Estimating the gas laws based on collissions
Starting with a simple one dimensional force calculation problem
An object is traveling back and forth between two walls at a constant speed called travel speed except during collissions. During collissions the object slows down to 0 then speeds up to it's original speed but in the opposite direction. The walls are parrellel to each other and a distance called travel length apart except during collissions where the wall hit by the object temporarily deforms then returns to it's original shape or is temporarily displaced then returns to it's original position after each collission ends. The object moves in a path perpendicular to the walls. The time the object is not in a collission is called travel time and the time the object is in a collission is called collision time. The length a wall is temporarily displaced in the direction parrelel to the objects path of travel during a collission is called collission length. The wall is moved or deformed during collission and the object is treated like a point mass with no deformation to make the math simpler.
Mass refers to the mass of the object in one dimensional force calculations below. Force below refers to a scalar being the absolute value of the force vectors magnitude and not to the direction of the force vector. That is I am not counting forces going in opposite directions as having opposite values in the calculations below. If forces were treated as vectors and no absolute value was taken there would be a zero average force when an equal number of complete collissions in each opposite direction occured.
The mean force the object would experience with respect to time using impulse = change in momentum = force * time
The absolute value of the change in momentum in a single collission is twice the absolute value of the objects momentum in this case because it switches to the same momentum in the opposite direction.
mean force with respect to time during travel = 0
mean force with respect to time during collission =
= 2*mass*(travel speed) / (collission time)
mean force with respect to time =
= 2*mass*(travel speed) / ([travel time] + [collission time])
(travel time) = (travel length) / (travel speed)
limit of mean force with respect to time as [(collision time) / (travel time)] approaches 0 is =
= 2*mass*(travel speed) / (travel time) =
= 2*mass*([travel speed]^2) / (travel length)
The mean force the object would experience with respect to distance using work = force * distance
Work occurs during a collission to slow the object down to zero speed then again a second time to return it to original speed. So the work done in a single collission is twice the traveling kinetic energy of the object in this case.
mean force with respect to distance during travel = 0
mean force with respect to distance during collission =
= 2*0.5*mass*([travel speed]^2) / (collission length)
mean force with respect to distance =
= 2*0.5*mass*([travel speed]^2) / ([collission length]+[travel length])
limit of mean force with respect to distance as [(collision length) / (travel length)] approaches 0 is =
= 2*0.5*mass*([travel speed]^2) / ([travel length])
= mass*([travel speed]^2) / ([travel length])
Start Side Note
the mean force with respect to distance is half the mean force with respect to time when you assume collission length and collision time are both insignificantly small compared to travel length and travel time.
See Article
http://teachingthenarrowway.blogspot.com/2021/12/average-with-respect-to-time-vs-with.html
http://web.archive.org/web/*/http://teachingthenarrowway.blogspot.com/2021/12/average-with-respect-to-time-vs-with.html
End Side Note
Now Let's instead assume there is a three dimensional object of a cube of length L and instead multiple particles collide into the walls in any one of six directions or three orthogonal / perpendicular directions and there opposites each of those six directions being perpendicular to one of the six faces of the cubes. In this case we will assume the particles do not move any other direction than those six directions to make the math easier so that in each case it will work mathematically similar to the one dimensional case above. That is we will assume the particles will only collide with the faces of the cube moving in a direction perpendicular to the face they hit making the collission calculatable as a one dimensional problem in each case. Let us also assume the particles do not collide into each other or do not effect each other if they were to collide into another particle but "phase through" each other or act as though having "no clipping" with respect to each other but only collide with cube faces/walls or only experience collission forces through cube faces/walls and not other particles. Additionally we shall assume all particles have the same speed except during collissions even though in reality their would be a distribution of different speeds at most temperatures people normally experience.
Temperature = Average Kinetic Energy per molecue
(Travel Speed)^2 = Kinetic Energy per molecue / mass per molecue =
= Temperature / (mass per molecue)
Limit as [(collision time) / (travel time)] approaches 0 of (Average Force per Molecue) with respect to time =
= 2* (mass per molecue) *([travel speed]^2) / (L)
= 2*(mass per molecue)*[Temperature / (mass per molecue)] / L
= 2*Temperature/L
From now on I shall assume collission time is negligible and refer to force as the force calculated using the calculus limit above and as measured as on average with respect to time
Total force from all molecues =
= Force per molecue * Density * Volume / (mass per molecue)
= Force per mole * Density * Volume / (mass per mole)
= (2*Temperature/L) * Density * Volume / (mass per mole)
Surface Area of a cube = 6*(L^2)
Volume of a cube = L^3
Pressure = Total Force / Surface Area =
(2*Temperature/L)*Density*(L^3)/[(mass per mole)*(6*[L^2])]
Pressure = Temperature*Density/[3*(mass per mole)]
N = Density*Volume/(mass per mole)
PV=NrT
P*V = r*Temperature*Density*V/(mass per mole)
P = r * Temperature*Density/(mass per mole)
Temperature*Density/[3*(mass per mole)] = r * Temperature*Density/(mass per mole)
r = 1 / 3
I remember getting a different answer of r = 2 / 3 when I calculated this in high school. I decide to search for the phrase unitless gas constant online and see if anyone actually calculated a unitless value or if I am the only person to ever do so. Apparently someone else also got 2/3 which is different than the 1/3 (or 1/6 because I calculated force two different ways based on average force as a function of length vs average force as a function of time) value I calculated this time but the same as I remember calculating in high school.
https://duckduckgo.com/?q=unitless+gas+constant&ia=about
https://en.m.wikipedia.org/wiki/Talk:Gas_constant#2%2F3_cut_-_ref_please
http://web.archive.org/web/20211215070129/https://en.m.wikipedia.org/wiki/Talk:Gas_constant#2%2F3_cut_-_ref_please
One might conclude that there should be different pressure depending on the shape of the container if it has a different volume to surface area ratio than that of a cube since the calculations were made using a cube shaped container but an assumption was made molecues do not experience force when they collide with each other only the container holding them in the calculations above. If you remove the assumption of lack of molecular collission force except against the container walls the molecues would travel a much shorter distance between collissions. You can imagine if there are N molecues in the container then imagine the container is made of N cubes, the molecues would each inhabit a average volume of a cube shaped like the volume of the container / N and travel on average a length of the cube root of the (volume of the container divided by N) and have a average surface area of the cube shaped zone they move around in equal to 6 times the square of the cube root of the (volume of the container divided by N) you would end up with the same results as calculated above for the average pressure and force from these molecular collissions by partititioning a non cubed shape into many cubes.
It might make more sense to assume the molecues typically inhabit a sphere shaped shape than a cube but a sphere which has a diameter equal to the length of a side of a cube has an equal surface area to volume ratio to that of a cube but would have a smaller surface area and volume than that cube requiring more spheres to fit in the same container if they each had a diameter equal to a side length of a cube as such it would result in a slightly different unitless constant number for R, especially if the molecues were considered to be able to move in more than six directions and a distribution of speeds rather than all moving at the same speed except when colliding.
Pressure * Volume = Force/Area * Length/Area = Force * Length = Energy
Number of Moles * Temperature = Number of moles * Kinetic Energy per mole = Energy
Some number of Joules = PV = rNT = r* some number of Joules
Some number of calories = PV = rNT = r* some number of calories
r = some number without units
Even though r is technically unitless it is listed as a ratio of several units which cancel out each other to allow people to convert units from one type to another.
Surface Area / Volume for a cube of side length L
(6 L^2) / (L^3) = 6 / L
Surface Area / Volume for a Sphere
(4*pi*r^2)/(4*pi*(r^3)/3) = 3 / r
(pi*D^2)/(0.5*pi*(D^3)/3) = 6 / D
Surface Area for a E*L by F*L by G*L rectangular solid
2*(EF+FG+EG)*(L^2)
Volume for a E*L by F*L by G*L rectangular prism
E*F*G*(L^3)
Surface Area / Volume for a E*L by F*L by G*L rectangular prism
2*(EF+FG+EG)/(E*F*G*L)
Make a rectangular solid that is not a cube with the same surface area / volume as a cube
Let 2*(EF+FG+EG)/(E*F*G) = 6
When E = 2 and F = 1
2*(2+G+2G) / 2G = 6
12 G = 4 + 6G
6 G = 4
G = 2 / 3
2*(2+[2/3]+[4/3]) / [4/3] = 2*4 / [4/3] = 6
Let L = 1 inch
1 inches by 2 inches by 2/3 inch
volume = (4/3) inches^3
surface area = 2*([2*1]+[2*2/3]+[1*2/3]) inches^2 = 8 inches^2
8 inches^2 / (4/3) inches ^3 = 6 inches ^ -1
1 inch by 1 inch by 1 inch cube
6 inches ^ 2 / 1 inch ^ 3 = 6 inches ^ -1
Copyright Carl Janssen 2021
In some video games, noclip mode is a video game cheat command that prevents the first-person player character camera from being obstructed by other objects and permits the camera to move in any direction, allowing it to pass through such things as walls, props, and other players.
Noclipping can be used to cheat, avoid bugs (and help developers debug), find easter eggs, and view areas beyond a map's physical boundary.
http://web.archive.org/web/20170520115201/https://en.m.wikipedia.org/wiki/Noclip_mode
https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-8b/v/antiderivative-of-x-1
http://web.archive.org/web/20190515181746/https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-8b/v/antiderivative-of-x-1
https://socratic.org/questions/what-is-the-antiderivative-of-ln-x-4
http://web.archive.org/web/20211216034307/https://socratic.org/questions/what-is-the-antiderivative-of-ln-x-4
In the kinetic theory of gases, the mean free path of a particle, such as a molecule, is the average distance the particle travels between collisions with other moving particles.
https://en.m.wikipedia.org/wiki/Mean_free_path
http://web.archive.org/web/20220629235820/https://en.m.wikipedia.org/wiki/Mean_free_path
https://en.m.wikipedia.org/wiki/Kinetic_theory_of_gases
http://web.archive.org/web/20220513084951/https://en.m.wikipedia.org/wiki/Kinetic_theory_of_gases