November 16, 2024, 06:47:21 PM
Forum Rules: Read This Before Posting


Topic: MCAT style problem to teach  (Read 4118 times)

0 Members and 1 Guest are viewing this topic.

Offline rancatmyea3

  • Very New Member
  • *
  • Posts: 2
  • Mole Snacks: +0/-0
MCAT style problem to teach
« on: December 16, 2015, 05:26:04 PM »
Hello all,

I have to teach the following problem this weekend and I am having trouble with one part of it. Any help would be greatly appreciated!!!!



2. In a reaction at equilibrium, it is concluded that helium gas behaves more ideally than carbon dioxide. Which of the following accurately explains why this is so?


I. CO2 exerts a greater pressure because its molecules have lesser volume.

II. The intermolecular forces between CO2 molecules are stronger than those in He.

III. Helium molecules have greater kinetic energy, and therefore behave more ideally.

A. I only

B. II only

C. II and III only

D. I, II, and III


First, I am not sure what "reaction" would allow one to discern this, but that is besides the point.

My thoughts:

I. Wrong because a molecule of CO2(g) would occupy more volume than a molecule of helium(g).

II. True. He is a noble gas and therefore has minimal intermolecular forces. CO2 on the other is more polarizable and will have more temporary dipoles/van der waals forces.

III. I believe this is False, but I cannot come up with a completely exhaustive reason why. For gases I think of kinetic energy as average kinetic energy, and therefore equal to 3/2RT. No temperature is implied in this problem. I assume if the temperature of the 2 different gases was the same then they would have the same Kinetic energy (if we can assume they are both ideal, which seems unlikely in this problem). I have always thought of temperature as relying on the conditions that the gas is subject to rather than an intrinsic property of the gas itself. Also Van Der Waals equation does not directly account for variations in Temperature:   (P+a(n/V)^2) X (V - nb) = nRT.

So I thought about what would happen if you took two 1L containers, filled one with one mol CO2 and the other with one mol of He. Hold all conditions constant (except T), and then measured the temperature of each gas? Thinking about deviations from ideality, I noted that since CO2 has more attractive forces and more volume (on a molecular basis), these factors would lead to an decreased pressure and slightly increased volume. If you plug those into PV/nR = T then I suppose there would be a net effect of increasing the temperature and thus the KE relative to the container of He.

I am pretty sure that my thinking is flawed-- could someone please help me find where I have gone wrong?

Thank you so much!!!!




Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27846
  • Mole Snacks: +1812/-412
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: MCAT style problem to teach
« Reply #1 on: December 17, 2015, 02:57:11 AM »
I see nothing wrong with your way of thinking.

By definition molecules of the ideal gas don't occupy any volume and interact only when colliding. Whenever we talk about non-ideality we need to think about how the real gas differs from an ideal one, and that's what you did.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline rancatmyea3

  • Very New Member
  • *
  • Posts: 2
  • Mole Snacks: +0/-0
Re: MCAT style problem to teach
« Reply #2 on: December 17, 2015, 08:25:18 AM »
Thank you so much for the reply!!!

I realize that I didn't specifically state my problem well enough. By my thinking CO2 would have LESS (error in first post) kinetic energy because it has more attractive forces and thus less pressure than He. Answer III states "He has more kinetic energy and therefore behaves more ideally than CO2"? I agree that Helium could have more kinetic energy despite having lower mass than CO2, but is this an accurate explanation for why He behaves more ideally?

Thanks so much again for your time!!!


Offline mjc123

  • Chemist
  • Sr. Member
  • *
  • Posts: 2071
  • Mole Snacks: +302/-12
Re: MCAT style problem to teach
« Reply #3 on: December 17, 2015, 10:28:55 AM »
I don't think that's right. I think (someone correct me if I'm wrong) that temperature is essentially a measure of kinetic energy, and so the average kinetic energy is the same for all gases at the same temperature, whether they are ideal or not. The stipulation of equilibrium in the initial problem means the temperature is the same, therefore so is the KE, and III is wrong. (They might have been teasing you with the fact that He atoms move faster, because they have less mass; but they have the same KE.)
At normal pressures, PVm < RT for CO2 so at constant pressure it would have less volume than He, or at constant volume it would have lower pressure, but they would always have the same average KE at the same temperature. In your thought experiment above, if you adjusted the temperatures so the pressures were equal in the two containers (which I think is what you were doing), then the CO2 would be at slightly higher T than the He, but then they would not be in equilibrium.

Offline Vidya

  • Full Member
  • ****
  • Posts: 839
  • Mole Snacks: +46/-62
  • Gender: Female
  • Online Chemistry and Organic Chemistry tutor
    • Online Chemistry Tutor
Re: MCAT style problem to teach
« Reply #4 on: December 17, 2015, 10:42:40 AM »
He is behaving ideally because it has less attractive forces due to less molar mass and occupies negligible volume because of small size.It is very true all gases have same KE at the same temperature irrespective of their molar masses or volume.

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27846
  • Mole Snacks: +1812/-412
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: MCAT style problem to teach
« Reply #5 on: December 17, 2015, 11:08:26 AM »
I have definitely answered way too early, before my morning coffee ;) Still, most of what you wrote is OK, the only thing that I missed was this part:

So I thought about what would happen if you took two 1L containers, filled one with one mol CO2 and the other with one mol of He. Hold all conditions constant (except T), and then measured the temperature of each gas?

There is absolutely no reason why anything should change (assuming the gas is isolated).

Quote
I noted that since CO2 has more attractive forces and more volume (on a molecular basis), these factors would lead to an decreased pressure and slightly increased volume.

First, those non-idealities were already present in the gas used to fill the container, second, you said the container volume is 1 L, so I  don't see why the volume should change.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Sponsored Links