Question

A consumer upset with the latest trend of postal rate increases has decided to try to send letters by balloon even though they may not reach their intended destination. A 79200 cm3 gas-filled balloon will provide enough lift for a 43.4 g package to be accelerated upward at a rate of 3.40 m/s2. For these circumstances, calculate the density of the gas the consumer fills the balloon with. The acceleration due to gravity is g = 9.81 m/s2 and the density of air is ?air = 1.16 kg/m3. Neglect the mass of the balloon material and the volume of the package.

Answer 1

Answer:

0.31338 kg/m³

Explanation:

v = Volume of gas inside balloon = 79200 cm³

a = Acceleration of the package = 3.4 m/s²

g = Acceleration due to gravity = 9.81 m/s²

$\rho$ = Density

m = Mass of gas + Package

The buoyant force and apparent weight balance each other

$F_b=W_a\\\Rightarrow \rho gv=m(g+a)\\\Rightarrow m=\frac{\rho gv}{g+a}\\\Rightarrow m=\frac{1.16\times 9.81\times 79200\times 10^{-6}}{9.81+3.4}\\\Rightarrow m=0.06822\ kg$

Mass of the gas in the balloon and package is 0.06822 kg

Mass of gas

$0.06822-0.0434=0.02482\ kg$

The mass of gas in the balloon is 0.02482 kg

Density = Mass / Volume

$\rho=\frac{m}{v}\\\Rightarrow \rho=\frac{0.02482}{79200\times 10^{-6}}\\\Rightarrow \rho=0.31338\ kg/m^3$

The density of the gas in the balloon is 0.31338 kg/m³