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The most probable kinetic energy of gas molecule is (a) \(k T / 2\) (b) \(3 \mathrm{kT} / 2\) (c) \(k T\) (d) \(k T / 4\)

Short Answer

Expert verified
The most probable kinetic energy of a gas molecule is (a) \(kT/2\).

Step by step solution

01

Understanding the Concept

To find the most probable kinetic energy of a gas molecule, we must refer to the Maxwell-Boltzmann distribution of molecular speeds in a gas. The most probable kinetic energy corresponds to the most probable speed (the peak of the Maxwell-Boltzmann distribution curve).
02

Application of the Maxwell-Boltzmann Distribution

The most probable speed, which corresponds to the peak of the Maxwell-Boltzmann distribution, is given by the formula: \(v_p = \sqrt{(2kT/m)}\), where \(k\) is the Boltzmann constant, \(T\) is the temperature, and \(m\) is the mass of a gas molecule. The most probable kinetic energy \(E_p\) is then obtained using the kinetic energy formula \(E = \frac{1}{2}mv^2\), replacing the speed with the most probable speed.
03

Calculating the Most Probable Kinetic Energy

Using the kinetic energy formula and substituting \(v_p\) we get, \(E_p = \frac{1}{2}m(v_p)^2 = \frac{1}{2}m(\sqrt{(2kT/m)})^2 = \frac{1}{2}m(\frac{2kT}{m}) = kT/2\). Therefore, the correct answer is \(kT/2\).

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Key Concepts

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

Maxwell-Boltzmann Distribution
Understanding the distribution of molecular speeds in a gas is crucial when studying the kinetic theory of gases. The Maxwell-Boltzmann distribution is a statistical distribution of the speeds of particles within a classical ideal gas. This distribution arises from the assumption that particles move independently and their velocities are randomly distributed in all directions.

Put simply, it describes how likely it is to find a gas molecule moving at a certain speed at a given temperature. Imagine plotting the number of molecules against their speed; you'll get a curve that rises sharply from zero, reaches a peak, and then gradually tails off. This peak represents the most probable speed—the speed at which you'll find the most molecules.

The mathematical representation of the Maxwell-Boltzmann distribution is quite complex, but for our purposes, it's enough to know that the most probable kinetic energy relates to the most probable speed. When dealing with gases, understanding this principle is key to grasping concepts like diffusion, effusion, and the general behavior of gases with changes in temperature.
Kinetic Energy Formula
Kinetic energy is the energy a body possesses due to its motion. It's a fundamental concept across physics, from the motion of cosmic bodies to microscopic particles. For a gas molecule, this is especially important as it relates to temperature and thermal energy.

The classical formula for kinetic energy (\(E\text{KE}\)) is expressed as \(E_\text{KE} = \frac{1}{2}mv^2\), where \(m\) represents the mass of the particle and \(v\) is its velocity. In the context of gases, this formula helps us calculate the kinetic energy of molecules zipping around in their container.

When you link kinetic energy to temperature, you begin to see the fundamental nature of thermal energy. It's essentially the collective kinetic energy of countless particles. In physics problems, we often use the kinetic energy formula to relate the speed of particles to other properties like temperature and mass, offering a bridge between the micro and macroscales of our universe.
Boltzmann Constant
The Boltzmann constant, denoted as \(k\) or \(k_B\), is a bridge between the macroscopic and microscopic worlds, relating temperature with energy on a particle scale. Its value is approximately \(1.38 \times 10^{-23}\) joules per kelvin (J/K).

When discussing kinetic theory, the Boltzmann constant allows us to compute the average kinetic energy per molecule in a gas. It can be viewed as the conversion factor that translates the macroscopic readings of temperature (what we feel and measure with a thermometer) into the microscopic motion of particles (the true nature of thermal energy).

Knowing the Boltzmann constant is essential when performing calculations involving gases. From calculating individual molecular energies to understanding entropy and the distribution of particle speeds, \(k_B\) pops up in various fundamental equations in thermodynamics and statistical mechanics, underscoring its significance in the realm of physics.

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Most popular questions from this chapter

If at top of a hill \(2000 \mathrm{~m}\) above sea level, the atmospheric pressure is \(50 \mathrm{~cm}\) of \(\mathrm{Hg}\) and at the sea level the atmospheric pressure is \(74.5 \mathrm{~cm}\) of \(\mathrm{Hg}\), and you need as much oxygen to breath at sea level as on the top of the hill, how much faster need you breathe on the hill top? (a) \(2.44\) times (b) \(1.49\) times (c) 5 times (d) 7 times

If helium and methane are allowed to diffuse out of the container under the similar conditions of temperature and pressure, then the ratio of rate of diffusion of helium to methane is (a) \(2.0\) (b) \(1.0\) (c) \(0.5\) (d) \(4.0\)

A manometer is connected to a gas containing bulb. The open arm reads \(53.3 \mathrm{~cm}\) whereas the arm connected to the bulb reads \(15.6 \mathrm{~cm}\). If the barometric pressure is \(763 \mathrm{~mm}\) mercury, what is the pressure of gas, in atm? (a) \(1.05 \mathrm{~atm}\) (b) \(1.5 \mathrm{~atm}\) (c) \(0.51 \mathrm{~atm}\) (d) \(1.91\) atm

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The \(P V-P\) isotherms of 1 mole of different gases at \(273 \mathrm{~K}\), if the limit of pressure tending to zero, converge to a value of \(P V=\) (a) \(11.21\) -atm (b) \(22.41\) -atm (c) zero (d) \(22.41\)

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