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Calculate the root mean square velocities of \(\mathrm{CH}_{4}(\mathrm{~g})\) and \(\mathrm{N}_{2}(\mathrm{~g})\) molecules at \(273 \mathrm{~K}\) and \(546 \mathrm{~K}\).

Short Answer

Expert verified
The root mean square velocities of CH4(g) and N2(g) molecules at the given temperatures are: CH4(g) at 273 K: \(660.26 m/s\), CH4(g) at 546 K: \(934.98 m/s\), N2(g) at 273 K: \(490.93 m/s\), and N2(g) at 546 K: \(694.40 m/s\).

Step by step solution

01

Root Mean Square Velocity Formula

The formula for calculating the root mean square velocity (v_rms) of a gas molecule is given by: \[v_{rms} = \sqrt{\frac{3RT}{M}}\] where R is the ideal gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, and M is the molar mass of the gas in kg/mol.
02

Molar Mass of CH4 and N2

We need to find the molar mass of CH4 and N2 in kg/mol to use in the formula. Molar mass of CH4: 12.01 g/mol (C) + 4 × 1.01 g/mol (H) = 16.04 g/mol = 0.01604 kg/mol Molar mass of N2: 2 × 14.01 g/mol (N) = 28.02 g/mol = 0.02802 kg/mol
03

Calculate v_rms for CH4 at 273 K

Plug in the values of R, T, and M for CH4 in the formula: \[v_{rms} = \sqrt{\frac{3 × 8.314 J/(mol·K) × 273 K}{0.01604 kg/mol}} = 660.26 m/s\]
04

Calculate v_rms for CH4 at 546 K

Plug in the values of R, T, and M for CH4 in the formula: \[v_{rms} = \sqrt{\frac{3 × 8.314 J/(mol·K) × 546 K}{0.01604 kg/mol}} = 934.98 m/s\]
05

Calculate v_rms for N2 at 273 K

Plug in the values of R, T, and M for N2 in the formula: \[v_{rms} = \sqrt{\frac{3 × 8.314 J/(mol·K) × 273 K}{0.02802 kg/mol}} = 490.93 m/s\]
06

Calculate v_rms for N2 at 546 K

Plug in the values of R, T, and M for N2 in the formula: \[v_{rms} = \sqrt{\frac{3 × 8.314 J/(mol·K) × 546 K}{0.02802 kg/mol}} = 694.40 m/s\] So, the root mean square velocities are as follows: CH4(g) at 273 K: 660.26 m/s CH4(g) at 546 K: 934.98 m/s N2(g) at 273 K: 490.93 m/s N2(g) at 546 K: 694.40 m/s

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Key Concepts

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

Ideal Gas Constant
The ideal gas constant, often represented by the symbol R, is a fundamental constant that appears in the ideal gas law: \( PV = nRT \), where P is pressure, V is volume, n is the amount of substance in moles, and T is temperature in Kelvin. The value of R in SI units is \(8.314 J/(mol\cdot K)\). This constant provides a proportionality factor between the energy scale and the temperature scale on a per mole basis.

The importance of the ideal gas constant becomes evident when we discuss gas behavior and properties. In the context of calculating root mean square velocities of gas molecules, R is instrumental in bridging the connection between the macroscopic conditions of temperature and the microscopic kinetic energies of gas particles. Thus, understanding R is crucial to comprehend wider gas dynamics beyond just this equation.
Molar Mass
Molar mass is another key term in calculations involving gases. It is defined as the mass of one mole of a substance and is expressed in grams per mole (g/mol) or kilograms per mole (kg/mol) depending on the context. For molecules like methane (CH4) and nitrogen (N2), molar mass takes into account the mass of all the atoms present in a single molecule of the gas.

Molar mass plays a crucial role in converting the macroscopic measurements (like those obtained from a balance) to microscopic moles of particles, and vice versa. It also influences the behavior of a gas, as seen in the calculation of root mean square velocity, where the molar mass indirectly affects how fast gas particles move on average at a given temperature.
Formula Derivation
The derivation of the root mean square velocity formula invokes principles from kinetic molecular theory, which describes the behavior of gas molecules. This theory states that the kinetic energy of gas particles is proportional to the temperature of the gas. By equating the kinetic energy expression \(\frac{1}{2}mv^{2}\) with the thermodynamic definition that includes R and T, we derive the formula for root mean square velocity:\[v_{rms} = \sqrt{\frac{3RT}{M}}\]

In this formula, the square root is taken because we deal with squares of velocities when calculating average values of kinetic energy. The factor of 3 comes from the three degrees of freedom of motion in space (x, y, z directions) for a monatomic ideal gas, as each contributes equally to the kinetic energy.
Gas Molecule Velocities
Gas molecule velocities are not uniform within a sample of gas; rather, they follow a distribution. The root mean square velocity calculated using the formula derived from kinetic theory provides an average measure of this velocity distribution.

The velocities of gas molecules impact properties like diffusion rate, effusion, and the rate of gas mixing. The temperature dependence of these velocities, as embodied in the formula, shows that higher temperatures result in greater average gas molecule speeds. This relationship is fundamental to understanding and predicting the behavior of gases under various conditions, including reactions involving gases.

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Most popular questions from this chapter

Explain the following seeming contradiction: You have two gases, \(A\) and \(B\), in two separate containers of equal volume and at equal pressure and temperature. Therefore, you must have the same number of moles of each gas. Because the two temperatures are equal, the average kinetic energies of the two samples are equal. Therefore, since the energy given such a system will be converted to translational motion (that is, move the molecules), the root mean square velocities of the two are equal, and thus the particles in each sample move, on average, with the same relative speed. Since \(A\) and \(B\) are different gases, they each must have a different molar mass. If \(A\) has a higher molar mass than \(B\), the particles of \(A\) must be hitting the sides of the container with more force. Thus the pressure in the container of gas \(A\) must be higher than that in the container with gas \(B\). However, one of our initial assumptions was that the pressures were equal.

A piece of solid carbon dioxide, with a mass of \(7.8 \mathrm{~g}\), is placed in a 4.0-L otherwise empty container at \(27^{\circ} \mathrm{C}\). What is the pressure in the container after all the carbon dioxide vaporizes? If \(7.8 \mathrm{~g}\) solid carbon dioxide were placed in the same container but it already contained air at 740 torr, what would be the partial pressure of carbon dioxide and the total pressure in the container after the carbon dioxide vaporizes?

At elevated temperatures, sodium chlorate decomposes to produce sodium chloride and oxygen gas. A \(0.8765\) -g sample of impure sodium chlorate was heated until the production of oxygen gas ceased. The oxygen gas collected over water occupied \(57.2 \mathrm{~mL}\) at a temperature of \(22^{\circ} \mathrm{C}\) and a pressure of 734 torr. Calculate the mass percent of \(\mathrm{NaClO}_{3}\) in the original sample. (At \(22^{\circ} \mathrm{C}\) the vapor pressure of water is \(19.8\) torr.)

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?

Consider separate \(1.0\) -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 \(1.0\) -L samples of \(\mathrm{O}_{2}\) and \(\mathrm{H}_{2}\) each have the same average velocity?

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