Question

A spherical balloon contains 10m^3 of helium at 1 atmosphere and 0C. What will be the diameter of the balloon at the altitute of 3.2 kilometer, where the temperature are -7C and 85000 N/m^2 respectively? What would be the diameter of the balloon if instead of the helium we had hydrogen? Find the weight difference for both cases.

Answer 1

By assumption
V1 = 10 m^3
p1 = 1 atmosphere = 101325
N/m^2,
T1 = 0C = 273 K
p2 = 85000 N/m^2
T2 = -7 C = 273-7
= 266 K

We have that

p1 * V1 / T1 = p2 * V2 /
T2,

whence

V2& =& p1 * V1 * T2 /(T1 * p2) =
& =
101325 * 10 * 266 / (273 * 85000) = 11.615

The volume of a ball of
diameter d=2r is equal to

V = 4/3 pi r^3 = 4/3 pi (d/2)^3 = pi
d^3 / 6,

whence

d2 = (6 V2 / pi)^(1/3) = 2.81
m.


Notice that the identity pV/T=const holds for any ideal gas, so if
instead the helium the balloon was filled with
hydrogen, the diameter of the
ball in the second case would be the same 2.81 m.


Recall that the
density of the helium is 0.17847 kg/m^3, while the density of the hydrogen is
0.0899 kg/m^3.
Therefore the weight of the balloon with helium is equal to

P(He) = m g = pho V g = 0.0899 * 10 * 9.8 =& 8.81 N,
and the weight
of the balloon with hydrogen is
P(H) = m g = pho V g = 0.17847 * 10 * 9.8
=& 17.49 N.

Therefore the weight difference is
P(H) - P(He) =
17.49 - 8.81 = 8.68 N.