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Chapter 5: Problem 25

Which noble gas has the smallest density at STP? Explain.

Short Answer

Expert verified
Helium (He) has the smallest density at STP with a value of 0.1786 g/L. This is due to its low molar mass of 4 g/mol compared to other noble gases, resulting in fewer mass units per unit volume.

Step by step solution

01

List the Noble Gases

The noble gases are a group of chemical elements in the 18th column of the periodic table. They are as follows: Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe), and Radon (Rn).
02

Find Molar Masses of the Noble Gases

Look up the molar masses of each noble gas from the periodic table: - Helium (He): 4 g/mol - Neon (Ne): 20 g/mol - Argon (Ar): 40 g/mol - Krypton (Kr): 84 g/mol - Xenon (Xe): 131 g/mol - Radon (Rn): 222 g/mol
03

Understand the Ideal Gas Law and Density Equation

We can use the Ideal Gas Law equation to find the density of each noble gas under STP conditions. The Ideal Gas Law equation is as follows: \(PV = nRT\) Where P is pressure, V is volume, n is the amount of gas in moles, R is the ideal gas constant, and T is the temperature. Under STP conditions, the pressure is 1 atm, and the temperature is 273.15 K (0°C). Since we're looking for the density (\(\rho\)), we must rearrange the equation: \(\rho = \frac{m}{V} = \frac{nM}{V}\) Where m is the mass of the gas, M is the molar mass, and V is its volume. Replacing n by combining the Ideal Gas Law equation and applying STP conditions, we have: \(\rho = \frac{MP}{RT}\)
04

Calculate the Density of Each Noble Gas at STP

Using the density equation above and the molar masses from Step 2, calculate the density for each noble gas at STP: - Helium (He): \(\rho = \frac{(4 g/mol)(1 atm)}{(0.0821 L/mol \cdot K)(273.15 K)} = 0.1786\, g/L\) - Neon (Ne): \(\rho = \frac{(20 g/mol)(1 atm)}{(0.0821 L/mol \cdot K)(273.15 K)} = 0.8995\, g/L\) - Argon (Ar): \(\rho = \frac{(40 g/mol)(1 atm)}{(0.0821 L/mol \cdot K)(273.15 K)} = 1.7990\, g/L\) - Krypton (Kr): \(\rho = \frac{(84 g/mol)(1 atm)}{(0.0821 L/mol \cdot K)(273.15 K)} = 3.6753\, g/L\) - Xenon (Xe): \(\rho = \frac{(131 g/mol)(1 atm)}{(0.0821 L/mol \cdot K)(273.15 K)} = 5.7695\, g/L\) - Radon (Rn): \(\rho = \frac{(222 g/mol)(1 atm)}{(0.0821 L/mol \cdot K)(273.15 K)} = 9.9196\, g/L\)
05

Comparing and Explaining the Smallest Density

As per the density values calculated above, Helium (He) has the smallest density at STP with a value of 0.1786 g/L. Helium has the smallest density because it has the lowest molar mass among all noble gases, which results in fewer mass units per unit volume. This is due to the atomic configuration of helium, which contains only two electrons and two protons. The low molar mass and the properties of the ideal gas law equation ensure that helium has the lowest density among noble gases at STP.

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Most popular questions from this chapter

An ideal gas is contained in a cylinder with a volume of \(5.0 \times\) \(10^{2} \mathrm{~mL}\) at a temperature of \(30 .^{\circ} \mathrm{C}\) and a pressure of 710 . torr. The gas is then compressed to a volume of \(25 \mathrm{~mL}\), and the temperature is raised to \(820 .{ }^{\circ} \mathrm{C}\). What is the new pressure of the gas?

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Consider separate \(1.0-\mathrm{L}\) samples of \(\mathrm{He}(g)\) and \(\mathrm{UF}_{6}(g)\), both at \(1.00\) atm and containing the same number of moles. What ratio of temperatures for the two samples would produce the same root mean square velocity?

Consider separate \(1.0-\mathrm{L}\) gaseous samples of \(\mathrm{H}_{2}, \mathrm{Xe}, \mathrm{Cl}_{2}\), and \(\mathrm{O}_{2}\) all at STP. a. Rank the gases in order of increasing average kinetic energy. b. Rank the gases in order of increasing average velocity. c. How can separate \(1.0-\mathrm{L}\) samples of \(\mathrm{O}_{2}\) and \(\mathrm{H}_{2}\) each have the same average velocity?

Trace organic compounds in the atmosphere are first concentrated and then measured by gas chromatography. In the concentration step, several liters of air are pumped through a tube containing a porous substance that traps organic compounds. The tube is then connected to a gas chromatograph and heated to release the trapped compounds. The organic compounds are separated in the column and the amounts are measured. In an analysis for benzene and toluene in air, a \(3.00-\mathrm{L}\) sample of air at 748 torr and \(23^{\circ} \mathrm{C}\) was passed through the trap. The gas chromatography analysis showed that this air sample contained \(89.6\) ng benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) and \(153 \mathrm{ng}\) toluene \(\left(\mathrm{C}_{7} \mathrm{H}_{8}\right) .\) Calculate the mixing ratio (see Exercise 121 ) and number of molecules per cubic centimeter for both benzene and toluene.
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