Kinetic Theory of Gases A2 level notes LOJ

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1 Data Sheet Extract The theory for ideal gases makes the following assumptions: The gas consists of very small atoms or molecules (spoken of as particles), each of which has an identical mass and are perfectly spherical in shape, and elastic in nature. The number of molecules in the gas is so large that the mathematics of statistics can be applied to it. The molecules are in constant, random motion. The rapidly moving particles constantly collide with each other and with the walls of the container with no preferred direction. This means that we have to ignore the force due to gravity and any electrostatic attraction between them so that the interactions between molecules are negligible. They exert no forces on one another except during collisions. The collisions of gas particles with the walls of the container holding them are perfectly elastic (kinetic energy is conserved). The total volume of the individual gas molecules added up is negligible compared to the volume of the container. Another way of saying this is that the average distance separating the gas particles is relatively large compared to their size. The average kinetic energy of the gas particles depends only on the temperature of the system. Relativistic effects are negligible (relativity comes into play when speed approach the speed of light - ignore this). In other words Newtonian mechanics apply! The time during collision of molecule with the container's wall is negligible as comparable to the time between successive collisions. An ideal gas does not exist - the nearest we have to it is helium remote from its liquefying point. cyberphysics.co.uk 1

2 The kinetic theory of gases (derivation of the equation relating pressure to mean square speed and density) Let s try to explain experimentally some observed properties of gases by considering the motion of the particles (molecules or atoms) which they are made up of. To do that we need to make a number of assumptions: The molecules of a particular gas are identical. Collisions between the molecules and with the container are (perfectly) elastic. That means there is no loss of kinetic energy on collision. The molecules only have kinetic energy (no potential energy) so they exert no forces on each other except during impacts (which are assumed to have negligible duration anyway) and the effect of gravity is ignored so that: o between collisions the molecules move in straight lines at constant speed, and o the motion is random. There are a sufficiently large number of molecules for the mathematics of statistics to be meaningfully applied. The size of the molecules is negligible compared to their separation. They are point particles rather than real atoms and molecules The laws of Newtonian mechanics apply. Some of these assumptions run through the entire analysis; others are used more specifically. up or average them! We need to think about the velocity of the particles. To make the mathematical treatment easier to understand we need to break all of those random velocities into components that are mutually at right angles to each other. That will make it easier later on to add them all Consider a fixed mass of gas enclosed in a cubical container of side L. Let each particle of the gas have mass m and let there be N particles cyberphysics.co.uk 2

3 Let s start with just looking at one of the millions of particles in the gas. Look at the diagram. The particle it represents is one of millions, but we will forget those for the moment! The diagram shows the u component of a single particle moving towards the wall shaded in grey. We will call this particle 1 The x-component of velocity is u 1 towards the wall The x-component of momentum is mu 1 towards the wall. When the particle hits the wall it will bounce back with the same speed but in the opposite direction (because kinetic energy is conserved the collision being elastic!). The x-component of momentum will then be -mu 1 The change in momentum - p - will therefore be mu 1 - (-mu 1 ) = 2mu 1 The particle has to travel a distance 2L (from the grey wall to the opposite face and back again) before it next collides with the grey wall. Now, velocity = distance/time so so the rate of change of momentum due to a collision with a wall will be: By Newton's second law, the rate of change of momentum is equal to force, so mu 1 2 /L is the force exerted on the particle by the wall during the collision. By Newton's third law, the particle exerts an equal but oppositely directed force on the wall, and so we can say that: The force on the wall during the collision = mu 1 2 /L Now it isn t only one particle that hits the wall there are N particles so their x-components of velocity are going to be u 1, u 2,. u N The total force, F, will therefore be = mu 1 2 /L + mu 2 2 /L + mu 3 2 /L.. + mu N 2 /L F = m/ L ( u u u u N 2 ) Now, pressure on that wall will be force/contact area = F/A, but A = L 2 Therefore pressure on the wall the pressure of the gas p = m/ L 3 ( u u u u N 2 ) We can work out the mean square velocity is the x direction from this: cyberphysics.co.uk 3

4 So now we know that the total pressure due to the u-components of the velocities of the particles will be: So far we have only looked at one of the component velocities. It is now time to look at the other two and to add them into the equation. By Pythagoras c 2 = u 2 + v 2 + w 2 So, it follows that this is true for the mean square velocities also. Since there are large numbers of molecules and they are moving randomly the mean square components are equal to each other (random movement means there is no preferred direction). We can therefore say that We can now replace our mean square velocity component with one related to the mean square velocity itself. and and as L 3 = V we get Now, density is mass/volume, Nm is the total mass of gas and L 3 is the volume; we can therefore say that: To investigate the relationship between molecular kinetic energy and temperature From the kinetic theory of gases proof above we have: Let us multiply both sides by V to get an expression for any volume V of gas: cyberphysics.co.uk 4

5 Now So, Where M is the total mass of the gas of volume V. Now, M = Nm ( where m is the mass of one particle in the gas and N is the number of particles in the gas) So A bit of mathematical magic can make that last part of the equation represent kinetic energy. To do that we need a ½ factor before the m to make it fair we have to put a factor of 2 in front of the N: Now the Equation of State or Ideal Gas Equation (which you should know by heart!) is where pv = nrt R = the universal molar gas constant, T = the temperature in kelvin, and n = number of moles of gas We can therefore put: Rearranging the equation we get that: Now, as N/n is the number of molecules per mole, i.e. N A, the Avogadro constant, so we can get rid of the factors that require us to know the details of the gas and replace them with a constant. cyberphysics.co.uk 5

6 Both R and N A are universal constants, and therefore so also is R/N A ; it is called Boltzmann's constant, k and it is the gas constant per particle. The left-hand term is the average translational kinetic energy of a single molecule, and therefore Notes T must be in kelvin remember to convert! The E k is in joules and is for one average particle the question often asks you for the energy of a number of moles of gas. cyberphysics.co.uk 6