Balloon in Earth's atmosphere

In summary: 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?In summary, the conversation discusses how to calculate the required mass of hydrogen gas to fill a balloon at a given altitude, temperature, and air pressure, as well as the required mass of total ballast for the balloon to float at a constant height. The conversation goes on to discuss the buoyancy force and how to calculate it, using the ideal
  • #1
cycrups
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0

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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  • #2
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?
 
  • #3
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?
 
  • #4
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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  • #5
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.
 
  • #6
Yes, so what must be true if the balloon is not supposed to accelerate?
 
  • #7
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.
 
  • #8
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?
 
  • #9
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?
 
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  • #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?
 

1. What is a balloon in Earth's atmosphere?

A balloon in Earth's atmosphere is a man-made object filled with a gas, typically helium or hydrogen, that is designed to float in the air. It is commonly used for scientific research, weather forecasting, and recreational purposes.

2. How do balloons stay afloat in Earth's atmosphere?

Balloons stay afloat in Earth's atmosphere because of the principle of buoyancy. The gas inside the balloon is less dense than the surrounding air, causing it to rise and stay aloft. The size and weight of the balloon also play a role in its ability to stay afloat.

3. What is the purpose of launching a balloon in Earth's atmosphere?

The purpose of launching a balloon in Earth's atmosphere is to collect data and information about the atmosphere at different altitudes. This can include measuring temperature, pressure, humidity, and other atmospheric conditions. It can also be used to carry equipment for scientific experiments and to capture images and videos of the Earth from above.

4. How high can a balloon in Earth's atmosphere go?

The altitude a balloon can reach in Earth's atmosphere depends on several factors, such as the size and weight of the balloon, the type of gas used, and weather conditions. On average, balloons can reach heights of 20-30 kilometers, but some specialized balloons have reached heights of over 50 kilometers.

5. What happens to a balloon in Earth's atmosphere once it reaches its maximum height?

Once a balloon reaches its maximum height, it will eventually burst due to the decreasing atmospheric pressure at higher altitudes. The remains of the balloon and its payload will then fall back to the ground. Some balloons are equipped with parachutes or other devices to safely return the equipment to the ground for retrieval and data analysis.

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