual calculation of mole fraction and rauolt's law to understand why the highest VP of methanol will be when methanol is the greatest in solution. The presence of ethanol decreases methanol's vapor pressure because it takes up space at the top of the solution so less methanol is breaking away from the solution at any temperature. Methanol will therefore have the highest vapor pressure when there is more methanol because there will be less ethanol taking up space. So then its just a matter of finding out which beaker has more methanol than ethanol, and if you just do some quick dimensional analysis of grams to moles for I, moles to grams for III, and mls to grams for II, you see than I and III have equal moles of ethanol and methanol, and II has more methanol than ethanol because methanol has a higher density per mL.

As for your second question, the calculation is actually required. You start off with 60 grams acetophenone (about .5 moles) and 60 grams 2-propanol (1 mole). The total moles of soln then is 1.5 moles. The mole fraction of acetophenone therefore is (.5)/(1.5) = .33, therefore the Raoult's law deviation from standard vapor pressure is (.33)(12) = 4 torr. Same for 2-propanol to get 32 torr. 4/32 = 1/8, or 1 part aceto to 8 parts prop, answer choice D.

 

[oasis67]

Ohh! So the greater the difference between ethanol and methanol present in the beaker, the more methanol will be in the vapor phase (provided the greater number is of methanol when calculating the difference) because the less ethanol there will be on the surface to hinder it?

For the second question: mole distribution percentage means vapor pressure percentage? Because I thought it was comparing the amount of moles so I chose B (33% to 66%)