TRANSLATIONAL MOTION OF GASES
The basic concept to focus here is - the gas molecules enjoy their freedom by moving from one place to another. This is known as translational motion. The effect of such motion is felt in the form of pressure and temperature.
Pressure is the force of bombardment of the gas molecules on the wall of the container. Its value should be a function of the frequency and intensity of bombardment. If the energy (or speed) is high and the number of molecules in the container is high, the impact on the wall will be more forceful and more often. This will result in a higher pressure on the container. The frequency of bombardment is of course dependent on the volume. (Just picture a swimmer in the pool swimming back a forth.) If the length is twice, the frequency should drop by half.
Of course pressure is relative. If I have a container with a piston and the amount of gas in the container is 1 bar pressure then the piston will remain stationary since the gas in the environment is also bombarding the wall of the container at the same intensity and frequency. If the bombardment on both sides of the container is not equal then the piston will move to a new volume until they are equal.
VAPOUR PRESSURE
When a compound (either in the liquid or solid phase) is left in an enclosed space it can give off gaseous molecules according to the Maxwell-Boltzmann distribution. Those molecules with sufficient energy will just leave the liquid (or solid) phase to become vapour. Like all processes, eventually an equilibrium will be reach where the rate of leaving to be vapour will be the same as the rate of molecules returning to the original phase. The molecules in the vapour then exert a pressure on the container. The value of the vapour pressure depends only on the temperature and is available from standard tables. As the temperature goes up, there are more and more molecules with the right combination of energy and direction to break free of the liquid's surface.
Tutorial 1
1.00 dm� of oxygen are collected over water at 26�C and 658.5 mmHg. What is the volume of dry oxygen at SATP?
(vapour pressure of H2O at 26�C = 25.2 mmHg) Answer
KINETIC THEORY of GASES
According to Newton's Second Law of motion, pressure is the rate of change of momentum. (Newton Physics can be used for atomic and molecular particles but not for sub-atomic particles.) Using this classical physics it is able to explain the ideal gas equation, and the transport properties of ideal gases: effusion, diffusion, thermal conductivity and viscosity.
So let us picture the molecule as a ball traveling towards the wall. Let the velocity be vx and its mass be m, the number of molecules bombarding an area of A within ∆t period of time would be (density of the molecules) x (volume). Or;
Statistically it should be logically to assume that on the average 50% of the molecules in this volume be moving towards the wall and 50% returning after hitting the wall. So the number of molecules bombarding the wall should be (� nNo / V) x Avx∆t
After hitting the wall the molecules would bounce directly back so the change in momentum of the molecule = (momentum when heading for the wall, mv) - (momentum after bouncing from the wall is, -mv) = 2mv.
| So the sum total change in momentum | = | (� nNo/V) x (Avx∆t) x 2mv |
| = | nNomAvx∆t / V | |
| = | nMAvx�∆t / V |
(Avogadro's number of molecules) x (the mass of one molecule) = mass of one mole of molecules, M. So;
Considering the molecules in all three directions. The "probable" speed for the molecules should be the root mean square speed.
| c | = | √ {vx� + vy� + vz�} |
where vx, vy and vz are the mean resolved speed along each direction. There is no good reason why the "vs" should not be equal in numerical value. So;
| c | = | √ {3vx�} | |
| Or | vx | = | c� / 3 |
| pV | = | nMc� / 3 |
Comparing with the ideal gas equation, pV = nRT ; then RT = Mc� / 3.
| Or | c | = | √ {3RT / M} |
One very important concept to note here is we will approach the study from a molecular level but in application it has to be at a mole level. In all cases we would use two equations to do this. Gas constant, R = Nok and molar mass, M = Nom. (k, is the Boltzmann constant; m, the mass of one molecule; and No is Avogadro's number)
Tutorial 2
Consider a cylinderical tube with radius d (where d is the diameter of a spherical molecule) and length L. Show that the number of collision per unit time by a molecule, travelling along the axis of this cylinder, with other molecules is directly proportional to the pressure and inversely proportional to the temperature. Answer
