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Which gas is most dense at \(1.00 \mathrm{~atm}\) and \(298 \mathrm{~K} ? \mathrm{CO}_{2}\), \(\mathrm{N}_{2} \mathrm{O}\), or \(\mathrm{Cl}_{2}\). Explain.

Short Answer

Expert verified
The most dense gas at 1.00 atm and 298 K among CO鈧, N鈧侽, and Cl鈧 is Cl鈧 with a density of 2.90 g/L.

Step by step solution

01

Write down the Ideal Gas Law

Recall the Ideal Gas Law equation: \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
02

Rearrange Ideal Gas Law for molar volume

We want to find density, which is mass per unit volume. So first, let's find the volume for each gas per mole. Rearrange the Ideal Gas Law equation for molar volume (V/n) and solve for it: \[\frac{V}{n} = \frac{RT}{P}\] Now, we can plug in the given temperature, T = 298 K, and pressure, P = 1.00 atm, and the value of R = 0.0821 L atm K鈦宦 mol鈦宦: \[\frac{V}{n} = \frac{(0.0821 \,\mathrm{L\,atm\,K^{-1}\,mol^{-1}})(298 \,\mathrm{K})}{1.00\, \mathrm{atm}}\]
03

Calculate molar volume

Simplify the expression to find the molar volume: \[\frac{V}{n} = 24.45\, \mathrm{L/mol}\]
04

Calculate densities of the gases

Now that we have the molar volume, we can calculate the density for each gas by dividing its molar mass by the molar volume. The molar masses of CO鈧, N鈧侽, and Cl鈧 are 44.01 g/mol, 44.02 g/mol, and 70.9 g/mol, respectively. For CO鈧: \[\rho_{CO_{2}} = \frac{44.01 \, \mathrm{g/mol}}{24.45 \, \mathrm{L/mol}} = 1.80 \, \mathrm{g/L}\] For N鈧侽: \[\rho_{N_{2}O} = \frac{44.02 \, \mathrm{g/mol}}{24.45 \, \mathrm{L/mol}} = 1.80 \, \mathrm{g/L}\] For Cl鈧: \[\rho_{Cl_{2}} = \frac{70.9 \, \mathrm{g/mol}}{24.45 \, \mathrm{L/mol}} = 2.90 \, \mathrm{g/L}\]
05

Compare the densities

Compare the calculated densities of CO鈧, N鈧侽, and Cl鈧: \(\rho_{CO_{2}} = 1.80 \, \mathrm{g/L}\) \(\rho_{N_{2}O} = 1.80 \, \mathrm{g/L}\) \(\rho_{Cl_{2}} = 2.90 \, \mathrm{g/L}\)
06

Determine the most dense gas

Since Cl鈧 has the highest density (2.90 g/L) at the given conditions (1.00 atm and 298 K), it is the most dense gas among CO鈧, N鈧侽, and Cl鈧.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ideal Gas Law
The Ideal Gas Law is a fundamental principle in chemistry that allows us to predict and understand the behavior of gases under various conditions. At the heart of this concept is the equation: \[PV = nRT\],where P represents the pressure, V stands for volume, n is the number of moles of the gas, R is the universal gas constant, and T denotes the temperature in Kelvin. The beauty of the Ideal Gas Law lies in its ability to model the state of an 'ideal' gas鈥攁 theoretical gas that perfectly follows the law's assumptions of internal energy depending only on temperature and no interactions between the gas particles aside from perfectly elastic collisions.
When dealing with real gases, this law can still provide reasonable approximations under many conditions, such as standard temperature and pressure. By manipulating this equation, scientists and students alike can explore various properties, such as the molar volume鈥攖hat is, the volume one mole of the gas occupies. In the given exercise, using this law provided the foundation for solving the density question, asserting its indispensable role in the study of gaseous substances.
Gas Density
Gas density, which represents the mass of a gas per unit volume, plays an instrumental role in determining how substances behave under different environmental conditions. It's crucial for many applications, including industrial processes and environmental science. Unlike liquids and solids, the density of gases is highly dependent on pressure and temperature. To find the density of a gas, the equation \[\rho = \frac{m}{V}\] is often used, where \(\rho\) is the density, m is the mass, and V is the volume.
In the context of the exercise, the molar volume obtained from the Ideal Gas Law was used to calculate the density of different gases, allowing for a direct comparison. Since all gases in the problem are at the same temperature and pressure, the gas with the greatest molar mass displayed the highest density. This fact illustrates the direct relationship between gas density and molar mass, making density a derived property that hinges not only on the environmental conditions but also on the intrinsic characteristics of the gas itself.
Molar Mass
Molar mass is the weight of 1 mole (6.022 x 10虏鲁 particles) of any chemical substances, expressed in grams per mole (g/mol). It's a cardinal quantity in chemistry because it bridges the microscopic world of atoms and molecules with the macroscopic world we interact with every day. Knowing the molar mass aids in converting between moles, which count particles, and grams, which measure mass. This is a pivotal step in chemical calculations for creating solutions, stoichiometry, and predicting yields.
In our exercise, the molar mass directly influenced the density calculations for each gas. Since the gases shared the same molar volume under the given conditions, it was simply a matter of dividing their molar mass by this common volume to find their respective densities. Here we can see that the molar mass forms the crux of determining a substance's density when the volume is standardized. This demonstration of molar mass influencing gas density is a vivid portrayal of the intrinsic link between these two critical concepts in chemistry.

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Most popular questions from this chapter

A gaseous mixture of \(\mathrm{O}_{2}\) and \(\mathrm{Kr}\) has a density of \(1.104 \mathrm{~g} / \mathrm{L}\) at 435 torr and \(300 \mathrm{~K}\). What is the mole percent \(\mathrm{O}_{2}\) in the mixture?

Consider the following reaction: $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g) $$ Imagine that this reaction occurs in a container that has a piston that moves to allow a constant pressure to be maintained when the reaction occurs at constant temperature. (a) What happens to the volume of the container as a result of the reaction? Explain. (b) If the piston is not allowed to move, what happens to the pressure as a result of the reaction? [Sections \(10.3\) and 10.5]

The molar mass of a volatile substance was determined by the Dumas-bulb method described in Exercise \(10.51\). The unknown vapor had a mass of \(0.846 \mathrm{~g} ;\) the volume of the bulb was \(354 \mathrm{~cm}^{3}\), pressure 752 torr, and temperature \(100^{\circ} \mathrm{C}\). Calculate the molar mass of the unknown vapor.

Ammonia, \(\mathrm{NH}_{3}(g)\), and hydrogen chloride, \(\mathrm{HCl}(g)\), react to form solid ammonium chloride, \(\mathrm{NH}_{4} \mathrm{Cl}(s)\) : $$ \mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(s) $$ Two 2.00-L flasks at \(25^{\circ} \mathrm{C}\) are connected by a stopcock, as shown in the drawing. One flask contains \(5.00 \mathrm{~g}\) \(\mathrm{NH}_{3}(g)\), and the other contains \(5.00 \mathrm{~g} \mathrm{HCl}(g)\). When the stopcock is opened, the gases react until one is completely consumed. (a) Which gas will remain in the system after the reaction is complete? (b) What will be the final pressure of the system after the reaction is complete? (Neglect the volume of the ammonium chloride formed.)

In the United States, barometric pressures are generally reported in inches of mercury (in. Hg). On a beautiful summer day in Chicago the barometric pressure is \(30.45\) in. \(\mathrm{Hg}\). (a) Convert this pressure to torr. (b) A meteorologist explains the nice weather by referring to a "high-pressure area." In light of your answer to part (a), explain why this term makes sense.

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