In research that required the careful measurement of gas densities, John Rayleigh, a physicist, found that the density of \(\mathrm{O}_{2}(\mathrm{g})\) had the same value whether the gas was obtained from air or derived from one of its compounds. The situation with \(\mathrm{N}_{2}(\mathrm{g})\) was different, however. The density of \(\mathrm{N}_{2}(\mathrm{g})\) had the same value when the \(\mathrm{N}_{2}(\mathrm{g})\) was derived from any of various compounds, but a different value if the \(\mathrm{N}_{2}(\mathrm{g})\) was extracted from air. In \(1894,\) Rayleigh enlisted the aid of William Ramsay, a chemist, to solve this apparent mystery; in the course of their work they discovered the noble gases. (a) Why do you suppose that the \(\mathrm{N}_{2}(\mathrm{g})\) extracted from liquid air did not have the same density as \(\mathrm{N}_{2}(\mathrm{g})\) obtained from its compounds? (b) Which gas do you suppose had the greater density: \(\mathrm{N}_{2}(\mathrm{g})\) extracted from air or \(\mathrm{N}_{2}(\mathrm{g})\) prepared from nitrogen compounds? Explain. (c) The way in which Ramsay proved that nitrogen gas extracted from air was itself a mixture of gases involved allowing this nitrogen to react with magnesium metal to form magnesium nitride. Explain the significance of this experiment. (d) Calculate the percent difference in the densities at \(0.00^{\circ} \mathrm{C}\) and 1.00 atm of Rayleigh's \(\mathrm{N}_{2}(\mathrm{g})\) extracted from air and \(\mathrm{N}_{2}(\mathrm{g})\) derived from nitrogen compounds. [The volume percentages of the major components of air are \(78.084 \% \mathrm{N}_{2}, 20.946 \% \mathrm{O}_{2}, 0.934 \% \mathrm{Ar},\) and \(0.0379 \% \mathrm{CO}_{2} .\)

Short Answer

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The nitrogen gas extracted from air has a higher density because it invariably contains traces of noble gases which have higher atomic masses. Nitrogen prepared from nitrogen compounds is largely free of such contaminants, leading to a lower density. Ramsay used this difference to aim to prove the presence of other gases in the nitrogen extracted from air by reacting this nitrogen with magnesium to form magnesium nitride. The percent difference in densities can be calculated using the volume percentages given for the different gas components of air.

Step by step solution

01

Analyzing N2(g) from Air and Compounds

The density of a gas is dependent on the type of gas particles involved. The density difference of nitrogen extracted from air and nitrogen derived from compounds can be attributed to an additional component in the nitrogen derived from air, which turned out to be the noble gases. These noble gases increased the overall density of the nitrogen extracted from air.
02

Determining which N2(g) is Denser

Given that nitrogen gas extracted from air also contains noble gases, it would have greater density than nitrogen gas prepared from nitrogen compounds. The reason is that the noble gases, being monoatomic and having larger atomic masses than nitrogen, contribute to an increased average mass of gas particles, resulting in an increased density.
03

Significance of Reaction with Magnesium Metal

When the nitrogen gas extracted from air reacts with magnesium to form magnesium nitride, it's evidence that the gas extracted from air is slightly different from pure nitrogen. The remaining unreacted gas is likely the additional component (noble gas) that caused the initial density variation. This experiment was critical in leading to the discovery of noble gases.
04

Calculating Percent Difference in Densities

To calculate the percent difference in densities, we first calculate the average density of Nitrogen in air using the given volume percentages. Since the density is proportional to the molar mass for a given pressure and temperature, the average molar mass (and hence the density) of air is the sum of the molar masses of its components, each multiplied by its volume percentage. The density of pure nitrogen can be obtained from the periodic table. Then, the percent difference in densities = \((Density_{air} - Density_{N_{2}}) / Density_{N_{2}} * 100\) . Note: Nitrogen from nitrogen compounds is expected to have the same density as pure nitrogen unless specified otherwise.

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Most popular questions from this chapter

A \(0.156 \mathrm{g}\) sample of a magnesium-aluminum alloy dissolves completely in an excess of \(\mathrm{HCl}(\mathrm{aq}) .\) The liberated \(\mathrm{H}_{2}(\mathrm{g})\) is collected over water at \(5^{\circ} \mathrm{C}\) when the barometric pressure is 752 Torr. After the gas is collected, the water and gas gradually warm to the prevailing room temperature of \(23^{\circ} \mathrm{C} .\) The pressure of the collected gas is again equalized against the barometric pressure of 752 Torr, and its volume is found to be \(202 \mathrm{mL}\). What is the percent composition of the magnesium-aluminum alloy? (Vapor pressure of water: \(6.54 \mathrm{mmHg}\) at \(5^{\circ} \mathrm{C}\) and \(21.07 \mathrm{mmHg}\) at \(\left.23^{\circ} \mathrm{C}\right)\)

If the van der Waals equation is solved for volume, a cubic equation is obtained. (a) Derive the equation below by rearranging equation (6.26). \(V^{3}-n\left(\frac{R T+b P}{P}\right) V^{2}+\left(\frac{n^{2} a}{P}\right) V-\frac{n^{3} a b}{P}=0\) (b) What is the volume, in liters, occupied by \(185 \mathrm{g}\) \(\mathrm{CO}_{2}(\mathrm{g})\) at a pressure of \(125 \mathrm{atm}\) and \(286 \mathrm{K} ?\) For \(\mathrm{CO}_{2}(\mathrm{g})\) \(a=3.61 \mathrm{L}^{2} \mathrm{atm} \mathrm{mol}^{-2}\) and \(b=0.0429 \mathrm{Lmol}^{-1}\) [Hint: Use the ideal gas equation to obtain an estimate of the volume. Then refine your estimate, either by trial and error, or using the method of successive approximations. See Appendix A, pages A5-A6, for a description of the method of successive approximations.

What is the pressure (in \(\mathrm{mmHg}\) ) of the gas inside the apparatus below if \(P_{\text {bar. }}=740 \mathrm{mm} \mathrm{Hg}, h_{1}=30 \mathrm{mm}\) and \(h_{2}=50 \mathrm{mm} ?\)

A 2.35 L container of \(\mathrm{H}_{2}(\mathrm{g})\) at \(762 \mathrm{mmHg}\) and \(24^{\circ} \mathrm{C}\) is connected to a 3.17 L container of \(\mathrm{He}(\mathrm{g})\) at \(728 \mathrm{mmHg}\) and \(24^{\circ} \mathrm{C}\). After mixing, what is the total gas pressure, in millimeters of mercury, with the temperature remaining at \(24^{\circ} \mathrm{C} ?\)

An 886 mL sample of \(\mathrm{Ne}(\mathrm{g})\) is at \(752 \mathrm{mmHg}\) and \(26^{\circ} \mathrm{C}\). What will be the new volume if, with the pressure and amount of gas held constant, the temperature is (a) increased to \(98^{\circ} \mathrm{C} ;\) (b) lowered to \(-20^{\circ} \mathrm{C} ?\)

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