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Chapter 10: Problem 81

(a) Place the following gases in order of increasing average molecular speed at $25^{\circ} \mathrm{C}: \mathrm{O}_{2}, \mathrm{Ar}, \mathrm{CO}, \mathrm{HCl}, \mathrm{CH}_{4} \cdot(\mathbf{b})$ Calculate the rms speed of CO molecules at \(25^{\circ} \mathrm{C} .(\mathbf{c})\) Calculate the most probable speed of an argon atom in the stratosphere, where the temperature is \(0^{\circ} \mathrm{C}\).

Short Answer

Expert verified
(a) The order of increasing average molecular speed at 25°C for the given gases is: Ar < HCl < CO < O₂ < CH₄. (b) The rms speed of CO molecules at 25°C is approximately 517 m/s. (c) The most probable speed of Argon atoms in the stratosphere at 0°C is approximately 490 m/s.

Step by step solution

01

(a) Ordering Gases based on Average Molecular Speeds

According to the Graham's law of effusion, the average molecular speed of a gas is inversely proportional to the square root of its molar mass. We can write this as: \(v_{1}/v_{2}=\sqrt{M_{2}/M_{1}}\) Given the gases: O₂, Ar, CO, HCl, CH₄ Let's find the molar masses for each gas: O₂: 32 g/mol Ar: 40 g/mol CO: 28 g/mol HCl: 36.5 g/mol CH₄: 16 g/mol Now order the gases based on their molar masses: CH₄ < O₂ < CO < HCl < Ar Since the average molecular speed is inversely proportional to the square root of the molar mass, the order of increasing average molecular speed is the reverse of the order of molar masses. Thus, the order of increasing average molecular speed is: Ar < HCl < CO < O₂ < CH₄
02

(b) Finding the RMS Speed of CO Molecules

The formula for rms speed of molecules in a gas is: \(v_{rms} = \sqrt{3RT/M}\) where: v_rms = rms speed R = gas constant = 8.314 J/(mol K) T = temperature in Kelvin (25°C = 298.15 K) M = molar mass of the gas (CO) = 28 g/mol = 0.028 kg/mol Now we can plug the values into the equation and solve for v_rms. \(v_{rms}(CO) = \sqrt{3(8.314\;\text{J}/\text{mol K})(298.15\;\text{K}) / (0.028\;\text{kg}/\text{mol})}\) \(v_{rms}(CO) = \sqrt{3(8.314)(298.15) / (0.028)}\) \(v_{rms}(CO) ≈ 517\;\text{m/s}\) The rms speed of CO molecules at 25°C is approximately 517 m/s.
03

(c) Calculating the Most Probable Speed of Argon Atoms

The formula for the most probable speed of molecules in a gas is: \(v_{p} = \sqrt{2RT/M}\) where: v_p = most probable speed R = gas constant = 8.314 J/(mol K) T = temperature in Kelvin (0°C = 273.15 K) M = molar mass of the gas (Ar) = 40 g/mol = 0.040 kg/mol Now we can plug the values into the equation and solve for v_p. \(v_{p}(Ar) = \sqrt{2(8.314\;\text{J}/\text{mol K})(273.15\;\text{K}) / (0.040\;\text{kg}/\text{mol})}\) \(v_{p}(Ar) = \sqrt{2(8.314)(273.15) / (0.040)}\) \(v_{p}(Ar) ≈ 490\;\text{m/s}\) The most probable speed of Argon atoms in the stratosphere at 0°C is approximately 490 m/s.

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Most popular questions from this chapter

In a "Kipp generator", hydrogen gas is produced when zinc flakes react with hydrochloric acid: $$2 \mathrm{HCl}(a q)+\mathrm{Zn}(s) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)$$ If \(30.0 \mathrm{~mL}\) of wet \(\mathrm{H}_{2}\) is collected over water at \(20^{\circ} \mathrm{C}\) and a barometric pressure of \(101.33 \mathrm{kPa}\), how many grams of \(\mathrm{Zn}\) have been consumed? (The vapor pressure of water is tabulated in Appendix B.)

Cyclopropane, a gas used with oxygen as a general anesthetic, is composed of \(85.7 \% \mathrm{C}\) and \(14.3 \% \mathrm{H}\) by mass. \((\mathbf{a})\) If $1.56 \mathrm{~g}\( of cyclopropane has a volume of \)1.00 \mathrm{~L}$ at 99.7 \(\mathrm{kPa}\) and \(50.0^{\circ} \mathrm{C}\), what is the molecular formula of cyclopropane? (b) Judging from its molecular formula, would you expect cyclopropane to deviate more or less than Ar from ideal-gas behavior at moderately high pressures and room temperature? Explain. (c) Would cyclopropane effuse through a pinhole faster or more slowly than methane, \(\mathrm{CH}_{4} ?\)

Many gases are shipped in high-pressure containers. Consider a steel tank whose volume is \(210.0 \mathrm{~L}\) 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 $15.2 \mathrm{MPa} ?\( (d) What would be the pressure of the gas, in \)\mathrm{kPa}$, if it were transferred to a container at \(24^{\circ} \mathrm{C}\) whose volume is \(55.0 \mathrm{~L} ?\)

Carbon dioxide, which is recognized as the major contributor to global warming as a "greenhouse gas," is formed when fossil fuels are combusted, as in electrical power plants fueled by coal, oil, or natural gas. One potential way to reduce the amount of \(\mathrm{CO}_{2}\) added to the atmosphere is to store it as a compressed gas in underground formations. Consider a 1000-megawatt coal-fired power plant that produces about \(6 \times 10^{6}\) tons of \(\mathrm{CO}_{2}\) per year. (a) Assuming ideal-gas behavior, $101.3 \mathrm{kPa}\(, and \)27^{\circ} \mathrm{C}$, calculate the volume of \(\mathrm{CO}_{2}\) produced by this power plant. (b) If the \(\mathrm{CO}_{2}\) is stored underground as a liquid at \(10^{\circ} \mathrm{C}\) and $12.16 \mathrm{MPa}\( and a density of \)1.2 \mathrm{~g} / \mathrm{cm}^{3},$ what volume does it possess? (c) If it is stored underground as a gas at \(30^{\circ} \mathrm{C}\) and \(7.09 \mathrm{MPa}\), what volume does it occupy?

Complete the following table for an ideal gas: $$\begin{array}{llll} P & V & n & T \\ \hline 303.98 \mathrm{kPa} & 3.00 \mathrm{~L} & 1.500 \mathrm{~mol} & ? \mathrm{~K} \\ 50.663 \mathrm{kPa} & 0.750 \mathrm{~L} & ? \mathrm{~mol} & 300 \mathrm{~K} \\\ 101.33 \mathrm{kPa} & ? \mathrm{~L} & 3.333 \mathrm{~mol} & 300 \mathrm{~K} \\\ ? \mathrm{kPa} & .750 \mathrm{~L} & 0.750 \mathrm{~mol} & 298 \mathrm{~K} \\ \hline \end{array}$$
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