KINETIC THEORY OF GASES

     KINETIC THEORY OF GASES

brief summary

Bernoulli's Picture: Daniel Bernoulli, in 1738, was the primary to know atmospheric pressure from a molecular point of view. He drew an image of a vertical cylinder, closed at rock bottom , with a piston at the highest , the piston having a weight thereon , both piston and weight being supported by the atmospheric pressure inside the cylinder. He described what went on inside the cylinder as follows: “let the cavity contain very minute corpuscles, which are driven hither and thither with a really rapid motion; in order that these corpuscles, once they strike against the piston and sustain it by their repeated impacts, form an elastic fluid which can expand of itself if the load is removed or diminished…” Sad to report, his insight, although essentially correct, wasn't widely accepted. Most scientists believed that the molecules during a gas stayed more or less in situ , repelling one another from a distance, held somehow within the ether. Newton had shown that PV= constant followed if the repulsion were inverse-square. In fact, within the 1820’s an Englishman, John Herapath, derived the connection between pressure and molecular speed given below, and tried to urge it published by the Royal Society . it had been rejected by the president, Humphry Davy, who acknowledged that equating temperature with motion, as Herapath did, implied that there would be an temperature of temperature, a thought Davy was reluctant to simply accept . And it should be added that no-one had the slightest idea how big atoms and molecules were, although Avogadro had conjectured that equal volumes of various gases at an equivalent temperature and pressure contained equal numbers of molecules — his famous number — neither he nor anyone else knew what that number was, only that it had been pretty big. The Link between Molecular Energy and Pressure It is not difficult to increase Bernoulli’s picture to a quantitative description, relating the pressure to the molecular velocities. As a warm up exercise, allow us to consider one perfectly elastic particle, of mass m, bouncing rapidly back and forth at speed v inside a narrow cylinder of length L with a piston at one end, so all motion is along an equivalent line. Obviously, the piston doesn’t feel a smooth continuous force, but a series of equally spaced impacts. However, if the piston is far heavier than the particle, this may have an equivalent effect as a smooth force over times long compared with the interval between impacts. So what's the worth of the equivalent smooth force? Using Newton’s law within the form force = rate of change of momentum, we see that the particle’s momentum changes by 2mv whenever it hits the piston. The time between hits is 2L/v, therefore the frequency of hits is v/2L per second. this suggests that if there have been no balancing force, by conservation of momentum the particle would cause the momentum of the piston to vary by 2mv×v/2L units in each second. this is often the speed of change of momentum, then must be adequate to the balancing force, which is therefore F=mv2/L. We now generalize to the case of the many particles bouncing around inside an oblong box, of length L within the x -direction (which is along a foothold of the box). the entire force on the side of area A perpendicular to the x -direction is simply a sum of single particle terms, the relevant velocity being the component of the speed within the x -direction. The pressure is simply the force per unit area, P=F/A. in fact , we don’t know what the velocities of the particles are in an actual gas, but it seems that we don’t need the small print . If we sum N contributions, one from each particle within the box, each contribution proportional to v2x for that particle, the sum just gives us N times the typical value of v2x. that's to mention , P=F/A where there are N particles during a box of volume V. Next we note that the particles are equally likely to be occupation any direction, therefore the average value of v2x must be an equivalent as that of v2y or v2z, and since v2=v2x+v2y+v2z, This is a surprisingly simple result! The macroscopic pressure of a gas relates on to the typical K.E. per molecule. Of course, we've not considered possible complications caused by interactions between particles, but actually for gases like air at temperature these interactions are very small. Furthermore, it's well established experimentally that the majority gases satisfy the Gas Law over a good temperature range: PV=nRT Introducing Boltzmann’s constant kB=R/NA, it's easy to see from our result for the pressure and therefore the perfect gas law that the typical molecular K.E. is proportional to absolute the temperature. Maxwell finds the speed Distribution By the 1850’s, various difficulties with the prevailing theories of warmth , like the caloric theory, caused some rethinking, and other people took another check out the kinetic theory of gases of Bernoulli, but little real progress was made until Maxwell attacked the matter in 1859. Maxwell worked with Bernoulli’s picture, that the atoms or molecules during a gas were perfectly elastic particles, obeying Newton’s laws, bouncing off one another (and the edges of the container) with straight-line trajectories in between collisions. (Actually, there's some inelasticity within the collisions with the sides—the bouncing molecule can excite or deexcite vibrations within the wall, this is often how the gas and container come to equilibrium .) Maxwell realized that it had been completely hopeless to undertake to research this technique using Newton’s laws, albeit it might be wiped out principle, there have been far too many variables to start writing down equations. On the opposite hand, a totally detailed description of how each molecule moved wasn't really needed anyway. What was needed was some understanding of how this microscopic picture connected with the macroscopic properties, which represented averages over huge numbers of molecules.