/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 131 A \(15.0-\mathrm{L}\) tank is fi... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(15.0-\mathrm{L}\) tank is filled with \(\mathrm{H}_{2}\) to a pressure of \(2.00 \times 10^{2}\) atm. How many balloons (each \(2.00 \mathrm{L}\) ) can be inflated to a pressure of 1.00 atm from the tank? Assume that there is no temperature change and that the tank cannot be emptied below \(1.00\) atm pressure.

Short Answer

Expert verified
The number of balloons that can be inflated to a pressure of 1 atm from the tank is 199.

Step by step solution

01

Find the number of moles of hydrogen gas in the tank

To find the number of moles of hydrogen gas in the tank, we can use the ideal gas law, which states: \[PV = nRT\] Where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the gas constant (\(0.0821 \frac{L \cdot atm}{K \cdot mol}\)), and \(T\) is the temperature. Since the temperature remains constant, we can rewrite the equation as: \[n = \frac{PV}{RT}\] Now, we can plug in the given values for the initial pressure, volume, and temperature of the hydrogen gas in the tank: \(P = 2.00 \times 10^2 atm\) \(V = 15.0 L\) \(T = constant\) \(n_{tank} = \frac{(2.00 \times 10^2 atm)(15.0 L)}{(0.0821 \frac{L \cdot atm}{K \cdot mol})T}\)
02

Find the number of moles of hydrogen gas needed for each balloon

Next, we need to find how many moles of hydrogen gas are needed to inflate each balloon to the desired pressure and volume. We can use the ideal gas law again with the given values for the balloons: \(P = 1.00 atm\) \(V = 2.00 L\) \(T = constant\) \(n_{balloon} = \frac{(1.00 atm)(2.00 L)}{(0.0821 \frac{L \cdot atm}{K \cdot mol})T}\)
03

Calculate the number of balloons that can be inflated from the tank

We are given that the tank cannot be emptied below a pressure of 1 atm. To find how many moles of gas will remain in the tank at a pressure of 1 atm, we can use the ideal gas law again: \(n_{remaining} = \frac{(1.00 atm)(15.0 L)}{(0.0821 \frac{L \cdot atm}{K \cdot mol})T}\) Now we can find the total number of moles that can be used to inflate balloons: \(n_{usable} = n_{tank} - n_{remaining}\) To find the number of balloons that can be inflated, we will simply divide this number by the number of moles needed for each balloon: \(number\_of\_balloons = \frac{n_{usable}}{n_{balloon}}\) Note that \(T\) and \(\frac{L \cdot atm}{K \cdot mol}\) will cancel out, and the equation simplifies to: \(number\_of\_balloons = \frac{((2.00 \times 10^2 atm)(15.0 L) - (1.00 atm)(15.0 L))/15.0 L}{(1.00 atm)(2.00 L)/2.00 L}\) \(number\_of\_balloons = \frac{(2.00 \times 10^2 - 1.00) atm}{1.00 atm}\) Now calculate the number of balloons: \(number\_of\_balloons = 200 - 1 = 199\) So, 199 balloons can be inflated to a pressure of 1 atm from the tank.

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Key Concepts

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

Gas Pressure and Volume
Understanding the relationship between gas pressure and volume is critical when working with gases in closed systems. This relationship is governed by Boyle's Law, which states that for a fixed amount of an ideal gas kept at a constant temperature, the pressure of the gas is inversely proportional to its volume.

To put it simply, when the volume of the gas decreases, its pressure increases, and vice versa, as long as the temperature does not change. In practical settings, this principle allows us to predict how gases will behave when confined in a container like a tank or a balloon.

For example, with the exercise provided, if we inflate a balloon from a tank, we are essentially transferring gas from a large volume (the tank) to a smaller volume (the balloon). Using Boyle's Law, we can understand that the tank's pressure drops as gas is removed to fill balloons—until a limiting pressure point is reached. It's this fundamental concept that forms the basis for solving the provided exercise.
Molar Volume of Gas
The molar volume of a gas is the volume occupied by one mole of the gas under specified conditions of pressure and temperature. Avogadro’s Law tells us that equal volumes of gases, at the same temperature and pressure, contain an equal number of molecules—or moles—of gas. For ideal gases at standard temperature and pressure (STP), which is 0 degrees Celsius and 1 atmosphere of pressure, the molar volume is approximately 22.4 liters.

This concept is particularly useful in stoichiometry calculations involving gases where we can use the molar volume to translate between the volume of gas and the amount of substance in moles. For instance, in the exercise, understanding that at constant temperature the volume of the gas directly relates to the number of moles, supports finding how much gas we can use to inflate a certain number of balloons.
Stoichiometry of Gases
Stoichiometry is the aspect of chemistry that relates to the calculation of the quantities of reactants and products involved in chemical reactions. When it comes to gases, stoichiometry involves using the ideal gas law and the molar volume concept to deal with reactions where gases are either reactants or products.

In practical terms, like in our exercise, stoichiometry allows us to determine how much gas (in moles) is required to achieve a certain action, such as inflating balloons. The ideal gas law equation provides a direct relationship between the pressure, volume, temperature, and number of moles of a gas which allows us to calculate the exact amounts of gas needed or produced in a chemical reaction.
Gas Constant (R)
The gas constant, denoted as R, is an essential figure in the ideal gas law equation. It represents the constant of proportionality that relates the pressure, volume, and temperature to the number of moles of an ideal gas. In other words, it's a bridge that unites these different units of measurement.

The value of R depends on the units used for pressure, volume, and temperature. A common value used for R is 0.0821 L·atm/(K·mol), which applies when the pressure is measured in atmospheres, the volume is in liters, and the temperature is in Kelvin.

When solving gas-related problems, it's important to consistently use the correct value of R to ensure that the units cancel properly and the correct outcome is obtained. As shown in the exercise, the value of R is crucial in calculating the number of moles present, which leads to determining how many balloons can be filled with the tank's hydrogen gas.

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Most popular questions from this chapter

A sealed balloon is filled with \(1.00 \mathrm{L}\) helium at \(23^{\circ} \mathrm{C}\) and 1.00 atm. The balloon rises to a point in the atmosphere where the pressure is \(220 .\) torr and the temperature is \(-31^{\circ} \mathrm{C}\). What is the change in volume of the balloon as it ascends from \(1.00\) atm to a pressure of \(220 .\) 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: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.

The total volume of hydrogen gas needed to fill the Hindenburg was \(2.0 \times 10^{8} \mathrm{L}\) at 1.0 atm and \(25^{\circ} \mathrm{C}\). Given that \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{H}_{2} \mathrm{O}(l)\) is \(-286 \mathrm{kJ} / \mathrm{mol},\) how much heat was evolved when the Hindenburg exploded, assuming all of the hydrogen reacted to form water?

Complete the following table for an ideal gas. $$\begin{array}{|lccc|}\hline P(\mathrm{atm}) & V(\mathrm{L}) & n(\mathrm{mol}) & T \\ \hline 5.00 & & 2.00 & 155^{\circ} \mathrm{C} \\\\\hline0.300 & 2.00 & & 155 \mathrm{K} \\ \hline 4.47 & 25.0 & 2.01 & \\\\\hline & 2.25 & 10.5 & 75^{\circ} \mathrm{C} \\\ \hline\end{array}$$

Natural gas is a mixture of hydrocarbons, primarily methane \(\left(\mathrm{CH}_{4}\right)\) and ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right) .\) A typical mixture might have \(\chi_{\text {mathane }}=\) 0.915 and \(\chi_{\text {ethane }}=0.085 .\) What are the partial pressures of the two gases in a \(15.00\) -\(\mathrm{L}\) container of natural gas at \(20 .^{\circ} \mathrm{C}\) and \(1.44\) atm? Assuming complete combustion of both gases in the natural gas sample, what is the total mass of water formed?

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