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Do all the molecules in a 1 -mole sample of \(\mathrm{CH}_{4}(g)\) have the same kinetic energy at 273 K? Do all molecules in a I-mole sample of \(\mathrm{N}_{2}(g)\) have the same velocity at \(546 \mathrm{K} ?\) Explain.

Short Answer

Expert verified
In conclusion, all molecules in a 1-mole sample of CH4(g) do not have the same kinetic energy at 273 K, and all molecules in a 1-mole sample of N2(g) do not have the same velocity at 546 K. This is because the distribution of velocities in both cases follows the Maxwell-Boltzmann distribution, which demonstrates that gas particles have a range of velocities rather than a single, uniform velocity.

Step by step solution

01

The kinetic energy of a single gas molecule is given by the equation: \[KE = \frac{1}{2}mv^2\] where \(KE\) is the kinetic energy, \(m\) is the mass of the molecule, and \(v\) is the velocity of the molecule. The root-mean-square velocity (v_rms) is a measure used to define the average velocity of gas molecules and is defined as: \[v_\text{rms} = \sqrt{ \frac{3kT}{m} }\] where \(k\) is the Boltzmann constant, \(T\) is the temperature in Kelvin, and \(m\) is the mass of the molecule. #Step 2: Explain the Maxwell-Boltzmann distribution#

The Maxwell-Boltzmann distribution is a probability distribution that describes the distribution of molecular speeds in an ideal gas. It shows that particles in a gas do not all have the same velocity; rather, they have a range of velocities that follow a bell-shaped curve. This means that even though there is an average velocity (v_rms), individual molecules in the gas may have velocities higher or lower than the v_rms. #Step 3: Determine the kinetic energy of CH4 molecules at 273 K#
02

Since the kinetic energies of the individual molecules depend on their velocities, and we know that the molecules in the CH4 gas have a range of velocities due to the Maxwell-Boltzmann distribution, we can conclude that all CH4 molecules do not have the same kinetic energy at 273 K. #Step 4: Determine and compare velocities of N2 molecules at 546 K#

Similarly, the velocities of the N2 molecules at 546 K will be distributed according to the Maxwell-Boltzmann distribution. As a result, we can conclude that not all molecules in a 1-mole sample of N2(g) have the same velocity at 546 K. In conclusion, all molecules in a 1-mole sample of CH4(g) do not have the same kinetic energy at 273 K, and all molecules in a 1-mole sample of N2(g) do not have the same velocity at 546 K. The distribution of velocities in both cases follows the Maxwell-Boltzmann distribution, which demonstrates that gas particles have a range of velocities rather than a single, uniform velocity.

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Key Concepts

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

Kinetic Energy of Gas Molecules
Kinetic energy is a measure of the energy that gas molecules possess due to their motion. It's an essential concept in understanding gas behavior. Each molecule in a gas has kinetic energy given by the equation:\[KE = \frac{1}{2}mv^2\]Here, \(KE\) represents kinetic energy, \(m\) stands for the mass of the molecule, and \(v\) symbolizes its velocity. This equation tells us that kinetic energy is directly proportional to the square of velocity and to the mass of the molecule. However, not all molecules have the same kinetic energy.- Gas molecules move with different velocities.- As velocity varies, so does kinetic energy.- The range of velocities is due to the distribution described by the Maxwell-Boltzmann curve.Thus, at a given temperature, while a group of molecules may share an average kinetic energy, individual molecules can still have various energies due to their differing speeds.
Root-Mean-Square Velocity
The root-mean-square velocity, often abbreviated as \(v_{\text{rms}}\), is a way to calculate the average speed of gas molecules. Unlike simple arithmetic mean, \(v_{\text{rms}}\) takes into account both the velocity and mass of particles. It is calculated using the formula:\[v_{\text{rms}} = \sqrt{ \frac{3kT}{m} }\]where \(k\) is the Boltzmann constant, \(T\) is the temperature in Kelvin, and \(m\) is the mass of the molecule. This formula highlights a few vital points:- As temperature increases, \(v_{\text{rms}}\) increases.- Heavier molecules move slower for a given temperature, resulting in a lower \(v_{\text{rms}}\).- The \(v_{\text{rms}}\) provides a statistical measure of molecular speed but doesn't mean all molecules travel at this speed.Speed distribution follows a range: some molecules move faster, and others slower than the \(v_{\text{rms}}\), reflecting the Maxwell-Boltzmann distribution, which describes the statistical nature of their motion.
Gas Molecules and their Motion
Gas molecules are in constant, random motion, colliding with each other and with the walls of their container. This movement is crucial for understanding how gases behave: - Their velocities are spread out over a range, not all molecules move at the same speed. - At a given temperature, a variety of molecular speeds exist. This distribution of molecular velocities in a gas is captured by the Maxwell-Boltzmann distribution, a key concept in gas physics. It predicts a bell-shaped curve that shows most molecules have moderate speeds, while fewer have very high or very low speeds. At higher temperatures, molecules gain kinetic energy, leading to higher speeds: - Gas molecules move faster, contributing to an overall increase in pressure. - The range of possible molecular velocities widens at elevated temperatures. Understanding these principles helps explain phenomena such as pressure, diffusion, and reaction rates in gases, demonstrating the kinetic theory of gases in action.

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Most popular questions from this chapter

Calculate the pressure exerted by \(0.5000\) mole of \(\mathrm{N}_{2}\) in a \(1.0000-\mathrm{L}\) container at \(25.0^{\circ} \mathrm{C}\) a. using the ideal gas law. b. using the van der Waals equation. c. Compare the results.

Xenon and fluorine will react to form binary compounds when a mixture of these two gases is heated to \(400^{\circ} \mathrm{C}\) in a nickel reaction vessel. A \(100.0\) -\(\mathrm{mL}\) nickel container is filled with xenon and fluorine, giving partial pressures of \(1.24\) atm and \(10.10\) atm, respectively, at a temperature of \(25^{\circ} \mathrm{C}\). The reaction vessel is heated to \(400^{\circ} \mathrm{C}\) to cause a reaction to occur and then cooled to a temperature at which \(\mathrm{F}_{2}\) is a gas and the xenon fluoride compound produced is a nonvolatile solid. The remaining \(\mathrm{F}_{2}\) gas is transferred to another A \(100.0\) -\(\mathrm{mL}\) nickel container, where the pressure of \(\mathrm{F}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(7.62\) atm. Assuming all of the xenon has reacted, what is the formula of the product?

Methane \(\left(\mathrm{CH}_{4}\right)\) gas flows into a combustion chamber at a rate of \(200 .\) L/min at \(1.50\) atm and ambient temperature. Air is added to the chamber at 1.00 atm and the same temperature, and the gases are ignited. a. To ensure complete combustion of \(\mathrm{CH}_{4}\) to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g),\) three times as much oxygen as is necessary is reacted. Assuming air is \(21\) mole percent \(\mathrm{O}_{2}\) and \(79\) mole percent \(\mathrm{N}_{2}\), calculate the flow rate of air necessary to deliver the required amount of oxygen. b. Under the conditions in part a, combustion of methane was not complete as a mixture of \(\mathrm{CO}_{2}(g)\) and \(\mathrm{CO}(g)\) was produced. It was determined that \(95.0 \%\) of the carbon in the exhaust gas was present in \(\mathrm{CO}_{2}\). The remainder was present as carbon in \(\mathrm{CO}\). Calculate the composition of the exhaust gas in terms of mole fraction of \(\mathrm{CO}, \mathrm{CO}_{2}, \mathrm{O}_{2}, \mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). Assume \(\mathrm{CH}_{4}\) is completely reacted and \(\mathrm{N}_{2}\) is unreacted.

Urea \(\left(\mathrm{H}_{2} \mathrm{NCONH}_{2}\right)\) is used extensively as a nitrogen source in fertilizers. It is produced commercially from the reaction of ammonia and carbon dioxide: $$2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{NCONH}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(g)$$ Ammonia gas at \(223^{\circ} \mathrm{C}\) and \(90 .\) atm flows into a reactor at a rate of \(500 .\) L/min. Carbon dioxide at \(223^{\circ} \mathrm{C}\) and \(45\) atm flows into the reactor at a rate of \(600 .\) L/min. What mass of urea is produced per minute by this reaction assuming \(100 \%\) yield?

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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