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.)

Short Answer

Expert verified
Using Boyle's law equation, \(P1V1 = P2V2\), we find that the final volume of the helium balloon will be approximately 3.04 L when it rises to an altitude where the atmospheric pressure is 500 mm Hg. Since this volume is greater than the maximum allowed volume of the balloon (2.5 L), the balloon will burst when it rises to an altitude where the atmospheric pressure is only 500 mm Hg, assuming constant temperature.

Step by step solution

01

List given information and convert to appropriate units

We are given the following information: - The maximum volume of the balloon is 2.5 L. - The volume of helium at sea level is 2.0 L. - The atmospheric pressure at sea level (P1) is 760 mm Hg. - The atmospheric pressure at high altitude (P2) is 500 mm Hg. - The temperature is constant. We'll need to make sure that all pressure values are in the same unit: atm. To do this, we will use the conversion 1 atm = 760 mm Hg.
02

Convert the given pressures to atm

We will convert P1 and P2 to atm using the conversion factor: P1 (atm) = P1 (mm Hg) × (1 atm / 760 mm Hg) P1 = 760 mm Hg × (1 atm / 760 mm Hg) = 1 atm P2 (atm) = P2 (mm Hg) × (1 atm / 760 mm Hg) P2 = 500 mm Hg × (1 atm / 760 mm Hg) ≈ 0.658 atm Now that both pressures are in atm, we can use Boyle's law equation.
03

Use Boyle's law to find the final volume

At constant temperature, Boyle's law states that the product of the initial volume and initial pressure is equal to the product of the final volume and final pressure: P1V1 = P2V2 To find the final volume (V2), we can rearrange the equation for V2: V2 = (P1V1) / P2 Now, plug in the values: V2 = (1 atm × 2.0 L) / 0.658 atm V2 ≈ 3.04 L
04

Compare the final volume to the maximum allowed volume

We found that the final volume of the helium balloon will be approximately 3.04 L when it rises to an altitude where the atmospheric pressure is 500 mm Hg. This volume is greater than the maximum allowed volume of the balloon, which is 2.5 L.
05

Determine if the balloon will burst

Since the final volume (3.04 L) exceeds the maximum allowed volume (2.5 L) of the balloon, we can conclude that the balloon will burst when it rises to an altitude where the atmospheric pressure is only 500 mm Hg, assuming a constant temperature.

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Key Concepts

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

Atmospheric Pressure
Atmospheric pressure is the force per unit area exerted on a surface by the weight of the air above that surface in the atmosphere of Earth. It is an important factor to consider when studying the behavior of gases, as it can affect their volume and pressure. At sea level, the standard atmospheric pressure is about 101,325 Pascals or 760 mm Hg (millimeters of mercury). As altitude increases, atmospheric pressure decreases because there are fewer air molecules above the surface.

This concept is crucial when we talk about gases expanding or contracting in different altitudes. The exercise shows a practical application of atmospheric pressure in predicting whether a helium balloon will burst when it ascends to higher altitudes. In the context of the balloon, as it rises, the external pressure around it decreases, which influences the gas inside to expand.
Gas Laws
Gas laws are the physical laws that describe the properties of gases, including their volume, temperature, pressure, and quantity. The most prominent of the gas laws are Boyle's Law, Charles's Law, Gay-Lussac's Law, and the Combined Gas Law. These laws enable us to predict the behavior of a gas when one of the properties changes, assuming all other properties remain constant or are controlled.

Boyle's Law

Boyle's Law, in particular, explains the relationship between the pressure and volume of a gas at constant temperature. According to Boyle’s Law, for a given amount of gas at constant temperature, the volume of the gas is inversely proportional to its pressure. This means if the pressure increases, the volume decreases, and vice versa, which is the focus of the exercise provided.
Volume and Pressure Relationship
Understanding the relationship between volume and pressure of a gas under constant temperature is key to solving problems that apply Boyle's Law. This inverse relationship is mathematically represented as:
\(P_1V_1 = P_2V_2\)
where \(P_1\) and \(V_1\) are the initial pressure and volume and \(P_2\) and \(V_2\) are the final pressure and volume, respectively. When volume increases, pressure decreases, and when volume decreases, pressure increases.

In the balloon exercise, this relationship helps to predict whether the balloon will burst or not. If the balloon’s volume at sea level is 2.0 L and the pressure decreases as it ascends, the volume will increase. We calculate its final volume under the reduced atmospheric pressure using Boyle’s Law formula, and by comparing it to the balloon’s maximum allowed volume, we conclude if it is going to burst. The exercise shows that the final volume of 3.04 L exceeds the safe limit, indicating the balloon would indeed burst.

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Most popular questions from this chapter

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: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.

Consider an equimolar mixture (equal number of moles) of two diatomic gases \(\left(A_{2} \text { and } B_{2}\right)\) in a container fitted with a piston. The gases react to form one product (which is also a gas) with the formula \(A_{x} B_{y}\). The density of the sample after the reaction is complete (and the temperature returns to its original state) is \(1.50\) times greater than the density of the reactant mixture. a. Specify the formula of the product, and explain if more than one answer is possible based on the given data. b. Can you determine the molecular formula of the product with the information given or only the empirical formula?

A gauge on a compressed gas cylinder reads \(2200 \mathrm{psi}\) (pounds per square inch; 1 atm \(=14.7\) psi). Express this pressure in each of the following units. a. standard atmospheres b. megapascals (MPa) c. torr

A \(20.0\) -\(\mathrm{L}\) nickel container was charged with \(0.859\) atm of xenon gas and \(1.37\) atm of fluorine gas at \(400^{\circ} \mathrm{C}\). The xenon and fluorine react to form xenon tetrafluoride. What mass of xenon tetrafluoride can be produced assuming \(100 \%\) yield?

Complete the following table for an ideal gas. $$\begin{aligned} &-\\\ &\begin{array}{|llll|} \hline {P} & {V} & {n} & {\boldsymbol{T}} \\ \hline 7.74 \times 10^{3} \mathrm{Pa} & 12.2 \mathrm{mL} & & 25^{\circ} \mathrm{C} \\ \hline & 43.0 \mathrm{mL} & 0.421 \mathrm{mol} & 223 \mathrm{K} \\ \hline 455 \text { torr } & & 4.4 \times 10^{-2} \mathrm{mol} & 331^{\circ} \mathrm{C} \\ \hline 745 \mathrm{mm} \mathrm{Hg} & 11.2 \mathrm{L} & 0.401 \mathrm{mol} & \\ \hline \end{array} \end{aligned}$$

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