Calculating Helium Balloon Buoyancy Force at Launch and 2km Altitude

[JWSiow]

Homework Statement

A helium meteorological balloon is made of a bag of impervious fabric that does not stretch, and when fully inflated forms a spherical shell of 1m diameter enclosing the He. At launch it is filled with He (at STP) to 15% capacity. The launch takes place in the Antarctic, in winter, at a temperature of 220K, so it is reasonable to assume that the temperature does not vary much with height. Note all assumptions and approximations made.

a)What volume does the balloon occupy at launch at sea level?
b)What mass of He is in the balloon?
c)What is the mass of air displaced by the balloon at sea level?
d)Estimate the buoyancy force at launch.
e)After launch the balloon rises. Estimate the buoyancy force when it has reached an altitude of 2km.

Homework Equations

PV = nRT
n=m/M
V=4/3$$\pi$$r³

The Attempt at a Solution

a)V=4/3$$\pi$$ x 0.5³ x 0.15 = 0.0785m³

b)No densities were given.

PV = nRT. Therefore n = PV/RT.
P = 1.013x10⁵Pa (given from formula sheet)
T = 273K <-- Should this be 220K?
V = 0.0785m³
M = 4 atomic mass units (given)

Therefore n = 3.504mol
m=nM = 3.504 x 4 =14.01g

c) Mass displaced = mass of Helium = 14.01g.

d)
m = 14.01/1000 = 0.014kg
Buoyancy force = weight of volume displaced = mg = 0.137N

e)From a) and b), density = m/V = 0.014/0.0785 = 0.178kg/m³
At 2km, P=75100Pa (given). n, R and T remain constant.
V = nRT/P = (3.504 x 8.314 x 220)/75100 = 0.0853m³

Assuming density remains constant as altitude changes,
m=density/Volume = 0.178/0.0853 = 0.0152kg.

Buoyance force = 0.0152 x 9.8 = 0.149N

Basically, I'm just not sure if this is correct and want someone to check if my working out is all right :) The part I'm most unsure of is part e) and the fact that the buoyancy forces are so small!.

 

[dulrich]

b) ... T = 273K <-- Should this be 220K?

Yes, it should be 220K.

c) Mass displaced = mass of Helium = 14.01g.

Mass displaced = mass of air, not helium.

...Assuming density remains constant as altitude changes...

Nope. Density will change. You've already got the new volume. Just use the method from parts (b), (c), and (d) to recalculate the bouyant force.

 

[JWSiow]

But, I used the number of moles found in part b) to calculate the new volume, which means if I used the new volume to calculate the mass using the same method, I'd get the same mass, and the buoyancy force would stay constant...

 

[dulrich]

You are right. I got the masses mixed up in my mind when I was responding to your question.

You can assume the density of the air doesn't change (the density of the helium does, but that doesn't matter to the question). Given the volume, recalculate the mass and weight of air displaced by the balloon.

 

[rwisz]

Your calculations seem to be correct. However, it is important to note that the buoyancy force is small because the balloon is only filled to 15% capacity at launch. If it were filled to full capacity, the buoyancy force would be much larger. Also, keep in mind that these calculations are based on various assumptions and approximations, so they may not be entirely accurate. It would be important to take into consideration other factors such as air resistance and the change in temperature with altitude to get a more precise estimation of the buoyancy force. Overall, your approach and calculations seem to be reasonable.