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A 15.0-L tank is filled with helium gas at a pressure of \(1.00 \times 10^{2}\) atm. How many balloons (each \(2.00 \mathrm{~L}\) ) can be inflated to a pressure of 1.00 atm, assuming that the temperature remains constant and that the tank cannot be emptied below 1.00 atm?

Short Answer

Expert verified
The number of balloons that can be inflated with the given helium tank and constraints is 0.

Step by step solution

01

Identify the given information

We are given the following information: - Tank volume: V₁ = 15.0 L - Initial tank pressure: P₁ = 1.00 × 10² atm - Final tank pressure: P₃ = 1.00 atm (as it cannot be emptied below this pressure) - Balloon volume: V₂ = 2.00 L - Balloon pressure: P₂ = 1.00 atm
02

Calculate the initial moles of helium

We can use the Ideal Gas Law, PV=nRT, to calculate n, which is the number of moles in the tank. However, we don't have the temperature or gas constant R. We can rearrange the Ideal Gas Law to determine the initial moles: n₁ = P₁V₁/RT Since the temperature is constant, the ratio P₁V₁/RT remains constant. Therefore, we can rewrite the equation as: n₁/n₃ = P₁V₁/P₃V₃ We have the values for P₁, V₁, and P₃; however, we do not have V₃. To find this, we will first find the volume of gas that remains in the tank at 1.00 atm pressure, which can be found using the Boyle's Law (P₁V₁ = P₂V₂): V₃ = P₁V₁/P₃ = (1.00 × 10² atm × 15.0 L) / 1.00 atm = 1500 L Now we can find the initial moles, n₁: n₁/n₃ = (1.00 × 10² atm × 15.0 L) / (1.00 atm × 1500 L) n₁ = n₃ This means that the moles of helium gas in the tank do not change as the pressure changes from 1.00 x 10² atm to 1.00 atm.
03

Calculate the moles available for inflating balloons

Now that we know the initial moles are equal to the remaining moles, we can calculate the moles available for inflating balloons: n₂ = n₁ - n₃ Since n₁ = n₃, the equation can be simplified to: n₂ = n₁ - n₁ = 0 This means that there are no additional moles of gas available for inflating balloons beyond maintaining the 1.00 atm pressure in the tank.
04

Determine the number of balloons that can be inflated

Since there are no additional moles of gas available for inflating balloons, the number of balloons that can be inflated is 0. Thus, the answer is 0 balloons can be inflated with the given constraints.

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Key Concepts

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

Boyle's Law
Boyle's Law is fundamental in understanding how gas pressures and volumes affect one another under a constant temperature. Essentially, this law implies that if a gas's temperature remains consistent, then its pressure and volume have an inverse relationship; as one increases, the other decreases. The law is mathematically expressed as:
\( P_1V_1 = P_2V_2 \).
The connection between these variables enables us to solve various problems, like determining how many balloons can be filled from a gas tank without changing the temperature. The calculation in the problem assumes that the pressure inside the tank cannot fall below a certain threshold (1.00 atm) which is crucial in maintaining the gas's volume. When applying Boyle's Law, one must cautiously track the volume and pressure to resolve the actual usable gas—an essential aspect for students to keep tabs on while they navigate through homework problems related to gas laws.
Pressure-Volume Relationship
Exploring the pressure-volume relationship of gases leads us to an intuitive understanding of how they behave under different scenarios. This movement towards understanding is not only about Boyle's Law but also about visualizing gases in a real-world context.
In the given exercise, the pressure-volume relationship is at the core of determining the capability of a helium tank to inflate a certain number of balloons. We see that if the pressure within the tank is reduced, more volume becomes available for use—however, the tank's minimal pressure creates a constraint. It's critical to acknowledge that these relationships don't occur in isolation; they can be affected by factors such as the surrounding temperature and the amount (in moles) of the gas. Proper application combined with careful calculation allows students to grasp how precisely these variables interact, particularly when dealing with labs or practical gas usage situations.
Calculating Moles of Gas
Moles are a key unit in chemistry, specifically when it comes to gauging amounts of substances. Regarding gases, calculating the number of moles can provide insightful information about the quantity of gas present. It is a central part of applying the Ideal Gas Law which is represented as:
\( PV = nRT \)
where 'P' stands for pressure, 'V' for volume, 'n' for the number of moles, 'R' for the gas constant, and 'T' for temperature. In practice, if you know the pressure, volume, and temperature of a gas, you can calculate the number of moles. For students tackling gas law problems, it is paramount to recognize that while the number of moles might remain unchanged in a reactive process, available moles for work such as inflating balloons may differ. This aspect was crucial in the textbook exercise—despite the initial and remaining moles being identical due to Boyle's Law, they could not be used to inflate balloons. Understanding how to calculate moles in different pressures and volumes firmly roots a student's competence in solving gas law exercises.

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Most popular questions from this chapter

An herbicide is found to contain only \(\mathrm{C}, \mathrm{H}, \mathrm{N},\) and \(\mathrm{Cl} .\) The complete combustion of a 100.0 -mg sample of the herbicide in excess oxygen produces \(83.16 \mathrm{~mL}\) of \(\mathrm{CO}_{2}\) and \(73.30 \mathrm{~mL}\) of \(\mathrm{H}_{2} \mathrm{O}\) vapor at STP. A separate analysis shows that the sample also contains \(16.44 \mathrm{mg}\) of Cl. (a) Determine the percent composition of the substance. (b) Calculate its empirical formula. (c) What other information would you need to know about this compound to calculate its true molecular formula?

Natural gas is very abundant in many Middle Eastern oil fields. However, the costs of shipping the gas to markets in other parts of the world are high because it is necessary to liquefy the gas, which is mainly methane and has a boiling point at atmospheric pressure of \(-164{ }^{\circ} \mathrm{C}\). One possible strategy is to oxidize the methane to methanol, \(\mathrm{CH}_{3} \mathrm{OH},\) which has a boiling point of \(65^{\circ} \mathrm{C}\) and can therefore be shipped more readily. Suppose that \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane at atmospheric pressure and \(25^{\circ} \mathrm{C}\) is oxidized to methanol. (a) What volume of methanol is formed if the density of \(\mathrm{CH}_{3} \mathrm{OH}\) is \(0.791 \mathrm{~g} / \mathrm{mL} ?\) (b) Write balanced chemical equations for the oxidations of methane and methanol to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\). Calculate the total enthalpy change for complete combustion of the \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane just described and for complete combustion of the equivalent amount of methanol, as calculated in part (a). (c) Methane, when liquefied, has a density of \(0.466 \mathrm{~g} / \mathrm{mL} ;\) the density of methanol at \(25^{\circ} \mathrm{C}\) is \(0.791 \mathrm{~g} / \mathrm{mL}\). Compare the enthalpy change upon combustion of a unit volume of liquid methane and liquid methanol. From the standpoint of energy production, which substance has the higher enthalpy of combustion per unit volume?

Many gases are shipped in high-pressure containers. Consider a steel tank whose volume is 55.0 gallons that contains \(\mathrm{O}_{2}\) gas at a pressure of \(16,500 \mathrm{kPa}\) at \(23{ }^{\circ} \mathrm{C}\). (a) What mass of \(\mathrm{O}_{2}\) does the tank contain? (b) What volume would the gas occupy at STP? (c) At what temperature would the pressure in the tank equal 150.0 atm? (d) What would be the pressure of the gas, in \(\mathrm{kPa},\) if it were transferred to a container at \(24^{\circ} \mathrm{C}\) whose volume is \(55.0 \mathrm{~L} ?\)

A quantity of \(\mathrm{N}_{2}\) gas originally held at 5.25 atm pressure in a 1.00-L container at \(26^{\circ} \mathrm{C}\) is transferred to a 12.5-L container at \(20^{\circ} \mathrm{C}\). A quantity of \(\mathrm{O}_{2}\) gas originally at \(5.25 \mathrm{~atm}\) and \(26^{\circ} \mathrm{C}\) in a 5.00-L container is transferred to this same container. What is the total pressure in the new container?

The planet Jupiter has a surface temperature of \(140 \mathrm{~K}\) and a mass 318 times that of Earth. Mercury (the planet) has a surface temperature between \(600 \mathrm{~K}\) and \(700 \mathrm{~K}\) and a mass 0.05 times that of Earth. On which planet is the atmosphere more likely to obey the ideal-gas law? Explain.

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