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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

An important process for the production of acrylonitrile \(\left(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}\right)\) is given by the following equation: $$2 \mathrm{C}_{3} \mathrm{H}_{6}(g)+2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ A \(150 .\)-\(\mathrm{L}\)reactor is charged to the following partial pressures at \(25^{\circ} \mathrm{C}:\) $$\begin{aligned}P_{\mathrm{C}, \mathrm{H}_{6}} &=0.500 \mathrm{MPa} \\\P_{\mathrm{NH}_{3}} &=0.800 \mathrm{MPa} \\\P_{\mathrm{O}_{2}} &=1.500 \mathrm{MPa}\end{aligned}$$ What mass of acrylonitrile can be produced from this mixture \(\left(\mathrm{MPa}=10^{6} \mathrm{Pa}\right) ?\)

Hydrogen cyanide is prepared commercially by the reaction of methane, \(\mathrm{CH}_{4}(g),\) ammonia, \(\mathrm{NH}_{3}(g),\) and oxygen, \(\mathrm{O}_{2}(g),\) at high temperature. The other product is gaseous water. a. Write a chemical equation for the reaction. b. What volume of HCN( \(g\) ) can be obtained from the reaction of \(20.0 \mathrm{L} \mathrm{CH}_{4}(g), 20.0 \mathrm{L} \mathrm{NH}_{3}(g),\) and \(20.0 \mathrm{LO}_{2}(g) ?\) The volumes of all gases are measured at the same temperature and pressure.

A compound contains only \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{N}\). It is \(58.51 \%\) C and \(7.37 \%\) H by mass. Helium effuses through a porous frit \(3.20\) times as fast as the compound does. Determine the empirical and molecular formulas of this compound.

Helium is collected over water at \(25^{\circ} \mathrm{C}\) and 1.00 atm total pressure. What total volume of gas must be collected to obtain \(0.586 \mathrm{g}\) helium? (At \(25^{\circ} \mathrm{C}\) the vapor pressure of water is 23.8 torr.)

A mixture of \(1.00 \mathrm{g} \mathrm{H}_{2}\) and \(1.00 \mathrm{g}\) He is placed in a \(1.00-\mathrm{L}\) container at \(27^{\circ} \mathrm{C}\). Calculate the partial pressure of each gas and the total pressure.

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