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Chapter 10: Problem 47

Rank the following gases from least dense to most dense at $101.33 \mathrm{kPa}\( and \)298 \mathrm{~K}: \mathrm{O}_{2}, \mathrm{Ar}, \mathrm{NH}_{3}, \mathrm{HCl}$.

Short Answer

Expert verified
The ranking of gases from least dense to most dense at 101.33 kPa and 298 K is: NH₃ (0.706 kg/m³) < O₂ (1.331 kg/m³) < HCl (1.517 kg/m³) < Ar (1.661 kg/m³).

Step by step solution

01

Find the molar mass of each gas

To rank the gases based on their densities, we must first determine their molar masses. Use the periodic table to find the atomic masses of each element in a molecule. - For O₂: Molar mass = 2 × 16.00 g/mol = 32.00 g/mol - For Ar: Molar mass = 39.95 g/mol (since it's a noble gas) - For NH₃: Molar mass = 14.01 g/mol (N) + 3 × 1.01 g/mol (H) = 17.03 g/mol - For HCl: Molar mass = 1.01 g/mol (H) + 35.45 g/mol (Cl) = 36.46 g/mol
02

Use the ideal gas law to find density

Now we will use the ideal gas law equation to find the density for each gas under the given conditions (pressure = 101.33 kPa and temperature = 298 K). The ideal gas law is given by: \(PV = nRT\) Where: - P = pressure - V = volume - n = number of moles - R = ideal gas constant (8.314 J/mol·K) - T = temperature The density of a gas is given by: \(\rho = \frac{mass}{volume} = \frac{molar~mass~ ×~ number~of~ moles}{volume}\) Combining the ideal gas law and the density formula, we get: \(\rho = \frac{molar~mass~ ×~ P}{R ~×~ T}\) Now plug in the values for pressure (101.33 kPa), temperature (298 K), and molar mass for each gas to find their respective densities.
03

Calculate density for each gas

Use the formula from the previous step to calculate the densities: - For O₂: \(\rho = \frac{32.00~ g/mol × 101.33~ kPa}{8.314~ J/mol·K × 298~ K} = 1.331~ kg/m³\) - For Ar: \(\rho = \frac{39.95~ g/mol × 101.33~ kPa}{8.314~ J/mol·K × 298~ K} = 1.661~ kg/m³\) - For NH₃: \(\rho = \frac{17.03~ g/mol × 101.33~ kPa}{8.314~ J/mol·K × 298~ K} = 0.706~ kg/m³\) - For HCl: \(\rho = \frac{36.46~ g/mol × 101.33~ kPa}{8.314~ J/mol·K × 298~ K} = 1.517~ kg/m³\)
04

Rank the gases by density

Now that we have calculated the densities for each gas, we can rank them from least dense to most dense: 1. NH₃ (0.706 kg/m³) 2. O₂ (1.331 kg/m³) 3. HCl (1.517 kg/m³) 4. Ar (1.661 kg/m³) Thus, the ranking of gases from least dense to most dense at 101.33 kPa and 298 K is: NH₃ < O₂ < HCl < Ar.

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Most popular questions from this chapter

In the contact process, sulfur dioxide and oxygen gas react to form sulfur trioxide as follows: $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)$$ At a certain temperature and pressure, \(50 \mathrm{~L}\) of \(\mathrm{SO}_{2}\) reacts with \(25 \mathrm{~L}\) of \(\mathrm{O}_{2}\). If all the \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) are consumed, what volume of \(\mathrm{SO}_{3}\), at the same temperature and pressure, will be produced?

Imagine that the reaction $2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)$ occurs in a container that has a piston that moves to maintain a constant pressure when the reaction occurs at constant temperature. Which of the following statements describes how the volume of the container changes due to the reaction: (a) the volume increases by \(50 \%,\) (b) the volume increases by \(33 \%\), (c) the volume remains constant, (d) the volume decreases by \(33 \%,(\mathbf{e})\) the volume decreases by $50 \%\(. [Sections 10.3 and 10.4\)]$

In the Dumas-bulb technique for determining the molar mass of an unknown liquid, you vaporize the sample of a liquid that boils below $100^{\circ} \mathrm{C}$ in a boiling-water bath and determine the mass of vapor required to fill the bulb. From the following data, calculate the molar mass of the unknown liquid: mass of unknown vapor, \(1.012 \mathrm{~g}\); volume of bulb, \(354 \mathrm{~cm}^{3}\); pressure, \(98.93 \mathrm{kPa} ;\) temperature, $99{ }^{\circ} \mathrm{C}$.

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

Table 10.3 shows that the van der Waals \(b\) parameter has units of $\mathrm{L} / \mathrm{mol}$. This implies that we can calculate the size of atoms or molecules from \(b\). Using the value of \(b\) for \(\mathrm{Xe},\) calculate the radius of a Xe atom and compare it to the value found in Figure \(7.7,\) that is, \(140 \mathrm{pm}\). Recall that the volume of a sphere is $(4 / 3) \pi r^{3}$.
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