PV=nRT question

[noobtech]

A decrease in the temperature of a gas sample in a closed container will always result in an increase in the:
C. gas pressure on the container.
D. attraction between gas molecules.

... My answer is C. Since the pressure inside the container is decreased, the pressure outside the container increased. This change is more dramatic than the attraction between gas molecues.
The right answer is D. WTF?

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[MedPR]

A decrease in the temperature of a gas sample in a closed container will always result in an increase in the:
C. gas pressure on the container.
D. attraction between gas molecules.

... My answer is C. Since the pressure inside the container is decreased, the pressure outside the container increased. This change is more dramatic than the attraction between gas molecues.
The right answer is D. WTF?

The question doesn't state that it is a rigid container. The volume could be decreasing and pressure remaining the same.

 

[DNA]

Temperature and Pressure are directly proportional. With volume being constant, pressure would have to decrease if temperature decreases. Gas behave most ideal at high temperatures and low pressures. Because the temperature is lower, kinetic energy and therefore, the velocity of the individual molecules have dropped as well.

Every molecule is attracted to one another in someway. This force of attraction is usually disregarded when these molecules are moving so fast, they just collide into one another but bounce back in the opposite direction. But at slow velocities (lower temperature), this attraction cannot be disregarded, which is why D is the best answer choice.

 

[SpecterGT260]

This question doesn't make sense... ideal gasses don't attract to begin with. The answer is neither. What were your other choices?

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[SpecterGT260]

Temperature and Pressure are directly proportional. With volume being constant, pressure would have to decrease if temperature decreases. Gas behave most ideal at high temperatures and low pressures. Because the temperature is lower, kinetic energy and therefore, the velocity of the individual molecules have dropped as well.

Every molecule is attracted to one another in someway. This force of attraction is usually disregarded when these molecules are moving so fast, they just collide into one another but bounce back in the opposite direction. But at slow velocities (lower temperature), this attraction cannot be disregarded, which is why D is the best answer choice.

I disagree. 1) this is the ideal gas law and ideal gases exert no forces on each other other than energetic repulsion through collision. 2) even if we allow an attractive force, it is independent of temp. One could argue that decreased repulsion equals increased attraction but that is like arguing that flapping your arms really fast decreases gravity. You can overcome an attraction or fail to overcome one, but this system does not describe a change in such a force. I suspect an oddly worded option A, B, or maybe E as the answer

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[DNA]

I disagree. 1) this is the ideal gas law and ideal gases exert no forces on each other other than energetic repulsion through collision. 2) even if we allow an attractive force, it is independent of temp. One could argue that decreased repulsion equals increased attraction but that is like arguing that flapping your arms really fast decreases gravity. You can overcome an attraction or fail to overcome one, but this system does not describe a change in such a force. I suspect an oddly worded option A, B, or maybe E as the answer

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The theories of the ideal gas law fall apart at lower temperatures and/or higher pressures. The question states that there's a decrease in temperature, which would indicate that the sample in question is not behaving ideally. So then you'd have to consider non-ideal forces acting on the gas molecule. A non-ideal (real gas) does have both attractive and repulsive forces and those forces are more significant when molecules are moving slower (lower temp) and/or bouncing more frequently into each other (high pressure).

I think what you seem to be saying is that the intrinsic value of attraction/repulsion (ie. the electric dipole charges) for a given molecule does not change, which is technically true. But for the sake of understanding this question, consider two pieces of a magnet that are attracted to each other. As you slowly bring both pieces of the magnets together, the magnetic force increases because the distance is decreasing. We all can understand this intuitively. But let's say that we're moving both magnets so rapidly in different directions at an extremely fast velocities. The attraction would still be there, but because the magnets are heavily driven by some external force, the force of the magnets themselves are not as significant, and for the most part can be disregarded.

 

[SpecterGT260]

The theories of the ideal gas law fall apart at lower temperatures and/or higher pressures. The question states that there's a decrease in temperature, which would indicate that the sample in question is not behaving ideally. .

No. This is test taking 101. Don't assume. A decrease could mean -0.0000001 degrees. No impact on level of "idealness". Don't reach to fit an answer. The question states "always". Is this always true? Id we are assuming why can't I assume a gas that naturally repels itself? All else left equal D really can't ever be the right answer. There are other choices the OP didn't give us. Wait for those.

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[DNA]

No. This is test taking 101. Don't assume. A decrease could mean -0.0000001 degrees. No impact on level of "idealness". Don't reach to fit an answer. The question states "always". Is this always true? Id we are assuming why can't I assume a gas that naturally repels itself? All else left equal D really can't ever be the right answer. There are other choices the OP didn't give us. Wait for those.

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I disagree. I think being overly critical of questions on the science section of the MCAT is a recipe for disaster. That type of approach would do you well in the verbal section of the exam, but for all intents and purposes, we should assume that a change in one variable is significant enough to effect other variables in someway. After all, we're being tested on understanding relationships between different variables. That's my take anyways, and it's proven to be very beneficial for me. But I do agree that without the other answer choices, you cannot be entirely certain which answer is correct.

Also, if like you said, the change was insignificant, then I'm sure it would be disregarded entirely.

 

[SpecterGT260]

I disagree. I think being overly critical of questions on the science section of the MCAT is a recipe for disaster. That type of approach would do you well in the verbal section of the exam, but for all intents and purposes, we should assume that a change in one variable is significant enough to effect other variables in someway. After all, we're being tested on understanding relationships between different variables. That's my take anyways, and it's proven to be very beneficial for me. But I do agree that without the other answer choices, you cannot be entirely certain which answer is correct.

Also, if like you said, the change was insignificant, then I'm sure it would be disregarded entirely.

ok....

The point is it says "always". You are introducing information not given in the question stem by assuming gas type (that attractive forces even exist) and deviation from ideal. You then land on an answer that is conditional which ignores the use of "always" and yes, ideal gasses have properties that always vary the same way whether it is a small or large change in the variable. They are hypothetical, so assuming we suddenly lose ideal behavior in the hypothetical gas is ridiculous to begin with. If you are comfortable with your answer, good for you. I think that answer is 100% incorrect as stated. But for the record, you are being overly critical of the question. I am simply reading it and answering what was asked. Don't do what you're doing on the mcat if you want to do well.

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[noobtech]

I'm sorry for not providing enough information for the addressed question.
This problem is from a passage describing gas not being ideal at extreme pressures. (10 to 1000 atm)

Here are the other 2 choices...neither of these is right.
A. value of the PV / (RT) ratio.
B. kinetic energy of individual gas molecules.

My problem was that...the volume is constant...decreasing temperature means decreasing internal pressure; hence pressure ON the container increases. And I assumed this change is more dramatic than intermolecular forces, which is wrong.

 

[SpecterGT260]

Using your same type of logic you could argue that the pressure inside the container decreases so relative pressure outside or "on" increases. This also makes several inappropriate assumptions. Both answers are wrong. Wait for the OP to give the other answer choices he excluded.

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[MedPR]

I'm sorry for not providing enough information for the addressed question.
This problem is from a passage describing gas not being ideal at extreme pressures. (10 to 1000 atm)

Here are the other 2 choices...neither of these is right.
A. value of the PV / (RT) ratio.
B. kinetic energy of individual gas molecules.

My problem was that...the volume is constant...decreasing temperature means decreasing internal pressure; hence pressure ON the container increases. And I assumed this change is more dramatic than intermolecular forces, which is wrong.

decreasing internal pressure means the pressure on the container decreases. Pressure "on the container" refers to how "hard" the gases inside the container are pushing on the inner walls.

In the case of a rigid container (constant volume), the pressure inside the container does not have any effect on the pressure outside the container. I think that's what you are missing/confused about.

 

[dsoz]

I'm sorry for not providing enough information for the addressed question.
This problem is from a passage describing gas not being ideal at extreme pressures. (10 to 1000 atm)

Here are the other 2 choices...neither of these is right.
A. value of the PV / (RT) ratio.
B. kinetic energy of individual gas molecules.

My problem was that...the volume is constant...decreasing temperature means decreasing internal pressure; hence pressure ON the container increases. And I assumed this change is more dramatic than intermolecular forces, which is wrong.

pressure ON the container does not change.

This question doesn't make sense... ideal gasses don't attract to begin with. The answer is neither. What were your other choices?

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It states it is a gas, not an ideal gas.

There are REAL attractive forces when you discuss REAL gasses. The ideal gas law gives you the conceptual basis to predict what will happen with real gasses, but is not perfect (or ideal).

When the temperature decreases, the kinetic energy of the molecules decreases and the attractive forces between the molecules increase. End of story.

 

[MedPR]

pressure ON the container does not change.



It states it is a gas, not an ideal gas.

There are REAL attractive forces when you discuss REAL gasses. The ideal gas law gives you the conceptual basis to predict what will happen with real gasses, but is not perfect (or ideal).

When the temperature decreases, the kinetic energy of the molecules decreases and the attractive forces between the molecules increase. End of story.

Attractive forces don't increase. Attraction increases because there is less kinetic energy to cancel out the attractive force. It may be a trivial point, but it's important considering the nature of the MCAT.

 

[Temperature101]

The answer was d then... People on sdn is so smart. I look at the problem as a proportional relationship, but the smart ...ss on sdn look at it as the behavior of gas molecules. You gotta love sdn for in depth thinking these members bring to these questions.

 

[MedPR]

The answer was d then... People on sdn is so smart. I look at the problem as a proportional relationship, but the smart ...ss on sdn look at it as the behavior of gas molecules. You gotta love sdn for in depth thinking these members bring to these questions.

You can use the relationship of variables to eliminate the other 3 incorrect answers too.

 

[Temperature101]

Attractive forces don't increase. Attraction increases because there is less kinetic energy to cancel out the attractive force. It may be a trivial point, but it's important considering the nature of the MCAT.

This. And all the other answers are false from a proportionality standpoint. Even if someone does not know the behavior of gas molecule, they should pick d.

 

[akim6890]

The problem has nothing to do with the pressure ON the container. The pressure is constant on the container because it's a closed system like the question says. Review deviations from the ideal gas law and conditions for the ideal gas law. Ok, when you approach an ideal gas problem, think about in a kind of yes/no diagram (although not all the questions might be yes or no; you get the idea). This is how I approach a lot of problems in the physical section especially with LAWS. Eventually you'll get very very fast.

1) Is it at 273K and 1 atm? No.
2) How does it deviate? Higher pressure, lower temperature.
3) What are the rules for deviations from the ideal gas law?
Lower volume, higher pressures, lower temperatures cause the gas to participate in more intermolecular interactions. (Look up why for each!).
4) Match it up, boom.

If you're not 100% after number 4, do process of elimination. You should still have time after being so fast.

a) PV/RT ratio increase? That might be tempting but when temperature goes down, so does pressure so it stays constant. Cross off.
b) KE of gas molecules increases? Intuitively, you should know this isn't correct but from the equation KE = 3/2(kT), you can cross it off.
c) Gas pressure on the container? Closed system, pressure inside doesn't effect outside so cross it off.
d) You're left with one answer, mark it. Move on.

 

[SpecterGT260]

pressure ON the container does not change.



It states it is a gas, not an ideal gas.

There are REAL attractive forces when you discuss REAL gasses. The ideal gas law gives you the conceptual basis to predict what will happen with real gasses, but is not perfect (or ideal).

When the temperature decreases, the kinetic energy of the molecules decreases and the attractive forces between the molecules increase. End of story.

Attraction is dependent on the nature of the gas. You can't just assume there are attractive forces because they don't say otherwise. What if its argon? You guys are making assumptions here that aren't valid. In terms of it acting as an ideal gas you can assume that because 1) the degree of temp change and the volume are not given. Nor is n. You have nothing to tell you that deviation occurs 2) you are give. The ideal gas equation for the basis of your question 3) it says always. I cannot stress this enough. Always means always. A single example of an answer foil that doesn't fit excludes the answer entirely.

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[SpecterGT260]

I'm sorry for not providing enough information for the addressed question.
This problem is from a passage describing gas not being ideal at extreme pressures. (10 to 1000 atm)

Here are the other 2 choices...neither of these is right.
A. value of the PV / (RT) ratio.
B. kinetic energy of individual gas molecules.

My problem was that...the volume is constant...decreasing temperature means decreasing internal pressure; hence pressure ON the container increases. And I assumed this change is more dramatic than intermolecular forces, which is wrong.

What is the source of this question (which qbank or book)? Its awful....

I would actually answer A. Its still a bad answer but technically speaking, if we talk about non ideal gases, temp varied directly with temp (duh), but pressure can't. It varies very closely but will have to be slightly behind because the atoms are space occupying. If that is the case, any decrease in temp will cause that ratio to increase, even slightly.

Is there an answer key for this? Its a terrible question either way and it wouldn't surprise me if they want D. But both answers require assumption of non ideal (which is ok given the passage) but A exploits a technicality and D is a result of poor wording at best, and inappropriate assumptions (an attractive non ideal gas and not a repulsive one) which cannot satisfy "always"

Does the passage give any other equations?

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[DNA]

I think you killed the thread, lol.

 

[SpecterGT260]

Hopefully. This is a testing technical thread. Prolonged discussion means something is wrong.

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[MedPR]

I disagree. I think being overly critical of questions on the science section of the MCAT is a recipe for disaster. That type of approach would do you well in the verbal section of the exam, but for all intents and purposes, we should assume that a change in one variable is significant enough to effect other variables in someway. After all, we're being tested on understanding relationships between different variables. That's my take anyways, and it's proven to be very beneficial for me. But I do agree that without the other answer choices, you cannot be entirely certain which answer is correct.

Also, if like you said, the change was insignificant, then I'm sure it would be disregarded entirely.

No, it's a recipe for success.

 

[SpecterGT260]

No, it's a recipe for success.

Disagree again
I just think he was confused on who was guilty of that. I've literally answered thousands of 3rd order questions at this point and I can't think of any time where "yeah but they didn't tell is it isn't_____" was sound test taking logic.

If an answer has to be rationalized by altering the question by adding or subtracting, shy away. The mcat is different than med school because you can't argue points back so that earlier poster who said D by process of elimination had the best advice for test taking. That said, this is a concept question. Who cares what the right answer here is? If you try to shoehorn the answer in by rationalizing it you set yourself up to get it wrong when it actually counts. Don't do that!

The concept is that ideal gasses are spaceless and exert no forces on each other. Non ideals are space occupying and do (BOTH attractive and repulsive). Apply that to a less crappy question and you'll be good.

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[MedPR]

Disagree again
I just think he was confused on who was guilty of that. I've literally answered thousands of 3rd order questions at this point and I can't think of any time where "yeah but they didn't tell is it isn't_____" was sound test taking logic.

If an answer has to be rationalized by altering the question by adding or subtracting, shy away. The mcat is different than med school because you can't argue points back so that earlier poster who said D by process of elimination had the best advice for test taking. That said, this is a concept question. Who cares what the right answer here is? If you try to shoehorn the answer in by rationalizing it you set yourself up to get it wrong when it actually counts. Don't do that!

The concept is that ideal gasses are spaceless and exert no forces on each other. Non ideals are space occupying and do (BOTH attractive and repulsive). Apply that to a less crappy question and you'll be good.

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I guess we have different ideas of what being overcritical of a question means. Sticking to "closed container" and "always" almost immediately leads you to answer D. Being less critical about the question stem leads you down a path that has you working out the relationships in PV=nRT and/or thinking through the list of gaseous qualities.

 

[Ragtime]

A decrease in the temperature of a gas sample in a closed container will always result in an increase in the:
C. gas pressure on the container.
D. attraction between gas molecules.

... My answer is C. Since the pressure inside the container is decreased, the pressure outside the container increased. This change is more dramatic than the attraction between gas molecues.
The right answer is D. WTF?

I think the answer is D only because of the process of elimination. When temperature is decreased, doesn't pressure also decrease? Sorry if I'm wrong, I have only taken an Intro Chem course.

 

[SpecterGT260]

I guess we have different ideas of what being overcritical of a question means. Sticking to "closed container" and "always" almost immediately leads you to answer D. Being less critical about the question stem leads you down a path that has you working out the relationships in PV=nRT and/or thinking through the list of gaseous qualities.

I guess I meant "overthink". Basically, there are just things about D that I am very uncomfortable with. Namely the increase in attractive force. This implies that volume contraction is due to increased attraction rather than decreased kinetics. This is why I am wondering if there was other info in the question stem. Type of gas, something about attraction or repulsion, or a coefficient to modify volume or pressure due to non-ideal behavior.

But in here we've been playing this "well IF it has x, and then IF it has Y, and then IF it has...." game which.... is just a really bad idea to do on a test question. Usually, only the information you are given can be assumed unless there is a standard out there somewhere. And if the question is inclusive in nature (using "always" in the stem) if your assumption is not a given and a situation where something with the opposite characteristic of your assumption is false then the assumption is false.

 

[MedPR]

I guess I meant "overthink". Basically, there are just things about D that I am very uncomfortable with. Namely the increase in attractive force. This implies that volume contraction is due to increased attraction rather than decreased kinetics. This is why I am wondering if there was other info in the question stem. Type of gas, something about attraction or repulsion, or a coefficient to modify volume or pressure due to non-ideal behavior.

But in here we've been playing this "well IF it has x, and then IF it has Y, and then IF it has...." game which.... is just a really bad idea to do on a test question. Usually, only the information you are given can be assumed unless there is a standard out there somewhere. And if the question is inclusive in nature (using "always" in the stem) if your assumption is not a given and a situation where something with the opposite characteristic of your assumption is false then the assumption is false.

Answer D doesn't say attractive force. It says attraction. If there's less kinetic energy, there's going to be greater attraction (same attractive force, just less KE to cancel it out) between molecules.

You can think of it like two magnets. The attractive force will be exactly the same if you move them over each other at 0.5m/s or at 50m/s. But when you do it at 50m/s, the attractive force won't be strong enough to make them stick together, whereas it will if you do it at 0.5m/s.

 

[cheesier]

This is an awful question. All gasses have attractive forces, even noble gases because of dipole dipole interactions, so D is correct. But like Specter said, A is right too if the gas is non-ideal. PV/RT will ALWAYS go up if you lower T, so two right answers (source: this is PV/RT with the van der waals expression for p substituted in.... look under alternate forms) .

 

[SpecterGT260]

This is an awful question. All gasses have attractive forces, even noble gases because of dipole dipole interactions, so D is correct. But like Specter said, A is right too if the gas is non-ideal. PV/RT will ALWAYS go up if you lower T, so two right answers (source: this is PV/RT with the van der waals expression for p substituted in.... look under alternate forms) .

Why would a noble gas have dipole interactions? There are momentary dipoles, but they are random and will repel as often as they attract. Average intermolecular force = 0 even when allowing for momentary dipoles.

 

[SpecterGT260]

Answer D doesn't say attractive force. It says attraction. If there's less kinetic energy, there's going to be greater attraction (same attractive force, just less KE to cancel it out) between molecules.

You can think of it like two magnets. The attractive force will be exactly the same if you move them over each other at 0.5m/s or at 50m/s. But when you do it at 50m/s, the attractive force won't be strong enough to make them stick together, whereas it will if you do it at 0.5m/s.

I wouldnt think of it as 2 magnets. There is nothing in the question stem to indicate that we need to assume there is any attraction. Think of it like you're supposed to with gas models. Lots of little bouncing balls. They they go faster they bounce harder and increase pressure. If you slow them down, are they now attracted to each other? or just banging against the container with less force causing decreased pressure?

 

[MedPR]

I wouldnt think of it as 2 magnets. There is nothing in the question stem to indicate that we need to assume there is any attraction. Think of it like you're supposed to with gas models. Lots of little bouncing balls. They they go faster they bounce harder and increase pressure. If you slow them down, are they now attracted to each other? or just banging against the container with less force causing decreased pressure?

That's only for ideal gas. Real gases do have attractive forces at short enough distances.

For ideal gases A, B, C, and D are all wrong. For real gases only A, B, and C are wrong.

 

[cheesier]

That's only for ideal gas. Real gases do have attractive forces at short enough distances.

For ideal gases A, B, C, and D are all wrong. For real gases only A, B, and C are wrong.

A isn't wrong. I linked the simplified pv/rt expression and you can see that lowering temperature will always increase the ratio.

Why would a noble gas have dipole interactions? There are momentary dipoles, but they are random and will repel as often as they attract. Average intermolecular force = 0 even when allowing for momentary dipoles.

I meant to say london forces, not dipole dipole. Think about it in terms of converting the gas to a liquid by lowering the temperature. A substance will only stay in the liquid phase if there are sufficient intermolecular interactions. When you lower the temperature, you are moving it closer to being a liquid, and therefore increasing the attraction.

 

[SpecterGT260]

That's only for ideal gas. Real gases do have attractive forces at short enough distances.

For ideal gases A, B, C, and D are all wrong. For real gases only A, B, and C are wrong.

I will believe you if you can show me something that says this, but my chem BS is screaming things at you that I can't in good conscience repeat here

A isn't wrong.

Right, for non-ideal gasses A is a true statement per the reasoning I outlined earlier. I still want to reject D as a conditional answer that can be made an outright falsehood depending either on interpretation or PR's relative ability to produce a source that discusses universal attraction among gas particles. Remember, these statements become utter falsehoods if only a single situation that fits the question stem is false by the foil. i.e. every gas must have net attraction between molecules for this to be a valid answer. For non-ideal gases, which the OP later said was the subject of the passage, A is always true, albeit seemingly beyond the scope of the question depending on what was in that passage.

 

[milski]

The corrections to the ideal gas law are for real molecules having volume and for molecules attracting each other. In that respect A. and D. are more or less the same.

"The next step is to allow for the attractions that occur among the particles in a real
gas. The effect of these attractions is to make the observed pressure Pobs smaller than it would be if the gas particles did not interact." Zumdahl & Zumdahl, Chemistry, 8th Ed, p 215

 

[SpecterGT260]

A isn't wrong. I linked the simplified pv/rt expression and you can see that lowering temperature will always increase the ratio.



I meant to say london forces, not dipole dipole. Think about it in terms of converting the gas to a liquid by lowering the temperature. A substance will only stay in the liquid phase if there are sufficient intermolecular interactions. When you lower the temperature, you are moving it closer to being a liquid, and therefore increasing the attraction.

ok...... I can buy that. I forgot about those guys (hate van der waals in general.....). I am still saying it is a toss up depending on what the passage said

 

[milski]

ok...... I can buy that. I forgot about those guys (hate van der waals in general.....). I am still saying it is a toss up depending on what the passage said

Supposedly he bought the farm almost a century ago - that's a long time to hold a grudge.

 

[gettheleadout]

Am I an idiot for immediately thinking that C can't possibly be right since decreasing the temperature inside the container would lower the pressure of the gas within and so the answer must be D?

Also where are choices A and B that you guys are talking about? I skimmed the thread but might have missed them.

 

[cheesier]

Am I an idiot for immediately thinking that C can't possibly be right since decreasing the temperature inside the container would lower the pressure of the gas within and so the answer must be D?

Also where are choices A and B that you guys are talking about? I skimmed the thread but might have missed them.

They're upthread. A said that decreasing temperature increases the PV/RT ratio. If you sub in the van der waals expression for p and simplify it, it is true as well.

 

[milski]

Am I an idiot for immediately thinking that C can't possibly be right since decreasing the temperature inside the container would lower the pressure of the gas within and so the answer must be D?

Also where are choices A and B that you guys are talking about? I skimmed the thread but might have missed them.

Post #10. http://forums.studentdoctor.net/showpost.php?p=13547034&postcount=10

Yes, increased pressure on the container made no sense to me either.

 

[SpecterGT260]

Supposedly he bought the farm almost a century ago - that's a long time to hold a grudge.

if he irritates me 100 years later in class, does that make it a haunting?

 

[milski]

if he irritates me 100 years later in class, does that make it a haunting?

That, or he is just a regular douche.

 

[SpecterGT260]

Am I an idiot for immediately thinking that C can't possibly be right since decreasing the temperature inside the container would lower the pressure of the gas within and so the answer must be D?

Also where are choices A and B that you guys are talking about? I skimmed the thread but might have missed them.

Not if this were a testing situation. However, as I stated earlier, this is a concept issue, not a test strategy issue. On game day, fly by exclusion if you wish as it is a good game plan. BUT it is a terrible thing to do on study prep. People in my class do this all the time with practice materials and it irritates me to no end. Force-fitting an explanation through assumptions, "what if's", and skewing of words may result in you having an understanding that allows you to answer that question, but in all likelihood the same concept will arise again on the real test in a different manner and you will have altered your understanding to fit a poorly worded pre-test question and you will very likely get it wrong.

A very good example of this happened in our micro course, where a practice test implied that gram negative bacterial cell walls (the PDG) was more porous than that of gram positives leading to loss of the violet coloring. We discussed it, and several people took to heart "gram negatives are more porous", and when discussing that question later on I had one girl try to tell me that the porins, transmembrane proteins found only in the outer membrane of gram negatives, actually traversed the entire cell membrane, from outer to inner and through the cell wall. A simple yet pretty vital fact learned wrong solely due to forcing an explanation on a crap question. Gm- PDG is not any more porous. It has fewer crosslinks than Gm+ does, but no actual pores in the wall itself. But the insistence that fewer crosslinks = pores led her to miss a very fundamental concept. Now think of when you are gearing up for your pharm test where there are 9 billion side effects and you can't possibly memorize them all and you fly by association. Suddenly a wrong generalization from forcing a bad answer on a question can creep through your understanding like cancer. That may be kinda extreme.... but it will definitely lead to missed points.

 

[gettheleadout]

Not if this were a testing situation. However, as I stated earlier, this is a concept issue, not a test strategy issue. On game day, fly by exclusion if you wish as it is a good game plan. BUT it is a terrible thing to do on study prep. People in my class do this all the time with practice materials and it irritates me to no end. Force-fitting an explanation through assumptions, "what if's", and skewing of words may result in you having an understanding that allows you to answer that question, but in all likelihood the same concept will arise again on the real test in a different manner and you will have altered your understanding to fit a poorly worded pre-test question and you will very likely get it wrong.

Okay here's the thing, I automatically associate the molecular interactions of non-ideal gases with attractive London dispersion / Van der Waals attractive forces, and assume that lowering temperature increases the effect of these attractive interactions on pressure. So, I understand that for both ideal and non-ideal gases, decreasing temperature increases pressure, right?

Now for the PV / RT ratio, decreasing temperature produces an equivalent decrease in pressure if the gas is ideal, right? So for non-ideal gases, according to what milksi quoted, the pressure is then lower than should it should be to keep a constant PV / RT ratio, as a result of attraction between the molecules. Shouldn't this decrease the ratio?

Edit: Or does the ratio increase because real (non-ideal) gases have molecular volume too and so V increases even though P decreases?

 

[SpecterGT260]

Okay here's the thing, I automatically associate the molecular interactions of non-ideal gases with attractive London dispersion / Van der Waals attractive forces, and assume that lowering temperature increases the effect of these attractive interactions on pressure. So, I understand that for both ideal and non-ideal gases, decreasing temperature increases pressure, right?

Now for the PV / RT ratio, decreasing temperature produces an equivalent decrease in pressure if the gas is ideal, right? So for non-ideal gases, according to what milksi quoted, the pressure is then lower than should it should be to keep a constant PV / RT ratio, as a result of attraction between the molecules. Shouldn't this decrease the ratio?

Edit: Or does the ratio increase because real (non-ideal) gases have molecular volume too and so V increases even though P decreases?

Non ideal gasses occupy space. This means that pressure cannot fall in a 1:1 ratio with temp. It lags slightly behind. Think in extremes. Non ideal gas made out of beach balls very far apart to the point of acting ideal. Start to compress them via pressure or temp (doesn't matter which). As they get closer they deviate from ideal because they occupy space. They hit each other more frequently than pinpoints would. Increased collisions is increased pressure. So pressure can't fall as fast as temp. The ratio always increases because T is reduced faster than P.

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[YouNeverKnow22]

the answer D makes sense, way too much over analyzation going on in here.

 

[Temperature101]

the answer D makes sense, way too much over analyzation going on in here.

You are right, but again this is SDN.

 

[Laureatebarrel]

Here is where you think too much: The question ask about the pressure on the inside of the container due to the gas being asked. You read the word "on" and you think it is due to the external pressure.
And also if the internal pressure decreases, the difference between outside and inside increases. There is no change of the external pressure.