Question

A balloon filled with Helium (M=4g/mol) has diameter of .5m at sea level. For this problem, we'll assume a balloon can expand or contract when the outside pressure decreases or increases, so that internal pressure is always equal to the outside pressure. A) when the balloon is immersed in water to a depth of 1.5m, what is the new diameter? B) If the same balloon is released, and floats up to an elevation of 1500m (rising through air with constant density of 1.2 kg/m^3), what is the new diameter?

Answer 1

A)intial pressure , Pi = 1.01 *10^5 Pa

let the final diameter is d2

d1 = 0.5 m

final pressure , Pf = 1.01 *10^5 + 1000 * 9.8 * 1.5

Pf = 1.157 *10^5 Pa

Using ideal gas equation

Pf * v2 = Pi * V1

1.157 *10^5 * (4/3) *pi*d2^3 = 1.01 *10^5 * (4/3) * pi *(0.5)^3

solving for d2

d2 = 0.48 m

the new diamter is 0.48 m

b) at the 1500 m height

final pressure , Pf = 1.01 *10^5 - 1.2 * 1500 * 9.8

Pf = 8.336 *10^4 Pa

Using ideal gas equation

Pf * v2 = Pi * V1

8.336 *10^4 * (4/3) *pi*d2^3 = 1.01 *10^5 * (4/3) * pi *(0.5)^3

solving for d2

d2 = 0.533 m

the new diamter is 0.533 m