Derivation of Average Speed of Gaseous Molecules

We can start by analyzing equation #15 from the Derivation of Ideal Gas Law below:

P=\frac{N}{V}\frac{m\overline{v^2}}{3}=\frac{N}{V}\frac{2E_{kinetic}}{3}=\frac{N}{V}kT=\frac{NkT}{V}

Because we want to derive the equation to the speed of gaseous molecules, the most important factor come to mind is the speed vx. Therefore, the equation will be

1. \frac{N}{V}\frac{m\overline{v^2}}{3}=\frac{N}{V}kT

Canceling out \frac{N}{V} from both side we get:

2. \frac{m\overline{v^2}}{3}=kT\Rightarrow{\overline{v^2}=\frac{3kT}{m}}

Since R = Nak. 

3. \overline{v^2}=\frac{3(R/N_a)(T)}{m}=\frac{3RT}{N_{a}m}

Finally, to calculate the avarage speed we find vrms(Root Meean Square of Speed).

4. v_{rms}=\sqrt{\overline{v^2}}=\sqrt{\frac{3RT}{N_{a}m}}=\sqrt{\frac{3RT}{M}}

M = Nam in the equation above is the mass of one mole of molecules (the molecular mass).

* The gas constant R must be expressed in correct units for the situation in which it is being used. In the ideal gas equation where pv=nRT, it is logical to use units  (L)(atm)/(mol)(K).

*In regard to speed, however, energy unit must be taken into account. Therefore, it is more appropriate to to convert it to (J)/(mol)(K). 

A. R=8.314\frac{J}{molK}

Derivation of PV=nRT, The Equation of Ideal Gas

According to the kinetic theory of gas,

– Gases are composed of very small molecules and their number of molecules is very large.
– These molecules are elastic.
– They are negligible size compare to their container.
– Their thermal motions are random.

To begin, let’s visualize a rectangular box with length L, areas of ends A1 and A2. There is a single molecule with speed vx traveling left and right to the end of the box by colliding with the end walls.

3D Demonstration of Ideal Gas

3D Demonstration of Ideal Gas

The time between collisions with the wall is the distance of travel between wall collisions divided by the speed.

1. t=\frac{2L}{v_x}

The frequency of collisions with the wall in collisions per second is

2. f=\frac{1}{t}=\frac{1}{2L/v_x}=\frac{v_x}{2L}

According to Newton, force is the time rate of change of the momentum

3. F=\frac{dp}{dt}=ma

The momentum change is equal to the momentum after collision minus the momentum before collision. Since we consider the momentum after collision to be mv, the momentum before collision should be in opposite direction and therefore equal to –mv.

4. \Delta{p}=mv_x-(-mv_x)=2mv_x

According to equation #3, force is the change in momentum \Delta{p} divided by change in time \Delta {t}.  To get an equation of average force \overline{F} in term of particle velocity v_x, we take change in momemtum \Delta{p} multiply by the frequency f from equation #2. 

5. \overline{F}=\Delta{p}(f)=2mv(\frac{v_x}{2L})=\frac{mv_x^2}{L}

The pressure, P, exerted by a single molecule is the average force per unit area, A. Also V=AL which is the volume of the rectangular box.

6. P_{1\:Molecule}=\frac{\overline{F}}{A}=(\frac{mv_x^2}{L})/A=\frac{mv_x^2}{LA}=\frac{mv_x^2}{V}

Let’s say that we have N molecules of gas traveling on the x-axis. The pressure will be

7. P_{N\:Molecules}=\frac{m}{V}(v_{x_1}^2+v_{x_2}^2+v_{x_3}^2....+v_{x_N}^2)=\sum_{a=0}^{N}\frac{mv_{x_a}^2}{V}

To simplify the situation we will take the mean square speed of N number of molecules instead of summing up individual molecules. Therefore, equation #7 will become

8. P_{N\:Particles}=\frac{Nm\overline{v_x^2}}{V}

Earlier we are trying to simplify the situation by only considering that a molecule with mass m is traveling on the x axis.  However, the real world is much more complicated than that. To make a more accurate derivation we need to account all 3 possible components of the particle’s speed, vx, vy and vz.

9. \overline{v^2}=\overline{v^2_x}+\overline{v^2_y}+\overline{v^2_z}

Since there are a large number of molecules we can assume that there are equal numbers of molecules moving in each of co-ordinate directions.

10. \overline{v^2_x}=\overline{v^2_y}=\overline{v^2_z}

Because the molecules are free too move in three dimensions, they will hit the walls in one of the three dimensions one third as often. Our final pressure equation becomes 

11. P=\frac{Nm\overline{v^2}}{3V}

However to simplify the equation further, we define the temperature, T, as a measure of thermal motion of gas particles because temperature is much easier to measure than the speed of the particle. The only energy involve in this model is kinetic energy and this kinetic enery is proportional to the temperature T. 

12. E_{kinetic}=\frac{mv^2}{2}\propto{T}

To combine the equation #11 and #12 we solve kinetic energy equation #12 for mv2. 

13. mv^2=2E_{kinetic}\Rightarrow\frac{mv^2}{3}=\frac{2E_{kinetic}}{3}

Since the temperature can be obtained easily with simple daily measurement like a thermometer, we will now replace the result of kinetic equation #13 with with a constant R times the temperature, T. Again, since T is proportional to the kinentic energy it is logical to say that T times k is equal to the kinetic energy E. k, however, will currently remains unknown.

14. kT=\frac{mv^2}{3}=\frac{2E_{kinetic}}{3}

Combining equation #14 with #11, we get:

15. P=\frac{N}{V}\frac{m\overline{v^2}}{3}=\frac{N}{V}\frac{2E_{kinetic}}{3}=\frac{N}{V}kT=\frac{NkT}{V}

Because a molecule is too small and therefore impractical we will take the number of molecules, N and divide it by the Avogadro’s number, NA= 6.0221 x 1023/mol to get n (the number of moles)

16. n=\frac{N}{N_a}

Since N is divided by Na, k must be multiply by Na to preserve the original equation. Therefore, the constant R is created.

17. R=N_ak

Now we can achieve the final equation by replacing N (number of melecules) with n (number of moles) and k with R. 

17. P=\frac{nRT}{V}\Rightarrow{PV=nRT}

Calculation of R & k

According to numerous tests and observations, one mole of gas is a 22.4 liter vessel at 273K exerts a pressure of 1.00 atmosphere (atm). From the ideal gas equation above:

A. R = \frac{PV}{nT}

B. R=\frac{(1 atm)(22.4L)}{(1 mol)(273K)}=0.082\frac{Latm}{molK}

C. k=\frac{R}{N_a}\Rightarrow{k=\frac{0.082 Latm/molK}{6.0221 x 10^{23}/mol}}=1.3806504 x 10^{-25}\frac{Latm}{K}

Sources:
1. Significant of PV=nRT
2. Kinetic Theory of Gas

Motion of Molecules

The idea that molecules are in constant motion was proposed by the kinentic theory of gases. The development of this theory in the 19th century are mostly based on the theory of atoms & molecules. Since there are no real experiments during that time, many leading physicists strongly opposed the idea. However, Brownian Motion, an observation done by botanist Robert Brown eliminated any opposition to the kinetic theory of gases.

Motion of Molecules

Motion of Molecules

According to the Theory:

1. The gas consists of very small particles, each of which has a mass or weight in SI units.

2. The number of molecules is large such that statistical treatment can be applied.

3. Molecules are in constant and random motion.

4. The rapidly moving particles constantly collide with each other and with the walls of the container.

5. The  collisions of gas particles with the walls of the container holding them are perfectly elastic.

6. The interactions among molecules are negligible. They exert no forces on one another except during collisions.

7. The total volume of the individual gas molecules added up is negligible compared to the volume of the container.

8.  The molecules are perfectly spherical in shape, and elastic in nature.

9. The average kinentic energy of the gas particles depends only on the temperature of the system.

10. The time during collision of molecule with the container’s wall is negligible as comparable to time between successive collisions.

11. The equation of motion of the molecules are time-reversible.

Sources:
1. Kinetic Theory
2. The Motion of Molecules

Concept of the Molecule

Although the atomic theory proposed by John Dalton created a basic structure of the atom, the general idea of molecules was not cleared. In 1809, Frech chemist Joseph-Louis Gay-Lussac and others began doing numerous experiments with gases by measuring the amounts of gass that actually reacted. They found that two volumes of hydrogen reacted with one volume of oxygen to form two volumes of water, and that one volume of hydrogen gas reacted with one volume of chlorine gas to form two volumes of hydrogen chloride gas. 

 2H_{2}+O_{2}\rightarrow2H_{2}O

 H_{2}+{CL}_{2}\rightarrow2HCL

In 1811, Avogadro proposed the following law:

“Equal volumes of ideal or perfect gases, at the same temperature and pressure, contain the same number of particles, or molecules.”

This Law is later confirmed experimentally. With the basis of Avogadro’s Laws, it became possible to compare the relative weights of various melecules and atoms. 

According to Avogadro’s Law:

\frac{V_1}{n_1}=\frac{V_2}{n_2}=Constant

n: Number of moles, V: Volume, T: Temperature (Constant), P: Pressure (Constant)

Example: The reaction in which hydrogen and oxygen combine to form water can be displayed as the following.

Water Molecules Formation

Water Molecules Formation

Avogadro’s Constant
The number of molecules in one mole, that is the number of atoms in exactly 12 grams of carbon-12.  

Avogadro's\:Constant=N_A=6.0221367\times10^{23}\:mol^{-1} 

Sources:
1. Molecules History
2. Avogadro’s Law
3. Introduction of the Concept of the Molecules

Atomic Proposal

The Idea of the atom were first proposed by the Greek philosophers Democritus and Leucippus around 400 B.C. At that time, there is absolutely no real evidence that support this proposal. Even after 20 centuries later, no experiment was strong enough to verify the existence of the atom. 

In the 18th Century, the first scientific data on the atom were gathered by A. L. Lavoisier and others from quantitative measurements of chemical reactions. From the experiment, he suggested that there exist some elements which could not be disintegrated into any smaller composition by usual chemical method. He defined this as chemical element. 

From the results of Lavoisier experiments, John Dalton proposed the first systematic atomic theory. This theory of the atom compose of two basic chemcial laws: the law of constant proportions and the law of multiple proportions. 

Law Of Constant Proportions
“The composition of a pure chemical compound is independent of its method of preparation”

Example: Water is a compound of hydrogen and oxygen. The ratio of the weight of hydrogen to oxygen in water is fixed at the value 1:8, independent of how it is formed. 

Law Of Constant Proportions
“When two elements A and B combine to form more than one compound, the weights of B which combine with a fixed weight of A are in the proportion of small whole numbers (integers)”.

Example: Carbon and oxygen react to form CO or CO2 but not CO1.1 or CO1.2.

Carbon Dioxide and Monoxide

Carbon Dioxide and Monoxide

Sources:
1. The discovery of the Atom

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