Question

Explain why: A balloon filled with helium gas does not rise in air indefinitely but halts after a certain height. Why?

Answer 1

Hint:The balloon rises up in the air because the weight of the helium gas is less than the weight of the displaced air by the balloon that is upthrust. As we rise in the atmosphere, the density of the air decreases. Recall the law of floatation to answer this question.

Complete answer:
We know that when we dip an empty container in the water, the buoyant force acts on the container to oppose the sinking. The direction of the buoyant force or upthrust is always upwards. The balloon rises up in the air because the weight of the helium gas is less than the weight of the displaced air by the balloon that is upthrust. We can see the upward force is greater than the downward (force of gravity or weight) force. We can express this case as,
\[mg < {F_b}\]
\[ \Rightarrow \left( {{\rho _h}{V_b}g} \right) \downarrow < \left( {{\rho _a}{V_a}g} \right) \uparrow \]
Here, \[{\rho _h}\] is the density of hydrogen, \[{V_b}\] is the volume of balloon, g is the acceleration due to gravity, \[{\rho _a}\] is the density of air and \[{V_a}\] is the volume of the displaced air.

In the above equation, the volume of the balloon is equal to the volume of the displaced air at any point. Now, we know that as we rise in the atmosphere, the density of the air decreases. Therefore, at a certain altitude, the density of the air equals the density of helium gas. At this point, we can see, the weight of the helium is equal to the buoyant force.
\[{\rho _h}{V_b}g = {\rho _a}{V_a}g\]

Since no net external force acts on the balloon at this point, the balloon will stop rising.

Note: In the expression for the buoyant force, the density term is the density of air or liquid and not the density of the body. We have assumed that no wind is flowing while the balloon is rising in the air. The external fluctuations will flow away the balloon since its weight is negligible.