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Journal of Pure and Applied Mathematics

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Tai-Choon Yoon*
 
Department of Thermodynamics and Energy, University of Science and Technology of China, Hefei, Anhui province, China, Email: tcyoon@hanmail.net
 
*Correspondence: Tai-Choon Yoon, Department of Thermodynamics and Energy, University of Science and Technology of China, Hefei, Anhui province, China, Email: tcyoon@hanmail.net

Received: 28-Jun-2023, Manuscript No. PULJMAP-23-6563; Editor assigned: 30-Jun-2023, Pre QC No. PULJMAP-23-6563 (PQ); Reviewed: 13-Jul-2023 QC No. PULJMAP-23-6563; Revised: 03-Mar-2025, Manuscript No. PULJMAP-23-6563 (R); Published: 10-Mar-2025

This open-access article is distributed under the terms of the Creative Commons Attribution Non-Commercial License (CC BY-NC) (http://creativecommons.org/licenses/by-nc/4.0/), which permits reuse, distribution and reproduction of the article, provided that the original work is properly cited and the reuse is restricted to noncommercial purposes. For commercial reuse, contact reprints@pulsus.com

Description

The traditional kinetic theory of gases, which states that PV=E=3/2kBT should be corrected to PV=E=kBT, as the process that leads to the coupling constants of 3 and 1/2 has changed and is no longer necessary.

Kinetic theory of gases

The kinetic theory of gases deals with the volume, pressure, and energy related to the thermal behavior of gases in a closed space [1]. The most basic theory that explains this is given as follows:

PV=Ek=3/2kBT                   (1)

Where,

P=Pressure, V=Volume, Ek=Kinetic energy, kB=Boltzmann constant, and T= Temperature.

Here, the coupling constant 3/2 is considered an essential coupling coefficient that appears when deriving the ideal gas law [2-5]. However, regarding the coupling coefficient of 3, the text on the “Boltzmann constant”, the energy Ek is given by:

Ek=1/2 mv2=1/2kBT              (2)

Therefore, considering translational motion as a one-dimensional motion along a straight path, the equation can be considered correct [6-10].

However, in reality, the velocity vector acting in a closed space acts perpendicular to the surface where the force acts. Therefore, the total acting area is not the cross-sectional area but the sum of the areas of the three surfaces [11-13]. This is why the coupling constant 3 needs to be removed and another reason is that the coupling constant 1/2 also needs to be removed [14].

Therefore, the kinetic theory of gases should be expressed as follows:

PV=Ek=kBT                          (3)

Derivation

In a closed space (V=L3), the momentum of a moving object is expressed by Newton’s law of inertia as follows:

equation

Where p is momentum, m0 is the rest mass in the space, and ∼V is the average velocity.

When external pressure is applied, the momentum changes in conjunction with the change in velocity as follows:

equation

The change in distance is the distance that the velocity V moves during the time Δt, which is given by:

equation

Where
equation
will be ignored due to very small quantity and t is given by:

equation

From this, the change in force (ΔF) can be obtained as follows:

equation

The pressure (P) is given as follows:

P=F/A                          (8)

Where A is the surface area of three sides of the closed space.

Rearranging equation (7), we get:

equation

Integrating equation (10), we obtain:

equation

=Ek/V

Where,

equation

And,

equation

As the result, the correct theory that satisfies the ideal gas law is as follows:

PV=E=kBT                          (14)

In consideration of relativistic time, by substituting the following equation into equation (6):

equation

From (6) and the above, we have

equation

We can obtain the following equation from equations (7), (10) and (12),

equation

Here, if <

equation

This is the exact expression for the traditional kinetic energy.

References

 
Google Scholar citation report
Citations : 83

Journal of Pure and Applied Mathematics received 83 citations as per Google Scholar report

Journal of Pure and Applied Mathematics peer review process verified at publons
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