AAMC 7 PS #20 : Ideal gases vs real gas conditions

[ihatebluescrubs]

The question: "A gas that occupies 10L at 1atm at 25C will occupy what volume at 500atm and 25C?

The answer "Somewhat more than 0.02L because of the space occpied by the individual gas molecules"


I don't get this because according to my PRHL physics review book, "at high pressure and low temp, real gases deviate from ideal gas behavior. In reality the actual volume and pressure for a real gas are less than those values obtained from the ideal gas law"

So because of that, I thought that the supposed real volume at 500atm is less than 0.02L ideal volume.

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[sdm33]

I am not sure , but because temp is constant we use p1v1=p2v2 = 1 . 10 = 500 . V2
V2 = 10/500= 0.02L. However due to the real gas high pressure , we forced all the gases next each other in a tight place so now they have volume so the over all is more than 0.02L , it is a silly concept if you ask me. it is like saying if cars are far a part they have no volume and when they are bumper to bumper of each other they have volume. 😕

 

[BerkReviewTeach]

The question: "A gas that occupies 10L at 1atm at 25C will occupy what volume at 500atm and 25C?

The answer "Somewhat more than 0.02L because of the space occpied by the individual gas molecules"


I don't get this because according to my PRHL physics review book, "at high pressure and low temp, real gases deviate from ideal gas behavior. In reality the actual volume and pressure for a real gas are less than those values obtained from the ideal gas law"

So because of that, I thought that the supposed real volume at 500atm is less than 0.02L ideal volume.

This is basically the same question as (or at least very similar to) one posed a few weeks ago.

This comes from example 6.8 in gen chem chapter 6 TBR

For an inert gas, if you were to reduce the pressure to half of its original value, then what is V_f relative to the initial V
A ¹/₂ V_i - a little bit
B ¹/₂ V_i + a little bit
C 2V_i - a little bit
D 2V_i + a little bit


Solution: when pressure is cut in half the ideal gas law predicts the volume should double. Because only the space between molecules increases, while the molecules remain the same size, the increase in volume is not as large as predicted by the ideal gas law. This makes choice C, 2V_i - a little bit, the best answer. "the little bit" term is attributed to the size of the molecues.... If the pressure were doubled, the volume wouldnt be reduced by exactly half either, the new volume would be ¹/₂ V_i + a little bit.


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The bolded part above is what I have the most problem with. I understand based on PV=nRT if the pressure were doubled the volume should be half its original value Then based on the van der walls equation I would expect that the volume would be half its value, minus a little bit, because of the size of the molecule. This made sense for how I solved the situation for reducing the pressure, but it doesnt work when the pressure is doubled?

On the other hand, if I actual USE the van der waals equation I get the exact opposite dilema
V_ideal = V_real - nb
2 V_ideal = V_real - nb
2V_ideal +nb = V_real
Now this logic works for when the pressure is doubled, but it DOESN'T work when the pressure is reduced by half!

Can someone help me out here?
Thank you!

The picture in the answer explanation (Figure 6-3) should help a little.

What they are saying is that when the volume of the container changes, the size of the molecules remains constant. So consider their system with a 3.0-L flask filled with enough gas that if you condensed it into a liquid it would occupy 0.1 liters. This means that the open space (volume where no molecules exist) is 2.9 liters. If the pressure is doubled, then the container will be squeezed down, but the molecules will not become any smaller. They would still need 0.1 liters if they were condensed. So the new volume could be considered to be ¹/₂(2.9) + 0.1 = 1.55. Rather than doing hypothetical math, the answer explanation is saying that the new volume is ¹/₂(V_initial) + a little, which fits with V_final = ¹/₂(3.0) + a little, where a little happens to be 0.05 (an arbitrary number for the example).

The key in both questions is to consider the size of the molecules. The intermolecular forces will change, but if the system is still a gas at the extreme conditions, then the attractive forces can't be that significant (otherwise it would have been compressed into a liquid).

In the question you asked, the pressure reduces the volume to what should be 0.02L if it were ideal. But because the molecules are the same size no matter what the external conditions may be, you can't compress them to become any smaller, so they will occupy the same space as before. Compressing the system reduced the distance between molecules (and thus the empty space between molecules), but the space in which the molecules exist remains the same. So the total volume of the system (empty space plus molecule-occupied space) is more than the 0.02L predicted by the ideal gas law.