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Derivation of Boyle's Law using calculus

Hi, I am new here. In my chemistry book, I came across a derivation of the Boyle's law using calculus which I did not understand. Well, actually, it is a small part that I did not understand. My question is not purely based on chemistry, it is mostly mathematics, and so I am posting here. Anyway, here is how it goes:

There is an imaginary cylinder of base A and slant height ct, where c is the molecular speed and t, a small interval of time. Θ is the angle between the axis of the cylinder and a perpendicular from the wall. And Φ is the angle on the surface of the wall. Molecules in it are moving parallel to the axis, and hence have a perpendicular component to the wall c cosΘ. So, the momentum imparted on the wall by one such molecule will be 2mc cosΘ (where m is the mass of the molecule).

The number of molecules that move parallel to the axis = (Act cosΘ) × (N/V) × (dΦ sinΘ dΘ/4π) [where N is the total number of molecules and V is the total volume, and π is pi though it does not quite look like it].

(dΦ sinΘ dΘ/4π) is found by dividing (r^{2} sinΘ dΘ dΦ) by the total surface area of the sphere (4πr^{2}). This is what I did not understand, how did they obtain (r^{2} sinΘ dΘ dΦ)?

Attachment 25524

Thank you :)

Re: Derivation of Boyle's Law using calculus

Hey FragrantTransition.

The sin(theta) component is the y-contribution of the particulars and the cos(theta) component is the x-component.

Think of the projection of each slice of the cylinder: you are adding up each individual slice (i.e. every cross section of the flattened tube) and this is weighted by the density of molecules per cross-sectional area (which is assumed constant) and then you add these up in the limit of the integral.

Now each slice is pi*r^2, but the length of all the slices will be ct*cos(theta) and the density of each individual slice will be density*sin(theta) of which the density is considered constant as (N/V) throughout the whole cylinder.

So we are calculating individual slice density * slice volume (infinitesimal) * length of cylinder or in other words, calculate each individual slice-density in first two terms and add up all these contributions. The density of each slice will be the total density * sin(theta) (just a projection of a vector running along the line ct projected to the the centre of that cylinder).

I don't know where the surface of the area comes in: maybe you could explain that argument but with regards to the other part, I've given my interpretation. Also I'm not sure why theta is being integrated with respect to (since it should be fixed) but the other one makes sense.

So in short (if it's not completely right, maybe you can build on it to make it right) you have a projection in both sin(theta) and cos(theta) where sin(theta) provides the projection of the density of a particular slice and cos(theta) provides the total length of the deformed tube and when slice volume * molecule density * number of slices through integral is computed you get total number of molecules travelling in that direction.

Re: Derivation of Boyle's Law using calculus

Quote:

Originally Posted by

**chiro** Hey FragrantTransition.

The sin(theta) component is the y-contribution of the particulars and the cos(theta) component is the x-component.

Think of the projection of each slice of the cylinder: you are adding up each individual slice (i.e. every cross section of the flattened tube) and this is weighted by the density of molecules per cross-sectional area (which is assumed constant) and then you add these up in the limit of the integral.

Now each slice is pi*r^2, but the length of all the slices will be ct*cos(theta) and the density of each individual slice will be density*sin(theta) of which the density is considered constant as (N/V) throughout the whole cylinder.

So we are calculating individual slice density * slice volume (infinitesimal) * length of cylinder or in other words, calculate each individual slice-density in first two terms and add up all these contributions. The density of each slice will be the total density * sin(theta) (just a projection of a vector running along the line ct projected to the the centre of that cylinder).

I don't know where the surface of the area comes in: maybe you could explain that argument but with regards to the other part, I've given my interpretation. Also I'm not sure why theta is being integrated with respect to (since it should be fixed) but the other one makes sense.

So in short (if it's not completely right, maybe you can build on it to make it right) you have a projection in both sin(theta) and cos(theta) where sin(theta) provides the projection of the density of a particular slice and cos(theta) provides the total length of the deformed tube and when slice volume * molecule density * number of slices through integral is computed you get total number of molecules travelling in that direction.

Thank you for your reply chiro.

Quote:

The density of each slice will be the total density * sin(theta) (just a projection of a vector running along the line ct projected to the the centre of that cylinder).

I didn't understand this. Why will the density of each slice be total density × sinΘ?

I am not sure about the significance of the surface area, but I have a hunch that it has got something to do with the direction in which the molecules are travelling. Initially, I thought that the first two terms of (Act cosΘ) × (N/V) × (dΦ sinΘ dΘ/4π) were only for the number of molecules in the cylinder, and the 3rd term was to pin point the direction of the moving molecules.

Re: Derivation of Boyle's Law using calculus

The way to visualize the sine part is think about the "flux" (for lack of a better word) in the direction of the ct line and sin(theta) represents the y-component (as opposed to the x-component which is cos(theta0 along the z or whatever axis you are measuring relative to).

So basically it's the projection of the density with respect to that angle theta where if we have a density for a slice say D then sin(theta)*D captures the projection of that density.

It might be easier to think of the density as a kind of flux since it has a directional aspect and multiplying it by sin(theta) gets the component of the flux (if its going along that straight line) in the direction of the centre of the transformed cylinder and this corresponds to the flux for the actual molecules moving through that cylinder.

I think density was the wrong word and I agree that in line with what you are describing from the book, the projection of the flux relative for the centre of the cylinder relative to the z-axis is a better description of what is going on.

Re: Derivation of Boyle's Law using calculus

Oh, I see. That explains the role of flux in the equation. But, from my physics lessons, I have learnt that the calculation of flux involves the dot product (which is essentially cosΘ). I am not intentionally trying to wrong you, just trying to understand this better.

Still, the flux that I am talking about is related to field lines and area. I don't know exactly how to visualize a density flux. What bothers me is the vector nature of density. Well, anyway, I understood the significance of density flux in finding the number of molecules moving in the direction of the cylindrical axis.

Re: Derivation of Boyle's Law using calculus

I thought, thought and thought, and I finally came to an answer that seemed to convince me. This is my interpretation of the term (r^{2} sinΘ dΘ dΦ)/(4πr^{2}). The whole point of this term is to account for the direction of the molecules as mentioned earlier. So, the value of r does not have any significance at all in the derivation (obviously because there is no r in the term). Basically, the idea behind finding the surface area on the sphere is like this. You have a sphere of radius r (which is arbitrary), and the angle subtended by a certain radial line with another radial line is Θ. When you say r sinΘ, you get the length opposite to angle Θ. When you multiply r sinΘ with dΦ, you get another length (this one is perpendicular to r sinΘ, as it cuts an infinitesimally small arc on the circle with radius r sinΘ). Now, the only terms remaining are r and dΘ (apart from 1/(4πr^{2})). If you join a radial line to the infinitesimally small length just obtained, and multiply it with dΘ, it will give another infinitesimal length perpendicular to the former. So there you have it, an infinitesimally small area. This divided by the total surface area gives you the fraction of the small area with the total surface area of the sphere. You would have probably guessed by now that if you extend the infinitesimally small area to the centre of the circle, you will get the cylinder which we had in the start, and A is that small area.