Determine all of the Physical Chemistry Variables of the System

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Homework Help Overview

The discussion revolves around a weather balloon filled with Helium gas that ascends to 18 km, expanding to a diameter of 15 m. Participants are tasked with determining the changes in various physical chemistry variables such as ΔV, ΔP, ΔT, ΔU, ΔH, and entropy changes for the system and surroundings.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of changes in volume, pressure, and temperature as the balloon ascends. There are attempts to define the system and its surroundings, with some questioning the assumptions about equilibrium and reversibility. Various interpretations of ΔH and its dependence on pressure conditions are also explored.

Discussion Status

Participants are actively questioning their assumptions and definitions, particularly regarding the nature of the system and the effects of altitude on temperature and pressure. Some guidance has been offered regarding the definitions of state functions and the concept of buoyant force, but there is no explicit consensus on the interpretations or outcomes.

Contextual Notes

There is uncertainty regarding the initial conditions of the balloon, such as its temperature and pressure at release. Participants are also grappling with the implications of reversibility and the driving forces behind the balloon's ascent.

Gwozdzilla
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Homework Statement


A weather balloon is filled with Helium gas and released from the ground. It goes up 18km and achieves a diameter of 15m. Determine if the following values are greater than zero, less than zero, or equal to zero: ΔV, ΔP, ΔT, ΔU, ΔH, Ssys, surr, Stot

Homework Equations


ΔU = CV,mΔT
ΔH = CP,mΔT
Ssys + Ssurr = Stot = 0 if process is reversible
Ssys + Ssurr = Stot > 0 if process is irreversible
"Reversible" means all of the forces are at equilibrium at all times throughout the process.
Ssys = CP,mln(Tf/Ti) + nRln(Vf/Vi) is a "state function."

The Attempt at a Solution


Obviously, ΔV increases because the balloon expands as the diameter gets to be large compared to what it presumably was at the time of release, even though the diameter at that time isn't mentioned.
ΔV > 0

ΔP < 0 because pressure decreases as altitude increases.

ΔT < 0 because temperature increases as altitude increases.

ΔU < 0 because ΔT < 0
ΔH < 0 because ΔT < 0

It seems that I have assumed that the external properties of T and P apply to the inside of the balloon, so I guess the system is reversible.
Stot = 0

Ssys = -Ssurr

I don't really know where to go from here...

I want to argue that because the balloon actually does manage to overcome gravity, it's change in volume must be pretty substantial (making the w = -PdV be what causes the balloon to go up). Maybe it's more substantial than the change in temperature, giving a large positive term due to volume in the entropy state equation listed above.
Ssys > 0

However, if the system is being looked at as the balloon itself rather than the air inside the balloon, then when the balloon isn't inflated, it has more freedom of movement than after it's fully inflated and therefore more entropy, meaning that the final entropy change of the system would be negative.

If I am correct in guessing that this process is reversible AND in guessing that Ssys > 0, then Ssurr < 0.

I would rather not have all of my answers to physical chemistry problems be guesses. How can I solve this problem without guessing? How can I know if my guesses are correct or not? How can I better approach this problem? It's possible that the correct answer to this problem is due to something that I have completely overlooked. How do I not overlook considerations while doing these types of problems?
 
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Gwozdzilla said:
It seems that I have assumed that the external properties of T and P apply to the inside of the balloon, so I guess the system is reversible.
Stot = 0
Really?
 
Gwozdzilla said:
ΔV > 0
Check.
ΔP < 0 because pressure decreases as altitude increases.
Check.
ΔT < 0 because temperature indecreases as altitude increases.
Check.
ΔU < 0 because ΔT < 0
Check.
ΔH < 0 because ΔT < 0
How is ΔH defined?

What's your definition of "system?" Balloon? Balloon plus atmosphere? Is it at equilibrium before being released?
 
Bystander said:
How is ΔH defined?

What's your definition of "system?" Balloon? Balloon plus atmosphere? Is it at equilibrium before being released?

Personally, I think that the system is the air inside the balloon and that the surroundings is the air outside of the balloon, and that the balloon itself is just a barrier. I think it's at equilibrium prior to being released. Why would the balloon start at a different temperature and pressure other than the ambient temperature and 1atm of earth?

DrClaude said:
Really?
I remembered that in general, as altitude increases temperature and pressure decrease. I think that this is a "general" property of the surroundings and not of the system, so I just concluded that since the only thing I know about temperature and pressure in this situation has to do with the surroundings, it must apply to the system as well. What is flawed in this logic? If I can't use the properties of the surroundings here to determine P and T of the balloon, how else would I determine them? Or am I just incorrect in the meaning of reversible?
 
Actually, thinking about it for longer, if all of the forces were at equilibrium, then the balloon wouldn't be able to go up at all. The only reason it goes up is because some force is causing an acceleration that is greater than gravity. So I guess it's irreversible. Is there any kind of systematic way for determining reversibility?
 
Gwozdzilla said:
temperature and pressure in this situation has to do with the surroundings, it must apply to the system as well.
Good. You're pretty much stuck with this much, or the problem becomes messy. Now, what happens to ΔHHe in the balloon as it rises?
 
Well, my actual initial thought was "hot air rises. If it's rising, ΔT > 0."

But given that the external temperature is decreasing as the balloon ascends, I guess inside the balloon ΔT < 0 as well.

ΔH = CP,mΔT
So ΔH < 0?
 
Gwozdzilla said:
So ΔH < 0?
Gwozdzilla said:
ΔH = CP,mΔT
What's the highlighted subscript indicate?

The questions aren't to be "picky" with you, but to get you looking at details that will take the "guesswork" out of your analysis, and future analyses of other problems.
 
The highlighted subscript says P for constant pressure. But I was taught that ΔH = CP,mΔT is a state function, and therefore "always true" regardless of whether or not the actual pressure is held constant. Did I misunderstand something here?
 
  • #10
Gwozdzilla said:
Personally, I think that the system is the air inside the balloon and that the surroundings is the air outside of the balloon, and that the balloon itself is just a barrier. I think it's at equilibrium prior to being released. Why would the balloon start at a different temperature and pressure other than the ambient temperature and 1atm of earth?
The problem text says the balloon is filled with He gas. What is the force driving it upward?
 
  • #11
ehild said:
The problem text says the balloon is filled with He gas. What is the force driving it upward?

I'm actually not really sure why the balloon is going upwards. It's probably expanding to have an increase in entropy. Maybe this is applying some additional pressure on the inside of the balloon. But if the overall change in pressure is negative due to the increase in altitude, can pressure still be the driving force that moves the balloon up?
 
  • #12
Maybe ΔPint = - ΔPext. In which case, ΔPsys > 0, and this pressure is what is driving the balloon upwards?
 
  • #13
Why does a helium balloon rise? Think of buoyant force...
 
  • #14
I'm sorry, but I don't know what a buoyant force is either. From searching the internet, does the balloon rise because the weight of the Helium is less than the weight of the O2 and N2 in the atmosphere? I don't really see how that would work...
 

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