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Chapter 8: Problem 11

If you release a helium balloon, it soars upward and eventually pops. Explain this behavior.

Short Answer

Expert verified
A helium balloon soars upward and eventually pops due to the combination of density differences between helium and air, the buoyancy force, and the expansion of the helium gas as the balloon ascends. Being lighter than air, the helium displaces denser air, resulting in a buoyancy force that causes the balloon to rise. As it rises, the external air pressure decreases, causing the helium gas to expand and increase the balloon's volume. When the balloon's material reaches its critical stress point, it can no longer withstand the expansion, leading to rupture and popping of the balloon.

Step by step solution

01

1. Understanding Density Differences

Helium (He) is a lighter gas than the air we breathe, which is primarily composed of nitrogen (N鈧�) and oxygen (O鈧�). The average molar mass of air is about 29 g/mol, while helium has a molar mass of 4 g/mol. This difference in molar mass leads to a significant difference in densities between helium and air (helium being less dense than air).
02

2. Buoyancy Force

A helium-filled balloon rises in the air due to the buoyancy force acting on it. The buoyancy force on an object submerged in a fluid (such as air) is equal to the weight of the fluid displaced by this object. This can be represented by the formula: \( F_{buoyancy} = V * (蟻_{air} - 蟻_{helium}) * g\) , where V is the volume of the balloon, 蟻 is the density of the respective gases, and g is the acceleration due to gravity. Since the surrounding air is denser than helium, the buoyancy force exceeds the weight of the helium plus the balloon and other materials, causing it to rise in the air.
03

3. Balloon Expansion

As the helium balloon rises in the atmosphere, the air pressure around it decreases. Helium gas inside the balloon obeys the ideal gas law, which states: \( PV = nRT\), where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature. When pressure (P) decreases, either volume (V) increases or temperature (T) decreases to maintain the equation's balance. In this case, since the temperature difference isn't significant enough to balance the pressure decrease, the volume of the balloon increases. This results in the expansion of the balloon as it rises.
04

4. Balloon Popping

The balloon material has a finite elasticity, which means it can only stretch to a certain limit. As the balloon rises and keeps expanding due to lower external pressure, it reaches its maximum stretching capacity, indicated by the balloon's critical stress or strain point. Once the stress due to the expansion surpasses this critical point, the balloon material can no longer withstand it, causing the balloon to rupture and pop.

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Most popular questions from this chapter

A particular balloon is designed by its manufacturer to be inflated to a volume of no more than 2.5 L. If the balloon is filled with 2.0 L helium at sea level, is released, and rises to an altitude at which the atmospheric pressure is only \(500 . \mathrm{mm}\) Hg, will the balloon burst? (Assume temperature is constant.)

Calculate the root mean square velocities of \(\mathrm{CH}_{4}(g)\) and \(\mathrm{N}_{2}(g)\) molecules at \(273 \mathrm{K}\) and \(546 \mathrm{K}\).

You have a helium balloon at \(1.00\) atm and \(25^{\circ} \mathrm{C}\). You want to make a hot-air balloon with the same volume and same lift as the helium balloon. Assume air is \(79.0 \%\) nitrogen and \(21.0 \%\) oxygen by volume. The "lift" of a balloon is given by the difference between the mass of air displaced by the balloon and the mass of gas inside the balloon. a. Will the temperature in the hot-air balloon have to be higher or lower than \(25^{\circ} \mathrm{C} ?\) Explain. b. Calculate the temperature of the air required for the hotair balloon to provide the same lift as the helium balloon at \(1.00\) atm and \(25^{\circ} \mathrm{C}\). Assume atmospheric conditions are \(1.00\) atm and \(25^{\circ} \mathrm{C}\).

A mixture of \(1.00 \mathrm{g} \mathrm{H}_{2}\) and \(1.00 \mathrm{g}\) He is placed in a \(1.00-\mathrm{L}\) container at \(27^{\circ} \mathrm{C}\). Calculate the partial pressure of each gas and the total pressure.

Metallic molybdenum can be produced from the mineral moIybdenite, MoS \(_{2}\). The mineral is first oxidized in air to molybdenum trioxide and sulfur dioxide. Molybdenum trioxide is then reduced to metallic molybdenum using hydrogen gas. The balanced equations are $$\begin{array}{l}\operatorname{MoS}_{2}(s)+\frac{7}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{MoO}_{3}(s)+2 \mathrm{SO}_{2}(g) \\\\\mathrm{MoO}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{Mo}(s)+3 \mathrm{H}_{2} \mathrm{O}(l)\end{array}$$ Calculate the volumes of air and hydrogen gas at \(17^{\circ} \mathrm{C}\) and \(1.00\) atm that are necessary to produce \(1.00 \times 10^{3} \mathrm{kg}\) pure molybdenum from MoS \(_{2}\). Assume air contains \(21 \%\) oxygen by volume, and assume \(100 \%\) yield for each reaction.
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