The Kinetic Model of Gases

Transcription

1 Franziska Hofann, Stephan Steinann February 0, 2007 The Kinetic Model of Gases Introduction The rando otion of perfect gases can be described by kinetic theory. Based on a siple odel pressure, igration, diusion and even soe reaction rates can be calculated fro basic physical properties such as the teperature of the gas, it's olecular weight and soe constants, e.g. the Boltzann constant. The kinetic odel of a perfect gas is based on the following assuptions []: The gas consits of discrete particles. The size of the particles is sall copared with the average distance between the and the diension of the container. The particles are taken to be rigid spheres and not to interact with each other except in copletely elastic collisions. The particles are in ceaseless rando otion. The energy distribution in the gas is given by a Boltzann distribution (Equation ) The Boltzann distribution is given by: n i N = e ɛ i ɛ i i e () Where n i is the nuber of particles with the energy ɛ i, N the total nuber of particles, k Boltzann's constant and T the absolute teperature. The Boltzann distribution describes the ost probable energy distribution of indistinguishable, independent particles with a given total energy. In this experient, a "two diensional gas" is studied. The subjects of interesst are: The variation of "pressure" with the hight, which should follow the baroetric forula: 2 p(h) = p(0)e gh (2) This can also be expresse through the nuber of particles (n): n = e gh n 0 (3) The distribution of speeds, which is expected to follow the Maxwell-Boltzann distribution of speeds: 3 f(v) = v That eans, the total translational kinetic energy of the particles is conserved. 2 The derivation can be found on page 6 3 The derivation can be found on page 7 e 2 v2 (4)

2 2 Experiental Part A quadratic table (08 c) with holes, where air is blown through and equiped with a shaking frae and 20 disks (=30.6 g) is used as a odel of a two diensional gas. The table is segented in eleven segents of about 0 c. The table is tilted by The shaking frequency is chosen to be Hz and.5 Hz. For each frequency the position of the disks is evaluated twenty ties by switching the strea of air o and counting the disks in each segent. In order to get the speeds of the disks, pictures of the disks hovering over the at, shaking table are taken with a digital caera and an exposure tie of s. Ten pictures for each frequency ( Hz and.5 Hz) are being evaluated, the velocities are rounded to c/s. 3 Results The raw data of the results is given in Appendix and Figure : Fit of the data for Hz according to Equation 3 Figure 2: Fit of the data for.5 Hz according to Equation Figure 3: Plot of the logariths of the nubers of disks at level h at Hz. Figure 4: Plot of the logariths of the nubers of disks at level h at.5 Hz. 2

3 3. Baroetric Forula The baroetric forula of Equation 3 is used, by setting n 0 to the nuber of the lowest level and the height of this level to 0. The subesequent levels have (because of the tilt) a height of 0.n sin(0.89 ) with n =, 2, 3... The two graphs are shown in Figure and 2. In addition, the data is plotted according to the logarithic for of Equation 3 4. In this case p(h) = n(h) and h = ( n)sin(0.89 ) with n = 0,, 2... is used. This equation has two paraeters (p(0) and T), hence, the data can be tted better to the forula. These plots are given in Figure 3 and 4. Fro the paraeter of the regression (A or the solpe), one can calculate the teperture according to: T = g k A For Hz, this gives (2.7±0.3)0 9 K or (4.5±0.9)0 9 K For.5 Hz, this gives (7.2±0.7)0 9 K or (3±2)0 9 K The two teperatures are those corresponding to the linear regresson and the exponential t. 3.2 Maxwell-Boltzann Distribution of Speeds The data is treated as follows: The relative nuber of each velocity is calculated and is onitored against it's velocity, as Equation 4 proposes Figure 5: Maxwell-Boltzann distribution of speeds at Hz. The red line gives the tted distribution of speeds according to Equation Mean, ost probable and root ean square speed The ean speed is given by: v = 4 See 0 on page 7 0 f(v) v dv 3

4 Figure 6: Maxwell-Boltzann distribution of speeds at.5 Hz. The red line gives the tted distribution of speeds according to Equation 4. The ost probable speed is evaluated by: ( ) f(v) v v=v! = 0 And the root ean square speed is calculated according to: v rs = f(v) v 2 dv By using f(v) fro Equation 4 it follows that: π v = v = 2 0 v rs = 4 The calculated values for the dierent teperatures are given in Table. Method fro T/(0 9 K) v /(/s) v/(/s) v rs /(/s) Figure Figure Figure Figure Figure Figure Table : The ost probable (v ), ean ( v) and root ean square (v rs ) speeds at dierent teperatures for the disks. 4

5 4 Discussion The used echanical odel for a gas has soe shortcoings: The otion is not absolutely frictionless, therefore the particles are always "heated up" at the walls. It's like olecules in a hot container, before the equilibriu is established. To iniise this eect, the container is relatively sall copared with the particles which is contrarely to the assuptions of the kinetic odel of gases. Collisions ay happen too often, especially with the (hot) walls. Consequently, in the experient for the baroetric forula, the assuption of an isotheric syste is not so good fulllled, since the lowest level is the hottest and therefore is probably less populated, this can be seen for exaple in Figure 4. The logarithic plots see to result in too low teperatures, as a coparison of the calculated velocities with the experiental values shows. The exponential t-teperature on the other hand is too high, which could be related to the suggestion above, that the lowest level has a higher teperature than the others and by a t according to Equation 3 the nuber of disks in the lowest level (n 0 ), is ore iportant than in the linear t. Fro nearly all the gures one can see, that there are quite enorous uncertenties of easureent: The high variability is seen at once. The relative errors are all around 0% as the tting progra states. Nontheless the overall shape of the distribution is not too bad represented. The resulting teperatures are very high. But we think, this has to be like this. A teperature, at which acroskopic objects like disks have enough kinetic energy to overcoe the potential energy in a way, that they do not siply fall down to earth and rest there, but do follow a Boltzann distribution, such a teperature has to be (absurdly) 5 high. 4. The Pressure Variation according to the Baroetric Forula The baroetric forula (Equation 3) shows a strong dependence on the weight of the particles, i.e. on the coponents in the gas: The higher the weight, the steeper the function. This can be rationalised by the fact, that the force downward is proportional to the weight. This eect is visualised in Figure 7. The other variable is the teperature: The function is inversely proportional to the teperture, i.e. the lower the teperature, the steeper the function. This can be explaind qualitatively by the fact, that the teperatur is directly related to the kinetic energy of the particles and at lower kinetic energies, the potential energy is not as readily available. This behaviour is shown in Figure 8. For the pressure dierence between 5'000 and 0'000 it would be sensible to use a teperature around -40 C or 230 K. As a teperature prole of the atosphere shows, the teperature varies ore or less linearly fro about -20 C to nearly -60 C fro 5'000 to 0'000. [2] Figure 7: The shape of p(h) functions for atospheres consisting of H 2, Air, Ar, Xe at 25 C. Figure 8: The eect of the teperature on the shape of p(h). 5 No acroscopic object is supposed to be stable at 0 9 K. 5

6 4.2 For which constant teperature do we reach a pressure dierence of 20'000 Pa by changing the altitude fro 0 to 2'000? p p(0) p p(0) ( ln p ) p(0) p = p(0) p(h) = p(0) ( e Mgh RT ) = e Mgh RT = e Mgh RT = Mgh RT Mgh T = ( ) R ln p p(0) With p= 20'000 Pa, p(0) = 0 5 Pa, h = 2'000 and an atosphere of air (M 29 g/ol) this results in: T = ln ( )K = 307 K = 34 C Derivation of the Baroetric Forula for a Two Diensional Gas [3] The density for the two diensional case is given by: ρ 2D = A (5) The pressure of a two diensional gas can be dened as: p 2D = F l = a l In the two diensional case, the perfect gas equation can be written as: (6) Fro coparison of Equation 5 with Equation 7 it results: p 2D A = nrt (7) ρ 2D = p 2D nrt = const. Of course, this expression is only valid, if the teperature is constant. In reality, the atosphere is not isotheric, therefore, the baroetric forula is only a siplicated relation. With the olecular weight M= n it follows: Fro height h to h + dh the pressure varies as: ρ 2D (h) = M RT p 2D (h) (8) dp 2D = a ρ 2D (h)dh (9) That's because the pressure varies by a d l (the inus sign indicates, that the acceleration points in the opposite direction of h). d can be expressed by Equation 5 as ρ 2D da. da equals l dh. By substituting Equation 8 in Equation 9 the following dierential equation results: dp 2D = a M RT p 2D (h) dh 6

7 Separation of the variables produces: dp 2D p 2D (h) = a M RT dh Integration 6 and accounting for the boundary condition of ln (p 2D (h)) for h=0 is equal to ln(p 2D (0)): ln(p 2D (h)) = a M RT h + C = a M RT h + ln(p 2D (0)) (0) This can be rearranged to the baroetric forula for a two diensional gas: p 2D (h) = p 2D (0)e a M RT h One can explain that Equation 3 is a special case of the Boltzann distribution (Equation ) as follows: The exponential part of the function is undoubtedly the sae, since the potential energy of the particles in the height h can be written by gh. Because M = N A and R = kn A the fraction can be thought to be expanded by Avogadro's constant. According to pv=nrt p(h) corresponds to n h and p(0) to n 0. Then Equation gives: n 0 N = e gh ɛ i i e h=0 = ɛ i i e () n N = e gh ɛ i i e (2) Division of Equation 2 by Equation : n N N = e n 0 gh ɛ i e i ɛ i i e = n = e gh n 0 p(h) = p(0)e gh 4.4 Derivation of the Maxwell-Boltzann Distribution of Speeds for a Two Diensional Gas [4] The translational kinetic energy of a two diensional gas can be expressed by: E = 2 v2 x + 2 v2 y (3) Fro the Boltzann distribution we know, that the probability distribution has to follow a exponential relation of the general for: f = Ke E (4) Substitution of Equation 3 in Equation 4 gives: f = Ke 2 v2 x + 2 v2 y = Ke 2 v2 x e 2 v2 y = K f(v x )f(v y ) Every olecule has to have a velocity in x- and in y direction, which eans: K f(v x )f(v y ) dx dy = K e 2 v2 x dx e 2 v2 y 2π 2π! dy = K = Therefore K = f = 2π 2π e 2 v2 x e 2 v2 y (5) However, the distribution of speeds splitted up into the dierent directions is not the nal point of interesst, one would rather like to know the probability for a given velocity v, with v 2 = vx 2 + vy. 2 This 6 The acceleration a is taken to be a constant, though this is strictly not true. However the variation is only very sall. 7

8 corresponds to a function of a circle around the origin with a radius of v in a coordinate syste with the axes v x and v y. In order to get the probability of the velocity v, all the probabilities of the vectors with the lenght of v have to be sued up. Since the probability for a given v is independent of v x and v y (because they have to obey the function of a circle) the total probability of a velocity v can be calculated by ultiplying the probability of a specic case (given by Equation 5) by the lenght of the circle: f(v) = f 2π vx 2 + vy 2 = 2πv f = 2πv 2π e 2 v2 x e 2 v2 y = v e 2 v2 References [] Wedler, G. Lehrbuch der Physikalischen Cheie; 3rd ed.; 988. [2] Barthazy, E. Teperatur der Atosphäre; [3] Rudin, H. Physik I, Leitfaden zur Einführungsvorlesung; 992. [4] Atkins, P.; de Paula, J. Physical Cheistry; 7th ed.;