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

Short Answer

Expert verified
In a 1-mol sample of CH4(g), not all molecules have the same kinetic energy at 273 K, although their average kinetic energy is the same due to the temperature. Similarly, in a 1-mol sample of N2(g), not all molecules have the same velocity at 546 K, despite their average kinetic energy being the same due to the temperature. This variation in kinetic energy and velocity among individual molecules occurs because of their unique motion and collisions within the gas sample.

Step by step solution

01

Understand the relationship between temperature and kinetic energy

The average kinetic energy of a molecule in a gas sample is directly proportional to the temperature, meaning that increasing the temperature will increase the average kinetic energy of the molecules. The formula to find the average kinetic energy of a molecule can be written as: \[K.E = \dfrac{3}{2} kT\] Where K.E is the average kinetic energy, k is the Boltzmann constant, and T is the temperature in Kelvin. It is essential to note that this formula provides us with the average kinetic energy of the molecules in a sample, not the kinetic energy of each molecule.
02

Kinetic energy of CH4 molecules at 273 K

Now, we know that the average kinetic energy of the sample depends on the temperature. However, each molecule's kinetic energy can be different due to their unique velocities and collisions that occur within the gas sample. Even though the average kinetic energy at 273 K can be the same for all molecules in a 1-mol sample of CH4(g), individual molecules can have a range of different kinetic energies.
03

Relationship between velocity, mass, and kinetic energy

The kinetic energy of a molecule is related to its mass and velocity by the following formula: \[K.E = \dfrac{1}{2} mv^2\] Where K.E is the kinetic energy, m is the mass of the gas molecule, and v is its velocity. In this exercise, we are interested in examining the individual molecular velocities of N2(g) at 546 K.
04

Velocity of N2 molecules at 546 K

It is important to remember that molecular velocity, temperature, mass, and kinetic energy are related. Therefore, even if the average kinetic energy of the N2 molecules at 546 K is the same, their individual velocities can vary because of their unique pathway through the sample and collisions with other gas molecules. Hence, all molecules in a 1-mol sample of N2 do not have the same velocity at 546 K. In conclusion, - All molecules in a 1-mol sample of CH4(g) do not have the same kinetic energy at 273 K, even though their average kinetic energy is the same due to the temperature. - All molecules in a 1-mol sample of N2(g) do not have the same velocity at 546 K, even though their average kinetic energy is the same due to the temperature.

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Key Concepts

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

Molecular Velocity
Molecular velocity refers to the speed at which individual gas molecules move within a sample. Although the term might suggest uniform movement, in reality, molecules exhibit a wide range of velocities due to collisions and different pathways they take within the gas. This kinetic behavior ensures that not all molecules move at the same speed, even in a well-defined environment.
For instance, when examining the velocity of nitrogen (\(N_2\) molecules) in a sample kept at a higher temperature, such as 546 K, some molecules will move faster or slower than others. This variation is due to statistical distribution known as the Maxwell-Boltzmann distribution, which describes how the speed of molecules is spread out.
  • Molecules constantly collide with each other, changing velocities.
  • The temperature of the gas affects the spread of these velocities; higher temperatures typically mean higher velocities.
In summary, while the average velocity might give an idea of general motion, individual velocities differ significantly within a sample.
Temperature
Temperature is a fundamental concept in understanding gas behavior, particularly in relation to kinetic energy and molecular motion. It can be thought of as a measure of the average kinetic energy of the molecules in a substance. It's crucial to note that temperature doesn't tell us the energy of individual molecules but rather an average across a collection.
In gas dynamics, as seen in the gas samples of methane (\(CH_4\)) and nitrogen, temperature plays a critical role in determining the average kinetic energy.
Higher temperatures generally increase the kinetic energy and thus the molecular velocity, leading to more energetic and rapid movement of gas molecules:
  • An increase in temperature generally results in an increase in molecular velocity.
  • Temperature is proportional to kinetic energy, as shown in the formula \(K.E = \frac{3}{2} kT\).
Understanding temperature helps elucidate why molecules within the same sample may have differing velocities and energy levels.
Gas Molecules
Gas molecules are tiny particles that are constantly in motion, colliding and rebounding off each other and the walls of their container. This constant motion is central to understanding concepts of gas behavior such as molecular velocity and kinetic energy.
Gas samples like methane (\(CH_4\)) and nitrogen (\(N_2\)) consist of different molecules that are influenced by both temperature and their inherent molecular properties like mass.
  • Heavier gas molecules will generally move more slowly compared to lighter ones at the same temperature.
  • Despite the average values like temperature or velocity, individual molecules will have slightly different kinetic energies and velocities.
This variation is due to the large number of molecules in a sample, enabling different energy exchanges and collisions, resulting in a distribution of velocities.
Boltzmann Constant
The Boltzmann constant (\(k\)) is a crucial element in the realm of thermodynamics and kinetic theory of gases. It acts as a bridge between macroscopic and microscopic phenomena by relating temperature to energy. Represented as a factor in the equation \(K.E = \frac{3}{2} kT\), it helps calculate the average kinetic energy of gas molecules based on the temperature of the system.
Understanding the Boltzmann constant is essential in molecular studies:
  • It allows us to connect the temperature (a macroscopic variable) with energy at the molecular level (microscopic).
  • The constant is essential for equations describing the motion of particles, thermal energy, and speed distribution.
With a value of approximately \(1.38 \times 10^{-23} \text{ J/K}\), the Boltzmann constant provides insight into the universality of energy changes as a response to temperature fluctuations.

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Most popular questions from this chapter

A bicycle tire is filled with air to a pressure of 75 . psi at a temperature of \(19^{\circ} \mathrm{C}\). Riding the bike on asphalt on a hot day increases the temperature of the tire to \(58^{\circ} \mathrm{C}\). The volume of the tire increases by \(4.0 \%\). What is the new pressure in the bicycle tire?

At room temperature, water is a liquid with a molar volume of \(18 \mathrm{~mL}\). At \(105^{\circ} \mathrm{C}\) and 1 atm pressure, water is a gas and has a molar volume of over \(30 \mathrm{~L}\). Explain the large difference in molar volumes.

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) can be produced by the following reaction: $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g) $$ Hydrogen at STP flows into a reactor at a rate of \(16.0 \mathrm{~L} / \mathrm{min}\) Carbon monoxide at STP flows into the reactor at a rate of \(25.0\) L/min. If \(5.30 \mathrm{~g}\) methanol is produced per minute, what is the percent yield of the reaction?

Consider separate \(1.0-\mathrm{L}\) samples of \(\mathrm{He}(g)\) and \(\mathrm{UF}_{6}(g)\), both at \(1.00\) atm and containing the same number of moles. What ratio of temperatures for the two samples would produce the same root mean square velocity?

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

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