ideal gas law deviation

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stevvo111

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So I'm super confused by this one question in EK 1001 chemistry, it's on page 21, question 217 if you wanna check it out.

The question states, according to the graph, which of the following would be true, if the ideal gas law, PV-nrt were used to calculate the volume of a sample of CH4 gas from measured variables at 200atm and 300K and again at 600 atm and 300k?

the answer is... the calculated volume would be greater than the real volume for 200 atm and less than real volume for the 600 atm sample..

The graph essentially shows at pressure on the x axis and (PV/RT) on the y axis. Ideally, the y axis should be at 1 all the time, but at 200 it is less than 1 and at 600 it is more than 1.


I though I had ideal gas down pat, but this question is killing me. Any help would be greatly appreciated!

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At 200 atm, PV is actually less than nRT (since PV/nRT < 1). Now let's say you try to use the ideal gas law to calculate V. The ideal gas law states that PV = nRT. In other words, the ideal gas law overestimates PV at this pressure. As a result, your calculation will overestimate the volume.

Similar logic for 600 atm - at this pressure, PV is actually GREATER than nRT. The ideal gas law underestimates PV at 600 atm, so the calculated volume will be smaller than actual.
 
As pressure increases, volume decreases but not as much as expected. You must leave some space for the volume of the particles: decrease of V_real = decrease of V_ideal + V_particles.

As pressure decreases, volume increases but not as much as expected. You must account for the space occupied by the particles: increase of V_real = increase of V_ideal - V_particles.
 
So I'm super confused by this one question in EK 1001 chemistry, it's on page 21, question 217 if you wanna check it out.

The question states, according to the graph, which of the following would be true, if the ideal gas law, PV-nrt were used to calculate the volume of a sample of CH4 gas from measured variables at 200atm and 300K and again at 600 atm and 300k?

the answer is... the calculated volume would be greater than the real volume for 200 atm and less than real volume for the 600 atm sample..

The graph essentially shows at pressure on the x axis and (PV/RT) on the y axis. Ideally, the y axis should be at 1 all the time, but at 200 it is less than 1 and at 600 it is more than 1.


I though I had ideal gas down pat, but this question is killing me. Any help would be greatly appreciated!

PV = nRT
n = PV/RT
@ P = 200...n<1
@ P = 600...n>1

V = nRT/P

The ideal gas law would tell you that n = 1 at all times (assuming that is the amount you sampled). However, the graph tells you that at P = 200, n<1. Therefore, your calculated volume would be greater than the actual volume. The graph also tells you that at P = 600, n>1. Therefore, your calculated volume would be less than the actual volume.
 
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PV = nRT
n = PV/RT
@ P = 200...n<1
@ P = 600...n>1

V = nRT/P

The ideal gas law would tell you that n = 1 at all times (assuming that is the amount you sampled). However, the graph tells you that at P = 200, n<1. Therefore, your calculated volume would be greater than the actual volume. The graph also tells you that at P = 600, n>1. Therefore, your calculated volume would be less than the actual volume.


For some reason im still not getting this guys...

so i know for a real gas it's (P+something)(V-something) to make a better approximation for idealness. In this particular example they didn't use this, instead they used the ideal gas law PV=nRT. With the way the question was setup n=PV/RT, where in ideal conditions, n=1.

if we rearrange this as above V=nRT/P. Where n<1 and >1 for the problem, RT is some constant value and P is either 200atm or 600atm.

The graph has it so n<1 at 200atm. Giving me the impression that V=nRT/P is setup as... V=(0.5ish)(constant RT)/200.... this clearly makes V a small number relative to if n was 1.

for 600atm I set it up so V=(1.5ish)(constant RT)/600....V is pretty much the same as above...

still not seeing where the calculated volume would be greater than the real volume for 200atm and less than at 600.

V calculated can be thought of as Videal right?
 
For some reason im still not getting this guys...

for 600atm I set it up so V=(1.5ish)(constant RT)/600....V is pretty much the same as above...

still not seeing where the calculated volume would be greater than the real volume for 200atm and less than at 600.

V calculated can be thought of as Videal right?

Ok...i think I see where your confusion is. When you set up these equations, you are correcting the ideal equation by plugging in values from the graph. This is NOT the correct approach. What they are asking is, what happens if you don't know that there are deviations from ideal behavior at these pressures? Will you underestimate or overestimate the volume?

I believe the point of this question is that if you ASSUME ideal behavior (PV = nRT), with n=1, you are going to get an incorrect volume (either higher or lower than it should be). In the case of 200 atm, the volume calculated will be too high because you are overestimating PV with the ideal gas equation. Similar logic applies to 600 atm (you are underestimating PV).

Does this make sense?
 
Ok...i think I see where your confusion is. When you set up these equations, you are correcting the ideal equation by plugging in values from the graph. This is NOT the correct approach. What they are asking is, what happens if you don't know that there are deviations from ideal behavior at these pressures? Will you underestimate or overestimate the volume?

I believe the point of this question is that if you ASSUME ideal behavior (PV = nRT), with n=1, you are going to get an incorrect volume (either higher or lower than it should be). In the case of 200 atm, the volume calculated will be too high because you are overestimating PV with the ideal gas equation. Similar logic applies to 600 atm (you are underestimating PV).

Does this make sense?

kind of, so If we assume n is 1 at 200atm, Since P is small V would be large, thus it is overestimating what we observe (which is V being smaller since n<1). Similarly with 600atm, V would be small, and thus we would be underestimating the volume (which is much larger since n>1).

I think I understand, the question was just really murky to me for some reason.


So, I'm gonna calculate it just to prove it to myself

at 200, 200*V/(.08*300)=1, this gives V calculated as 1.something

at 600, 600*V/(.08*300)=1, gives V calculated as 0.something

According to the graph, at 200, actual n value is less than 1, meaning V is overestimated through equation above.

according to the graph, at 600, actual n value is more than 1, meaning V is underestimated, and should be more than calculated above.

Finally understanding it, haha. I think you were right jets, i was assuming the graph was ideal (and thus n values too), when i had no reason to assume this. Moral of the story, read the question carefully, and solve it out when all else fails (provided you can do it fast

Thanks doods, hope this thread can help others in need

EK chem 1001 page 21 question 217.
 
it appears you guys are all over thinking this.

we know that a gas behaves ideally at low pressure and high temperature. Therefore, without having to do any calculations, you know that increasing the pressure by 3X will cause the gas to deviate from ideal behavior and behave more like a real gas.

Next, we know that real gases have a greater volume and lesser pressure when compared to ideal gases.

Therefore, you should be able to reach your answer, given the fact that you understand that the ideal gas law describes an ideal gas.
 
it appears you guys are all over thinking this.

we know that a gas behaves ideally at low pressure and high temperature. Therefore, without having to do any calculations, you know that increasing the pressure by 3X will cause the gas to deviate from ideal behavior and behave more like a real gas.

Next, we know that real gases have a greater volume and lesser pressure when compared to ideal gases.

Therefore, you should be able to reach your answer, given the fact that you understand that the ideal gas law describes an ideal gas.

In general, I agree with you. However, the question specifically references a graph. Assuming that the graph will follow standard convention will get you in trouble. It is best to understand how to utilize the information the graph gave in order to answer the question. In my opinion, you should never, ever, make an assumption when given a graph with relative data.
 
sorry to bring up a dead thread, but i just noticed what I believe to be an inconsistency in TPR review book in regards to volume of a real gas; it states:

"V (real) < V(ideal) because molecules of real gases DO have volume that reduces the effective volume of the container (since the molecules take up space, there is less space in the container for all the other particles to occupy)" p. 524 of the 2009 edition

This is wrong (or at least not completely correct), right?

That is, if my understanding is correct, at moderately high pressures, V (real) will be less than V (ideal) due to intermolecular forces between gas particles, but at extremely high pressures, the size of the gas particles will result in a V (real) that is greater than V (ideal) because under the ideal gas law the size of gas particles are insignificant.

But the volume of a real gas won't be less than the ideal gas for the reason TPR book states.
 
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