How Much Hydrogen Is Needed for a Balloon to Float at 10km Altitude?

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Homework Help Overview

The discussion revolves around calculating the amount of hydrogen gas needed for a balloon to float at an altitude of 10 km, where the temperature is approximately -50°C and the air pressure is around 16 kPa. The problem involves using the ideal gas law and understanding the forces acting on the balloon, including buoyancy and weight.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the ideal gas law to determine the mass of hydrogen needed for a given volume. There are attempts to clarify the notation used in equations and to explore the conditions under which the balloon would not accelerate. Questions arise about the forces acting on the balloon and how buoyancy relates to the weight of the gas inside.

Discussion Status

Participants are actively engaging with the problem, questioning assumptions, and exploring different interpretations of the forces involved. Some guidance has been provided regarding the equations to use for calculating buoyancy and the mass of air displaced by the balloon. There is a recognition of the need to equate forces for the balloon to remain at a constant height.

Contextual Notes

Participants note the challenge of working with non-ideal gas behavior for air and the implications of varying pressure and temperature at high altitudes. There is also mention of the need to consider the molecular masses of hydrogen and air in the calculations.

cycrups
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Homework Statement



  1. At an altitude of about 10 km, the temperature of the Earth’s atmosphere is roughly -50°C and the air pressure is around 16 kPa. (Assume ideal gas)
    1. How many kilograms of hydrogen gas (H2, molecular mass 2g/mol) should be put in a balloon to fill it to 2000 m3?
    2. What is the required mass of total ballast that the balloon will float at constant height (molecular mass of air 29 g/mol)?

Homework Equations


PV = nRT

The Attempt at a Solution



For 1. I used V = (P*A*N)/(R*T) = ( 2g/mol * 16000 Pa * 2000m^2 ) / (8.314 * 223K) = 34.51 Kg I don't know if it's correct way to do it or not

For the second part I have no idea how to even start.
 
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cycrups said:
For 1. I used V = (P*A*N)/(R*T) = ( 2g/mol * 16000 Pa * 2000m^2 ) / (8.314 * 223K) = 34.51 Kg

Your numbers make sense, although the notation in your equation could be clearer. V is generally used to denote a volume, not a mass.

cycrups said:
For the second part I have no idea how to even start.

What must be fulfilled in order for the balloon not to accelerate either up or down?
 
Orodruin said:
Your numbers make sense, although the notation in your equation could be clearer. V is generally used to denote a volume, not a mass.
What must be fulfilled in order for the balloon not to accelerate either up or down?

certain weight?
 
Naturally, but what determines this weight? What determines whether any object is accelerating or not? Hint: There was this guy in the 17th century who got hit by an apple ...
 
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Orodruin said:
Naturally, but what determines this weight? What determines whether any object is accelerating or not? Hint: There was this guy in the 17th century who got hit by an apple ...

Force. Hope it's correct.
 
Yes, so what must be true if the balloon is not supposed to accelerate?
 
Orodruin said:
Yes, so what must be true if the balloon is not supposed to accelerate?
The forces in both directions must equal the same.
 
cycrups said:
I used V = (P*A*N)/(R*T)
I guess you mean M = (molecular mass)*n = (molecular mass)*PV/(RT)
cycrups said:
The forces in both directions must equal the same.
Right. And without the ballast, what two forces do you have?
 
haruspex said:
I guess you mean M = (molecular mass)*n = (molecular mass)*PV/(RT)

Right. And without the ballast, what two forces do you have?

I'd say gravity and the force of the helium pushing up the balloon?
 
  • #10
cycrups said:
I'd say gravity and the force of the helium pushing up the balloon?
How does the helium push the balloon up? Does it have negative mass??
 
  • #11
haruspex said:
How does the helium push the balloon up? Does it have negative mass??
because Helium is lighter than air that's why it tends to push up
 
  • #12
It would be more accurate to say that the surrounding air is pushing the helium up - with a force greater than the helium weight. If you block the helium from escaping, the surrounding air is providing the upward force on the balloon helium system.

To get back to the problem: What is the buoyancy force on the balloon + helium system? What forces act in the opposite direction? Does the weight of the system compensate for the buoyancy? If not, how much weight do you need to add?
 
  • #13
Orodruin said:
It would be more accurate to say that the surrounding air is pushing the helium up - with a force greater than the helium weight. If you block the helium from escaping, the surrounding air is providing the upward force on the balloon helium system.

To get back to the problem: What is the buoyancy force on the balloon + helium system? What forces act in the opposite direction? Does the weight of the system compensate for the buoyancy? If not, how much weight do you need to add?
Should I use this equation to get the Buoyant force F (b) = (air density)x(9.81 m/sec2)x(volume of the gas filled balloon) ?
 
  • #14
cycrups said:
Should I use this equation to get the Buoyant force F (b) = (air density)x(9.81 m/sec2)x(volume of the gas filled balloon) ?
Yes.
 
  • #15
haruspex said:
Yes.
or should I use Fb= mass of air * g = P * Vb * g because I don't have air density OR do I get the air density from

density = P / R * T ?
 
  • #16
cycrups said:
or should I use Fb= mass of air * g = P * Vb * g because I don't have air density
mass = pressure * volume? Where does that come from?
OR do I get the air density from density = P / R * T ?
Sort of. You know air density at standard surface temperature and pressure, right? So use that to figure out what it would be at the given temperature and pressure.
 
  • #18
haruspex said:
mass = pressure * volume? Where does that come from?

Sort of. You know air density at standard surface temperature and pressure, right? So use that to figure out what it would be at the given temperature and pressure.
Okay so the density would be → density of air = 16000 Pa / 8.31 * 223 K = 8.63 Kg/m3

and then to find the Buoyant force Fb = 8.63 * 10 m/s3 * 2000 m3 = 172600

Is that correct, and if it is what's the next step? Do I equate Fb = Fg to get the Maximum altitude?
 
  • #19
cycrups said:
Okay so the density would be → density of air = 16000 Pa / 8.31 * 223 K = 8.63 Kg/m3

and then to find the Buoyant force Fb = 8.63 * 10 m/s3 * 2000 m3 = 172600

Is that correct, and if it is what's the next step? Do I equate Fb = Fg to get the Maximum altitude?
Sorry, I wasn't paying enough attention to exactly what data you are given.
Here's the easiest way:
- you have already calculated the mass of H2 for the given volume, temperature and pressure
- you know the molecular mass of H2 and the molecular mass of air
- how can you simply combine those facts to obtain the mass of air at the given volume, temperature and pressure?
 
  • #20
haruspex said:
Sorry, I wasn't paying enough attention to exactly what data you are given.
Here's the easiest way:
- you have already calculated the mass of H2 for the given volume, temperature and pressure
- you know the molecular mass of H2 and the molecular mass of air
- how can you simply combine those facts to obtain the mass of air at the given volume, temperature and pressure?
I equate both of them together so the PV/nRT (of H2) = PV/nRT of air? I am sorry I am really having a hard time solving this part.
 
  • #21
cycrups said:
I equate both of them together so the PV/nRT (of H2) = PV/nRT of air? I am sorry I am really having a hard time solving this part.
PV/nRT is always 1, so that doesn't seem very profitable.
You have two equations of the form PV=nRT, one for air, one for H2: PaVa=naRTa, PhVh=nhRTh. You know that some of the variables are the same in the two equations. Which ones? What does that tell you about the remaining variables?
 
  • #22
haruspex said:
PV/nRT is always 1, so that doesn't seem very profitable.
You have two equations of the form PV=nRT, one for air, one for H2: PaVa=naRTa, PhVh=nhRTh. You know that some of the variables are the same in the two equations. Which ones? What does that tell you about the remaining variables?
Temperature and R should be the same. The thing I don't know would be the air Pressure and Volume but I doubt the Air would have volume + it's not an ideal gas because it's air. Am I making sense?
 
Last edited:
  • #23
cycrups said:
Temperature and R should be the same. The thing I don't know would be the air Pressure and Volume but I doubt the Air would have volume + it's not an ideal gas because it's air. Am I making sense?
We're talking about the mass of air displaced by the balloon. By definition, that will have the same volume as the balloon.
It is true that the pressure of the H2 will be slightly greater than that of the surrounding air, but only slightly or the balloon would rip. So if the pressures, volumes and temperatures all match, what do you deduce about na and nh?
 

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