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Chapter 6: Problem 43

In order for a gas-filled balloon to rise in air, the density of the gas in the balloon must be less than that of air. (a) Consider air to have a molar mass of \(28.96 \mathrm{g} / \mathrm{mol}\) determine the density of air at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{atm},\) in g/L. (b) Show by calculation that a balloon filled with carbon dioxide at \(25^{\circ} \mathrm{C}\) and 1 atm could not be expected to rise in air at \(25^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The density of air at \(25^{\circ} C\) and \(1 atm\) is approximately \(1.18 g/L\) while the density of carbon dioxide under the same conditions is approximately \(1.80 g/L\). Thus, the balloon filled with carbon dioxide is denser than air and will not rise.

Step by step solution

01

Calculating Air Density

First we need to calculate the density of air at \(25^{\circ} C\) and \(1 atm\). The equation for the density of an ideal gas is \(D = \dfrac{PM}{RT}\), where \(P\) is the pressure, \(M\) is the molecular mass (or molar mass), \(R\) is the ideal gas constant, and \(T\) is the temperature.\n\n Substituting the provided values \(P = 1 atm, M = 28.96 g/mol\), \(R = 0.0821 L.atm/(mol.K)\) (value for ideal gas constant in these units), \(T = 25^{\circ} C\) (changed to kelvin by adding 273.15 which results in \(T = 298.15 K\)), we have: \n\n \( D_{air} = \dfrac{1 atm * 28.96 g/mol}{0.0821 L.atm/(mol.K) * 298.15 K}\)
02

Density Comparison

Next, we need to calculate the density of carbon dioxide under the same conditions and compare it with the calculated air density. The carbon dioxide has a molar mass of \(44 g/mol\), so we substitute this mass into the density formula and keep all the other values (\(P, R, T\)) the same as before: \n\n \(D_{CO2} = \dfrac{1 atm * 44 g/mol}{0.0821 L.atm/(mol.K) * 298.15 K}\)
03

Interpretation of Results

Once we have both densities, we can compare if the density of carbon dioxide is greater or lesser than air. If it's greater, then it implies that a balloon filled with carbon dioxide cannot be expected to rise in air at \(25^{\circ} C\) as it is heavier than air.

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Most popular questions from this chapter

A 0.7178 g sample of a hydrocarbon occupies a volume of \(390.7 \mathrm{mL}\) at \(65.0^{\circ} \mathrm{C}\) and \(99.2 \mathrm{kPa}\). When the sample is burned in excess oxygen, \(2.4267 \mathrm{g} \mathrm{CO}_{2}\) and \(0.4967 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) are obtained. What is the molecular formula of the hydrocarbon? Write a plausible structural formula for the molecule.

Convert each pressure to an equivalent pressure (a) 736 mmHg; (b) 0.776 bar; in atmospheres. (c) 892 Torr; (d) 225 kPa.

How many liters of \(\mathrm{H}_{2}(\mathrm{g})\) at STP are produced per gram of \(\mathrm{Al}(\mathrm{s})\) consumed in the following reaction? \(2 \mathrm{Al}(\mathrm{s})+6 \mathrm{HCl}(\mathrm{aq}) \longrightarrow 2 \mathrm{AlCl}_{3}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{g})\)

To establish a pressure of 2.00 atm in a 2.24 L cylinder containing \(1.60 \mathrm{g} \mathrm{O}_{2}(\mathrm{g})\) at \(0^{\circ} \mathrm{C},\) (a) add \(1.60 \mathrm{g} \mathrm{O}_{2} ;(\mathrm{b})\) add \(0.60 \mathrm{g} \mathrm{He}(\mathrm{g}) ;(\mathrm{c})\) add \(2.00 \mathrm{g} \mathrm{He}(\mathrm{g})\) (d) release \(0.80 \mathrm{g} \mathrm{O}_{2}(\mathrm{g})\)

A mixture of \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g})\) is prepared by electrolyzing \(1.32 \mathrm{g}\) water, and the mixture of gases is collected over water at \(30^{\circ} \mathrm{C}\) and \(748 \mathrm{mmHg} .\) The volume of "wet" gas obtained is 2.90 L. What must be the vapor pressure of water at \(30^{\circ} \mathrm{C} ?\) $$2 \mathrm{H}_{2} \mathrm{O}(1) \stackrel{\text { electrolysis }}{\longrightarrow} 2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$
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