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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

A certain laser uses a gas mixture consisting of \(9.00 \mathrm{~g} \mathrm{HCl}, 2.00 \mathrm{~g} \mathrm{H}_{2}\) and \(165.0 \mathrm{~g}\) of Ne. What pressure is exerted by the mixture in a \(75.0\) -L tank at \(22^{\circ} \mathrm{C}\) ? Which gas has the smallest partial pressure?

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