Ideal Gas Vs Real Gas HELP!

[KiaBia]

Here's a question from my TPR ICC General Chem Passage:

Ammonia is not an ideal gas, primarily because it experiences moderate hydrogen bonding. Therefore, compared to the pressure and volume predicted by the ideal gas law, which relation is true for 1 mole of ammonia at 300K?

A. P ideal < P ammonia and V ideal < V ammonia
B. P ideal < P amm. and V ideal > V amm.
C. P ideal > P amm. and V ideal < V amm.
D. P ideal > P amm. and V ideal > V amm.

I automatically eliminated C and D, because I learned that real gases deviate from ideal gases due to high pressures and low temperatures. Therefore, I chose B because I know that P and V are inversely related.

However, the answer is D. I am a tad confused! Can anyone help explain? Maybe it's got something to do with the hydrogen bonding.

THANKS!

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[slz1900]

Pressure is force/area, so if there is hydrogen bonding, the molecules won't be htiting the container as hard since they are sticking together and pressure goes down. Volume is less than predicted volume for the same reason. With hydrogen bonding, molecules are kept closer to each other than they would be if they weren't interacting. This means they occupy less space than an ideal gas.

 

[dsoz]

You are correct in that it has to do with the hydrogen bonding.

Think of the hydrogen bonds as being a little extra attractiveness to the other molecules. They stick a little close (smaller volume), and they don't bounce around on the walls as much (lower pressure).

Your thought about the high pressure/low temperatures should not be "due to" but "when conditions of." When gasses are at low temperatures or at high pressures, most will deviate from ideal gas behavior. But Ammonia does this because of the hydrogen bonds.

dsoz

 

[LoLCareerGoals]

You can also use Van der Waals equation.
Focus on the second term since there is no mention of finite molecule volume (which is what 1st "-nb" term is responsible for):
P = nRT/(V-nb) - n^2*a/V^2 is the correction for molecular attraction (regardless of what kind H-bond/dipole/LDF).
Clearly it is reducing pressure -> eliminate A&B.
Now pretend that 2nd term is a constant and solve for V: V = nRT/(P+const). V got smaller due to const -> eliminate C.

 

[Meredith92]

Does hydrogen bonding occur in gases though? I thought so much of what we know about waters high BP is cuz h bonds need to be broken for it to evaporate. Are we saying h bonds can also affect the gas phase?

 

[KiaBia]

Okay, I'm understanding this now. Thank you for your replies! 🙂

 

[MedGrl@2022]

Here's a question from my TPR ICC General Chem Passage:

Ammonia is not an ideal gas, primarily because it experiences moderate hydrogen bonding. Therefore, compared to the pressure and volume predicted by the ideal gas law, which relation is true for 1 mole of ammonia at 300K?

A. P ideal < P ammonia and V ideal < V ammonia
B. P ideal < P amm. and V ideal > V amm.
C. P ideal > P amm. and V ideal < V amm.
D. P ideal > P amm. and V ideal > V amm.

I automatically eliminated C and D, because I learned that real gases deviate from ideal gases due to high pressures and low temperatures. Therefore, I chose B because I know that P and V are inversely related.

However, the answer is D. I am a tad confused! Can anyone help explain? Maybe it's got something to do with the hydrogen bonding.

THANKS!

I think it is D because if Hydrogen bonding is occuring then the molecules will be clumping together rather than disorderly expanding and taking up all the space that it can. Therefore, the volume would be less than ideal and the pressure would be less than ideal too because it is taking up less space therefore it would have less force per area. If I am wrong... please correct me.

Best,

Verónica

 

[brood910]

I think it is D because if Hydrogen bonding is occuring then the molecules will be clumping together rather than disorderly expanding and taking up all the space that it can. Therefore, the volume would be less than ideal and the pressure would be less than ideal too because it is taking up less space therefore it would have less force per area. If I am wrong... please correct me.

Best,

Verónica

About V, Videal is bigger because we assume that particles do not have any V in ideal gas (dont take up the V). But real gases dont act like that.

For P, for ideal gas, we expect that there's no intermolecular forces btwn particles, but they do exist in reality. As intermolecular forces go up, particles do not collide much, and thus, less P.

 

[sdm33]

due to high pressures and low temperatures.,but if you calculate for pressure , it is going to be very low, keyword (1 mole of ammonia at 300K?)

 

[brood910]

Ok I found the true answer to this today.

P of ideal gas is larger than P of real gas only if attraction is larger.

V of ideal gas is larger ONLY IF attractive forces are larger than repelling force.
V of idea gas is lower if repelling is stronger.


NOTE: This is when there is NO external pressure applied.

 

[circulus vitios]

How can the volume of an ideal gas ever be greater than the volume of a real gas? Ideal gas laws assume that gas molecules take up no space. No matter how strong intermolecular attraction may be, how does this ever negate the volume that the actual gas molecules occupy?

 

[circulus vitios]

I found this post: http://forums.studentdoctor.net/showpost.php?p=8196350&postcount=5

Looks like it's another case of a ****ty, poorly-defined question. Yay MCAT!

 

[brood910]

It is very unlikely that you will see a question like this where they require you to know the effects of intermolecular forces that cause deviations from the ideal gas law. Why? Because intermolecular forces are usually considered only under low T.

Under low T, we have less KE => slower V => more time between molecules => more time for intermolecular forces to act on them => more effects on their paths => takes longer to collide with the wall => less P than expected.

Also, we consider V deviation only under HIGH pressure.
HIgh pressure makes the volume of the flexible container smaller => the volume of the gas particles is not "negligible" anymore since it is significant compared to the entire volume => V is higher than expected.


For this particular question, they are not askin about the deviations under low T or high P.
They are merely askin for the effects of intermolecular forces.
If you have attractive forces, particles spend more time on sticking with each other => less volume than expected.

 

[LoLCareerGoals]

How can the volume of an ideal gas ever be greater than the volume of a real gas? Ideal gas laws assume that gas molecules take up no space. No matter how strong intermolecular attraction may be, how does this ever negate the volume that the actual gas molecules occupy?

Nobody has asked you to compare ideal gas to a real gas. You were only asked to compare ideal gas with ammonia that has H-bonds only.
Does such gas as ammonia with H-bonds only exist? No. Do we care? No.

Besides the passage clearly says primarily because.... It admits there might be something else but asks you to ignore it for now. Stop focusing on the real gas and focus on the fictional non-ideal gas pretty well defined in the passage.

 

[circulus vitios]

Nobody has asked you to compare ideal gas to a real gas. You were only asked to compare ideal gas with ammonia that has H-bonds only.
Does such gas as ammonia with H-bonds only exist? No. Do we care? No.

Besides the passage clearly says primarily because.... It admits there might be something else but asks you to ignore it for now. Stop focusing on the real gas and focus on the fictional non-ideal gas pretty well defined in the passage.

Can you walk me through your reasoning for the volume of the ammonia gas with h bonding? It still makes no sense.

 

[brood910]

Can you walk me through your reasoning for the volume of the ammonia gas with h bonding? It still makes no sense.

Think of squeezing an elastic ball with your hand.
It shrinks a bit.