Problem 47 - Chemistry: The Central Science

Question

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

Step-by-Step Solution

Answer

The most dense gas at 1.00 atm pressure and 298 K temperature among CO₂, N₂O, and Cl₂ is Cl₂, with a density of approximately 3.19 \(\frac{\mathrm{g}}{\mathrm{L}}\). This is determined using the Ideal Gas Law and molar mass to calculate the densities of the gases and comparing the results.

Step 1: Recall the Ideal Gas Law and Density formula

First, let's recall the Ideal Gas Law formula, which states: \[PV = nRT\] Where: - P represents the pressure (atm) - V represents the volume (L) - n represents the number of moles of gas - R is the ideal gas constant (\(0.08206\frac{\mathrm{L}\cdot\mathrm{atm}}{\mathrm{mol}\cdot\mathrm{K}}\)) - T represents the temperature (K) We will also need the formula to calculate the density of a gas (ρ), given by: \[\rho = \frac{m}{V} = \frac{n \cdot M}{V}\] Where: - ρ represents the density of the gas (\(\frac{\mathrm{g}}{\mathrm{L}}\)) - m represents the mass of the gas (g) - n represents the number of moles of gas - V represents the volume (L) - M represents the molar mass of the gas (\(\frac{\mathrm{g}}{\mathrm{mol}}\))

Step 2: Obtain molar masses

We will need the molar masses of each gas to calculate their densities. The molar masses are as follows: - \(CO_{2}\): \(12.01 + 2 \cdot 16.00 = 44.01 \frac{\mathrm{g}}{\mathrm{mol}}\) - \(N_{2}O\): \(2 \cdot 14.01 + 16.00 = 44.02 \frac{\mathrm{g}}{\mathrm{mol}}\) - \(Cl_{2}\): \(2 \cdot 35.45 = 70.90 \frac{\mathrm{g}}{\mathrm{mol}}\)

Step 3: Derive the density formula from Ideal Gas Law

By isolating the volume (V) in the Ideal Gas Law formula, we find: \[V = \frac{nRT}{P}\] We can now substitute the expression for V into the density formula and simplify: \[\rho = \frac{n \cdot M}{\frac{nRT}{P}}\] \[\rho = \frac{M}{\frac{RT}{P}}\] Physical conditions (pressure and temperature) are constant for all the gases, therefore, we can compare the values of Moles Molar of each gas according to its molar mass. The most significant value represents the most dense gas

Step 4: Calculate and compare densities

We substitute the molar masses and given values for temperature and pressure into the derived formula: For \(CO_{2}\): \[\rho_{CO_{2}} = \frac{44.01}{\frac{0.08206 \cdot 298}{1.00}} \approx 1.98 \frac{\mathrm{g}}{\mathrm{L}}\] For \(N_{2}O\): \[\rho_{N_{2}O} = \frac{44.02}{\frac{0.08206 \cdot 298}{1.00}} \approx 1.98 \frac{\mathrm{g}}{\mathrm{L}}\] For \(Cl_{2}\): \[\rho_{Cl_{2}} = \frac{70.90}{\frac{0.08206 \cdot 298}{1.00}} \approx 3.19 \frac{\mathrm{g}}{\mathrm{L}}\] Comparing the calculated densities, we can conclude that Cl₂ is the most dense gas among the three at 1.00 atm pressure and 298 K temperature, with a density of approximately 3.19 \(\frac{\mathrm{g}}{\mathrm{L}}\).

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry that characterizes the behavior of many gases under various conditions. It relates four critical physical properties of a gas: pressure (P), volume (V), number of moles (n), and temperature (T), through the formula
\[PV = nRT\]
Here, R represents the ideal gas constant, which integrates these variables so they can be compared and manipulated across different gases and conditions. The beauty of the Ideal Gas Law is in its ability to predict the behavior of a gas when subjected to changes in pressure, volume, temperature, or amount. When dealing with gas density problems, the Ideal Gas Law becomes essential as it allows us to relate these factors to the molar mass of the gas, which is a key element in determining the density.

Molar Mass

Molar mass is the weight of one mole of a substance, typically expressed in grams per mole (g/mol). It is a compound-specific value that amounts to the sum of the atomic masses of all the atoms in the molecule. For instance, the molar mass of carbon dioxide (CO₂) is derived by adding the atomic mass of one carbon atom (12.01 g/mol) to those of two oxygen atoms (16.00 g/mol each), giving us
\[ m_{CO_{2}} = 12.01 + 2 \times 16.00 = 44.01 \frac{\mathrm{g}}{\mathrm{mol}}\]
Knowing the molar mass allows us to relate the mass of a gas to the amount of substance (number of moles), which is a critical step in calculating a gas's density.

Gas Density Calculation

To calculate the density of a gas, we need to know both its mass and the volume it occupies. The density formula is given by
\[\rho = \frac{m}{V}\]
where ρ is the density, m is the mass, and V is the volume. To connect this with the Ideal Gas Law, we can manipulate the law to solve for volume, as seen in the textbook solution, and then substitute back into the density formula. This merged formula for density in terms of the Ideal Gas Law variables looks like
\[\rho = \frac{n \cdot M}{V} = \frac{M}{\frac{RT}{P}}\]
This equation shows that, under constant pressure and temperature, the density of a gas is directly proportional to its molar mass. Hence, a higher molar mass would mean a denser gas, assuming that the temperature and pressure remain unchanged. This is crucial in comparing the densities of different gases, as seen in the exercise where Cl₂, with its higher molar mass, turned out to be the densest of the three.