# Buoyancy Force

1. Jun 30, 2010

### JWSiow

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
PV = nRT
n=m/M
V=4/3$$\pi$$r3

3. 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!.

2. Jun 30, 2010

### dulrich

Yes, it should be 220K.
Mass displaced = mass of air, not helium.
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.

3. Jun 30, 2010

### 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...

4. Jun 30, 2010

### 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.