# Solving Flux of Molecules Problem

• Pushoam
In summary: Since there is no reason why I should take the probability of hitting at one place more than the other place on the...In summary, the flux of molecules is given by the number of molecules that strike the surface per unit area per unit time.
Pushoam

## Homework Statement

Derive the flux of molecules i.e. no. of molecules striking a surface per unit area per unit time.

## The Attempt at a Solution

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Let's say that there are n molecules per unit volume.

The no. of molecules which will touch the surface in the +ve x direction in time ##d t ## is
dN = n A dx = n A <vx> dt .
So, the flux should be ##\phi = n <v_x>##.
##< v_x> = < v_y> = < v_z> \neq \frac {<v>} 3##
The probability of getting a particle with speed between ##v_x## and ## v_x + dv_x## is given by
##g(v_x) dv_x ## ## \propto ## ## e ^ \left( - \frac {m{v_x}^2}{2 k_B T} \right) ##.
Then,
## <v_x> = \int_{0 }^{\infty } v_x C## ##e^{ \left( - \frac { m{v_x}^2 } {2 k_BT } \right) }## ## dv_x ##
Where C is the appropriate constant.
I have to express ## < v_x >## into ## <v>##. For this I should express ##<v_x>## as ## v \sin { \theta } \cos { \phi } ## and then take integration.

Is this correct?

Last edited:
Pushoam said:
dN = n A dx
Where dx is what, exactly?

The particle is moving in the x - direction. So, dx is the distace it moves in the x - direction in time dt before hitting the surface with area A.

Pushoam said:
The particle is moving in the x - direction. So, dx is the distace it moves in the x - direction in time dt before hitting the surface with area A.
Ok.
I don't think you need to do quite such a messy integral. If a particle has speed v but arbitrary direction, what, on average, is its speed in the x direction?

haruspex said:
Ok.
I don't think you need to do quite such a messy integral. If a particle has speed v but arbitrary direction, what, on average, is its speed in the x direction?

Since the body has speed v in any arbitrary direction, then the body will have speed v when it is moving in the x - direction only and speed v ##\cos \theta## when it is moving with speed v along a direction whose projection upon x is ##\cos \theta##.
Using ##< f(x)> = \int_{x_i}^{x_f} f(x) P(x) dx##
##<v_x> = \int_0 ^{\frac {\pi}{2} }v \cos \theta P(\theta) d \theta##
Now, should I take ## P(\theta) d\theta = \frac{d \theta}{2 \pi}##?

Pushoam said:
Now, should I take ## P(\theta) d\theta = \frac{d \theta}{2 \pi}##?
No. Consider the area on a sphere that would be hit by leaving the centre in a direction forming an angle to to the x-axis between θ and θ+dθ.

##P(\theta ) d \theta ## is the probability that the particle is moving in the direction making an angle ## \theta~ and~ \theta + d \theta ## with x – axis.

Since ## \theta ## goes from ## \frac { - \pi } {2 } ## to ## \frac { \pi } {2 } ## and each value is equally probable, the probability that the particle is moving in the direction making an angle ## \theta ~and ~\theta + d \theta ## with x – axis is ## \frac {d\theta } { \pi } ##. Is this correct?

Another way is:

##P(\theta ) d \theta ## = ## \frac { \text { The area between the angle } \theta ~and~ \theta + d \theta

} {\text { The area between the angle } \frac { - \pi } {2 } ~ to ~ \frac { \pi } {2 } } = \frac { 2 \pi R^2 \sin \theta d\theta } {4 \pi R^2 } = \frac {\sin \theta d\theta } { 2 } ##How to decide which one is correct?

Pushoam said:
Since ## \theta ## goes from ## \frac { - \pi } {2 } ## to ## \frac { \pi } {2 } ## and each value is equally probable,
Not so. If you consider a particle going in an arbitrary direction from the centre of a sphere the probability distribution for where it hits the surface should, by spherical symmetry, be uniform.

haruspex said:
Not so. If you consider a particle going in an arbitrary direction from the centre of a sphere the probability distribution for where it hits the surface should, by spherical symmetry, be uniform.
Since there is no reason why I should take the probability of hitting at one place more than the other place on the spherical surface, I have to take the probability of hitting any place on the surface to be uniform. This I understand.

In a similar way, I have no reason to take the probability of velocity in a particular direction more than other. So, the probability of a particle moving in any direction should also be uniform. This gives me ## P(\theta) d\theta = \frac{ d \theta} {\pi}##. What is wrong with this logic? I didn't get it.

Pushoam said:
the probability of a particle moving in any direction should also be uniform. This gives me ## P(\theta) d\theta = \frac{ d \theta} {\pi}##.
But that is not treating all directions equally. It is treating all angular deviations from a given direction equally, which is not the same thing. The choice of that reference direction creates a bias.

haruspex said:
It is treating all angular deviations from a given direction equally
From now on I will write x as s (cylindrical coordinate ) as I am considering spherical coordinate system. Then ##\theta ##is the angle which the velocity vector makes with z- axis.

And I have to consider the particles which hit the y- x surface. So, I have to calculate ##<v_z>##.

So, the probability that the particle's velocity will be in a direction between ## \theta ~ and ~ \theta + d\theta ## from z- axis is P(## \theta ##) = ## \frac {d \theta } { \pi } ## in s-z plane.

But, here s- axis is the projection of velocity vector upon x – y plane. Since, the velocity vector is changing, this axis, too, is changing. So, I unknowingly assumed that velocity vector is changing in such a way that its projection upon x-y plane remains constant in direction. This is where I am wrong. Right?

Pushoam said:
Since there is no reason why I should take the probability of hitting at one place more than the other place on the spherical surface, I have to take the probability of hitting any place on the surface to be uniform. This I understand.
The above tells me the probability of the particle hitting the surface.
It doesn't tell me the probability of the particle moving in a certain direction.Another way is:

##P(\theta ) d \theta ## = ## \frac { \text { The area between the angle } \theta ~and~ \theta + d \theta

} {\text { The area between the angle } \frac { - \pi } {2 } ~ to ~ \frac { \pi } {2 } } = \frac { 2 \pi R^2 \sin \theta d\theta } {4 \pi R^2 } = \frac {\sin \theta d\theta } { 2 } ##I don't see the connection between the above and the idea that the above is the way one can treat all directions equally.
haruspex said:
But that is not treating all directions equally.

Pushoam said:
The above tells me the probability of the particle hitting the surface.
It doesn't tell me the probability of the particle moving in a certain direction.Another way is:

##P(\theta ) d \theta ## = ## \frac { \text { The area between the angle } \theta ~and~ \theta + d \theta

} {\text { The area between the angle } \frac { - \pi } {2 } ~ to ~ \frac { \pi } {2 } } = \frac { 2 \pi R^2 \sin \theta d\theta } {4 \pi R^2 } = \frac {\sin \theta d\theta } { 2 } ##I don't see the connection between the above and the idea that the above is the way one can treat all directions equally.
As I wrote, if you consider a particle emitted from the centre of a sphere, all directions equally likely must be the same as the probability of hitting a given patch of the surface being proportional to the area of the patch.

haruspex said:
As I wrote, if you consider a particle emitted from the centre of a sphere, all directions equally likely must be the same as the probability of hitting a given patch of the surface being proportional to the area of the patch.
Is this an experimental fact which could not be proved but verified experimentally?

Pushoam said:
Is this an experimental fact which could not be proved but verified experimentally?
I would say it is the definition of "all directions equally likely". Actually, that is not a proper statement anyway since there are infinitely many directions and each by itself has probability zero. It needs to be expressed in terms probability densities, and I do not see any way to do that other than referring to solid angles, and solid angles are defined in terms of areas projected from a point.

Pushoam

## 1. What is the flux of molecules problem?

The flux of molecules problem refers to the movement of molecules across a boundary or through a specific area. It is a fundamental concept in physics and chemistry that is used to describe the rate at which molecules are transported or diffused in a given system.

## 2. How is the flux of molecules calculated?

The flux of molecules is calculated by multiplying the concentration gradient (difference in concentration between two points) by the diffusion coefficient and the surface area of the boundary. This can be represented mathematically as J = -D∇C, where J is the flux, D is the diffusion coefficient, and ∇C is the concentration gradient.

## 3. What factors affect the flux of molecules?

The flux of molecules can be affected by various factors, including the concentration gradient, temperature, molecular size, and the presence of barriers or obstacles. Additionally, the type of medium through which the molecules are moving (e.g. gas, liquid, or solid) can also impact the flux.

## 4. How is the flux of molecules related to diffusion?

The flux of molecules and diffusion are closely related concepts. The flux describes the rate at which molecules are transported, while diffusion is the process by which molecules move from an area of higher concentration to an area of lower concentration. In other words, the flux is a measure of the diffusion rate.

## 5. What applications does the flux of molecules have in science?

The flux of molecules is a critical concept in various fields of science, including biology, chemistry, and physics. It is used to understand and predict the movement of substances in living organisms, chemical reactions, and physical processes. For example, the flux of oxygen and carbon dioxide in and out of cells is essential for cellular respiration, and the flux of molecules in and out of the Earth's atmosphere affects climate change.

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