What is the approximate diameter of the latex balloon after inflation?

  • Thread starter Thread starter Dannbr
  • Start date Start date
  • Tags Tags
    Diameter Volume
Click For Summary
SUMMARY

The discussion centers on calculating the diameter of a latex balloon inflated with 1500 liters of helium for a high-altitude experiment. Using the formula for the volume of a sphere, the initial calculation yielded a radius of 0.710 meters, resulting in a diameter of 1.42 meters. However, the expected correct diameter is 3.1 meters, indicating a potential misunderstanding regarding the conditions of the balloon at high altitude or the initial state of the helium. The discrepancy highlights the importance of considering environmental factors in physics problems.

PREREQUISITES
  • Understanding of the volume formula for a sphere: V = (4/3)(π)(r^3)
  • Basic knowledge of unit conversions, specifically from liters to cubic meters
  • Familiarity with the behavior of gases under varying pressure and altitude
  • Proficiency in algebraic manipulation to solve for radius and diameter
NEXT STEPS
  • Research the ideal gas law and its application at high altitudes
  • Learn about the effects of atmospheric pressure on balloon volume
  • Study the principles of buoyancy and how they affect balloon ascent
  • Explore advanced topics in fluid dynamics related to gas expansion
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone involved in high-altitude balloon experiments or studying gas laws and buoyancy principles.

Dannbr
Messages
9
Reaction score
0

Homework Statement



In a recent high altitude experiment, some physics students used 1500 liter of helium to inflate a latex sounding balloon that would ascend to an altitude of over 30 km. what was the approximate diameter of the (assumed spherical) balloon after inflation?


Homework Equations



volume of a sphere= (4/3)(pi)(r^3)

1 liter = 1.0E-3

The Attempt at a Solution



V=(4/3)(pi)(r^3)

so,

r = [(3V)/(4(pi))]^(1/3)

(1500liter)(1.0E-3m^3/1liter)= 1.5m^3

[((3)(1.5m^3))/((4)(pi))]^(1/3) = .710m

2r = 1.42m

Correct answer should be 3.1 m

Could someone help me out..

Thanks
 
Physics news on Phys.org
The supposed correct answer seems out of line for the question posed. Are you sure the problem didn't involve the size of the balloon at 30km altitude? Or that the helium was in some initially compressed condition?
 
No, copied question word for word. Thanks for taking a look at it.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 3 ·
Replies
3
Views
8K
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
9K
Replies
1
Views
10K
  • · Replies 5 ·
Replies
5
Views
3K