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Chapter 5: Problem 27

The gas in the discharge cell of a laser contains (in mole percent) \(11 \% \mathrm{CO}_{2}, 5.3 \% \mathrm{~N}_{2}\), and \(84 \% \mathrm{He}\) (a) What is the molar mass of this mixture? (b) Calculate the density of this gas mixture at \(32^{\circ} \mathrm{C}\) and \(758 \mathrm{~mm} \mathrm{Hg}\). (c) What is the ratio of the density of this gas to that of air \((\mathrm{MM}=29.0 \mathrm{~g} / \mathrm{mol})\) at the same conditions?

Short Answer

Expert verified
Answer: The ratio of the density of the gas mixture to that of air at the same conditions is approximately 0.361.

Step by step solution

01

Calculate individual molar masses

For the gas mixture, we require the molar masses of CO2, N2, and He. Carbon has molar mass 12.01 g/mol, Oxygen has 16.00 g/mol, and Nitrogen has 14.01 g/mol. Helium has molar mass 4.00 g/mol. Now, we can calculate: Molar mass (CO2) = 12.01 + 2 * 16.00 = 44.01 g/mol Molar mass (N2) = 2 * 14.01 = 28.02 g/mol Molar mass (He) = 4.00 g/mol
02

Calculate the average molar mass

Using the mole percentages given in the exercise, we can find the average molar mass of the gas mixture: Average molar mass = (11% * 44.01) + (5.3% * 28.02) + (84% * 4.00) = 0.11 * 44.01 + 0.053 * 28.02 + 0.84 * 4.00 ≈ 10.44 g/mol
03

Use ideal gas law to compute the density

Using the ideal gas law, PV=nRT, where n/V represents the density in mol/L, we will first convert the given pressure and temperature into appropriate units. Temperature in Kelvin: 32°C + 273.15 = 305.15 K Pressure in atm: 758 mmHg * (1 atm / 760 mmHg) ≈ 0.997 atm Now, we can compute the density at these conditions: Density (mol/L) = P / (R * T) = (0.997 atm) / (0.0821 L atm/mol K * 305.15 K) ≈ 0.0400 mol/L
04

Obtain the density in g/L

To find the density in g/L, multiply the density in mol/L with the average molar mass of the gas mixture: Density (g/L) = Density (mol/L) * Average molar mass = 0.0400 mol/L * 10.44 g/mol ≈ 0.418 g/L
05

Calculate the density ratio to air

To find the ratio of the density of the gas mixture to that of air, divide the density of the gas mixture by the density of air at the same conditions. First, compute the density of air using its molar mass and ideal gas law: Density of air (g/L) = (0.0400 mol/L) * (29.0 g/mol) ≈ 1.16 g/L Now, we can find the density ratio: Density ratio = Density of gas mixture / Density of air = (0.418 g/L) / (1.16 g/L) ≈ 0.361 The ratio of the density of the gas mixture to that of air at the same conditions is approximately 0.361.

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Most popular questions from this chapter

Dichlorine oxide is used as bactericide to purify water. It is produced by the chlorination of sulfur dioxide gas. $$ \mathrm{SO}_{2}(g)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SOCl}_{2}(l)+\mathrm{Cl}_{2} \mathrm{O}(g) $$ How many liters of \(\mathrm{Cl}_{2} \mathrm{O}\) can be produced by mixing \(5.85 \mathrm{~L}\) of \(\mathrm{SO}_{2}\) and \(9.00 \mathrm{~L}\) of \(\mathrm{Cl}_{2}\) ? How many liters of the reactant in excess are present after reaction is complete? Assume \(100 \%\) yield and that all the gases are measured at the same temperature and pressure.

Some chambers used to grow bacteria that thrive on \(\mathrm{CO}_{2}\) have a gas mixture consisting of \(95.0 \% \mathrm{CO}_{2}\) and \(5.0 \% \mathrm{O}_{2}\) (mole percent). What is the partial pressure of each gas if the total pressure is \(735 \mathrm{~mm} \mathrm{Hg}\) ?

When hydrogen peroxide decomposes, oxygen is produced: $$ 2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}(g) $$ What volume of oxygen gas at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) is produced from the decomposition of \(25.00 \mathrm{~mL}\) of a \(30.0 \%\) (by mass) solution of hydrogen peroxide \((d=1.05 \mathrm{~g} / \mathrm{mL}) ?\)

A sample of oxygen gas is collected over water at \(25^{\circ} \mathrm{C}\) (vp \(\mathrm{H}_{2} \mathrm{O}(l)=23.8 \mathrm{~mm} \mathrm{Hg}\) ). The wet gas occupies a volume of \(7.28 \mathrm{~L}\) at a total pressure of \(1.25\) bar. If all the water is removed, what volume will the dry oxygen occupy at a pressure of \(1.07 \mathrm{~atm}\) and a temperature of \(37^{\circ} \mathrm{C}\) ?

A tire is inflated to a gauge pressure of \(28.0\) psi at \(71^{\circ} \mathrm{F}\). Gauge pressure is the pressure above atmospheric pressure, which is \(14.7\) psi. After several hours of driving, the air in the tire has a temperature of \(115^{\circ} \mathrm{F}\). What is the gauge pressure of the air in the tire? What is the actual pressure of the air in the tire? Assume that the tire volume changes are negligible.
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