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Chapter 5: Problem 11

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

Short Answer

Expert verified
The behavior of a helium balloon rising and eventually popping can be explained by the buoyancy force and the pressure difference between the inside and outside of the balloon. The buoyancy force comes from the difference in density between the helium and air, causing the balloon to rise as it displaces air. As it rises, the atmospheric pressure decreases while the pressure inside the balloon increases, causing the helium to expand. Additionally, lower temperatures at higher altitudes weaken the balloon's material. Eventually, the stress on the weakened material due to the increased pressure difference and expanding helium becomes too great, causing the balloon to rupture and release the helium into the atmosphere.

Step by step solution

01

Understand the buoyancy force acting on a helium balloon

The buoyancy force is the reason why a helium balloon floats. This force is due to the difference in density between the gas within the balloon (helium) and the surrounding gas (air). Helium is lighter than air, so as per Archimedes' principle, the buoyancy force acting on a helium balloon is equal to the weight of the air displaced by the balloon. Since the weight of the helium within the balloon is less than the weight of the displaced air, the net force (buoyancy force - weight of the helium) acts in the upward direction, causing the balloon to rise.
02

Examine the pressure difference between the inside and the outside of the balloon

The pressure inside the balloon is determined by the volume of the helium and the elasticity of the balloon's material. The pressure inside the balloon is usually greater than the atmospheric pressure outside the balloon. As the balloon rises, the atmospheric pressure decreases, causing the pressure difference between the inside and outside of the balloon to increase. This increasing pressure difference causes the helium within the balloon to expand.
03

Consider the decreasing atmospheric pressure and temperature with altitude

As the helium balloon rises, it experiences a decrease in atmospheric pressure and temperature. The decrease in atmospheric pressure allows the helium within the balloon to expand. The lower temperature causes the material of the balloon to become more brittle and less elastic. This combination of the helium expansion and the weakened material leads to an increased likelihood of the balloon rupturing.
04

Understand the popping of the helium balloon

The helium balloon will continue to rise due to the buoyancy force, and the helium inside will continue to expand as the atmospheric pressure decreases. Eventually, the increased pressure difference between the inside and outside of the balloon causes too much stress on the weakened, cold material of the balloon, leading to its rupture. When the balloon pops, the helium inside quickly disperses into the atmosphere due to the pressure difference.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Buoyancy Force
The buoyancy force is a fascinating concept that explains why objects like helium balloons float in the air. This force is the upward push exerted by a fluid (like air) on an object submerged in it. For a helium balloon, the gas inside (helium) is much lighter than the air outside. This difference in density creates a situation where the air can lift the balloon upward. According to Archimedes' Principle, the buoyancy force equals the weight of the air displaced by the balloon.
This means if the weight of the air displaced is more than the weight of the helium and the balloon combined, the balloon will rise.
Helium balloons are a perfect example because helium is not only lighter but also non-flammable and safe. This buoyancy force is crucial in allowing the helium balloon to float and ascend higher into the sky.
Atmospheric Pressure
Atmospheric pressure is a key factor in understanding why the helium balloon behaves the way it does as it rises. Atmospheric pressure is the weight of the air in the atmosphere pressing down on the Earth's surface. As the helium balloon ascends, the surrounding atmospheric pressure decreases.
This decrease means there's less air pressing on the balloon from the outside. Inside the balloon, the helium is still exerting a pressure that might be greater than or equal to when it was on the ground.
As the balloon climbs higher, the lower atmospheric pressure allows the helium inside to exert greater outward force, thus expanding the balloon. Eventually, this increasing volume and constant pressure difference contribute to the balloon stretching to its limit, which can cause it to pop. Understanding atmospheric pressure can help explain both the rise of the balloon and its eventual popping.
Archimedes' Principle
Archimedes' Principle is at the heart of why a helium balloon gets to soar into the sky. This principle states that the buoyant force on an object in a fluid is equal to the weight of the fluid it displaces. In the case of our helium balloon, the fluid is the air.
  • The balloon displaces a certain amount of air as it is introduced into the atmosphere.
  • Since helium inside the balloon is less dense than the air displaced, the buoyant force pushes the balloon upward.
  • As the balloon continues to displace more air, the buoyancy force remains greater than the weight of the helium inside, ensuring the balloon's rise.
Archimedes' Principle is a powerful explanation and is the reason behind not just helium balloons but also why ships float or why you feel light in water. Understanding this principle can illuminate why a seemingly heavy object can, in fact, be lifted by water or air.

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Most popular questions from this chapter

A tank contains a mixture of 52.5 g oxygen gas and 65.1 \(\mathrm{g}\) carbon dioxide gas at \(27^{\circ} \mathrm{C}\) . The total pressure in the tank is 9.21 atm. Calculate the partial pressures of each gas in the container.

Consider the unbalanced chemical equation below: $$\mathrm{CaSiO}_{3}(s)+\mathrm{HF}(g) \longrightarrow \mathrm{CaF}_{2}(a q)+\mathrm{SiF}_{4}(g)+\mathrm{H}_{2} \mathrm{O}( )$$ Suppose a 32.9 -g sample of \(\mathrm{CaSiO}_{3}\) is reacted with 31.8 \(\mathrm{L}\) of \(\mathrm{HF}\) at \(27.0^{\circ} \mathrm{C}\) and 1.00 \(\mathrm{atm}\) . Assuming the reaction goes to completion, calculate the mass of the \(\mathrm{SiF}_{4}\) and \(\mathrm{H}_{2} \mathrm{O}\) produced in the reaction.

At elevated temperatures, sodium chlorate decomposes to produce sodium chloride and oxygen gas. A 0.8765-g sample of impure sodium chlorate was heated until the production of oxygen gas ceased. The oxygen gas collected over water occupied 57.2 \(\mathrm{mL}\) at a temperature of $22^{\circ} \mathrm{C}$ and a pressure of 734 torr. Calculate the mass percent of \(\mathrm{NaClO}_{3}\) in the original sample. (At \(22^{\circ} \mathrm{C}\) the vapor pressure of water is 19.8 torr.)

One of the chemical controversies of the nineteenth century concerned the element beryllium (Be). Berzelius originally claimed that beryllium was a trivalent element (forming \(\mathrm{Be}^{3+}\) ions) and that it gave an oxide with the formula \(\mathrm{Be}_{2} \mathrm{O}_{3}\) . This resulted in a calculated atomic mass of 13.5 for beryllium. In formulating his periodic table, Mendeleev proposed that beryllium was divalent (forming \(\mathrm{Be}^{2+}\) ions) and that it gave an oxide with the formula BeO. This assumption gives an atomic mass of \(9.0 .\) In \(1894,\) A. Combes (Comptes Rendus \(1894,\) p. 1221 ) reacted beryllium with the anion \(C_{5} \mathrm{H}_{7} \mathrm{O}_{2}^{-}\) and measured the density of the gaseous product. Combes's data for two different experiments are as follows: $$\begin{array}{lll}{\text { Mass }} & {0.2022 \mathrm{g}} & {0.2224 \mathrm{g}} \\ {\text { Volume }} & {22.6 \mathrm{cm}^{3}} & {26.0 \mathrm{cm}^{3}} \\ {\text { Temperature }} & {13^{\circ} \mathrm{C}} & {17^{\circ} \mathrm{C}} \\ {\text { Pressure }} & {765.2 \mathrm{mm} \mathrm{Hg}} & {764.6 \mathrm{mm}}\end{array}$$ If beryllium is a divalent metal, the molecular formula of the product will be \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{2} ;\) if it is trivalent, the formula will be \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{3} .\) Show how Combes's data help to confirm that beryllium is a divalent metal.

A 2.50-L container is filled with 175 g argon. a. If the pressure is 10.0 atm, what is the temperature? b. If the temperature is 225 K, what is the pressure?
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