Stat_Phys_2nd Chapter.pdf

Statistical Physics Mr. P.R.S.Tissera Department of Physical Sciences & Technology Faculty of Applied Sciences Sabaragamuwa University of Sri Lanka  Mean Free Path  The mean free path is the average distance traveled by molecu...

Statistical Physics Mr. P.R.S.Tissera Department of Physical Sciences & Technology Faculty of Applied Sciences Sabaragamuwa University of Sri Lanka  Mean Free Path  The mean free path is the average distance traveled by molecules between collisions Figure 1 : A molecule traveling through a gas, colliding with other molecules in its path. Of course, the other molecules are themselves moving and experiencing collisions 2  In particular measure the distance it travels between collisions, and repeat the experiment many times  Wish to determine the average distance the molecule travels between collisions  We are thus continuing our efforts to relate macroscopic and microscopic properties of the gas  The average distance traveled by molecules between collisions r   r f r  i i i n ri  f ri    n ri  Where; i 3  First assume that we have no underlying theory of molecular collisions  We form a distribution based only on the results of our many measurements of the distance traveled by a molecule between collisions  After making a large number N of measurements of the distance traveled by a molecule, we sort those distances into B bins of equal widths r 4 Figure 2 : The statistical distribution of the distance traveled by a molecule between collisions. The mean value rҧ gives the mean free path of the molecules 5  Macroscopic Calculation of the Mean Free Path  Cause a beam of molecules to be incident at a rate or intensity I0 on a thin layer of gas  Measure the rate or intensity I at which the molecules emerge after passing through a thickness r of the gas Figure 3 : A beam of molecules is incident on a thin layer of gas of thickness r 6  The intensity I of the beam passing through the layer is a measure of the number of molecules of the beam that experience no collisions over the distance r  Assume that no molecule in the beam is scattered more than once by a target molecule  This assumption is justified if the density of target molecules is not too great  Now increase the thickness of the layer by an amount dr and find the resulting change dI in the intensity I 7  Expect that the decrease in intensity is proportional to I and to the additional thickness dr of the gas layer  Thus, dI   c I dr dI   c dr I I r dI  I0 I   c  dr 0 I r   I 0 e  c r I ln   cr I0 8  The calculation suggests an exponential form for f(r) f r   A e  c r r   r f r  i i i  Since rҧ is the mean free path 𝜆, we can rewrite the above equation    r f r  i i i    r f r  i i i  r nr  i i   i  nr  i i 9  Since n ri   N f ri   r N f r   r f r   r i i i i   i  i  N f r  i  f r   r i i i  Now make the widths r very small, so that we can write the above equation    r f r  dr  r A e  c r dr 1   0   0   c  f r  dr  A e  c r dr 0 0 10  Probability distribution can be written as; r  f r   A e   The total probability having the distance between collisions over 0 to ∞ should be unity   r   f r  dr  A  e  dr  1 0 0 r    e 1 A 0 1 A   1      r  f r   1   Thus the probability distribution is; e  11  Microscopic Calculation of the Mean Free Path  The mean free path is related to the size of the molecules and to their number per unit volume  Consider the molecules of gas to be sphere of diameter d  A collision will take place when the centers of two molecules approach within a distance d of one another  As equivalent description of collisions made by any one molecular is to regard that molecule as having a diameter 2d and all other molecules as point particles 12 Figure 4 : (a) A collision occurs when the centers of two molecules come within a distance d of each other (b) As equivalent but more convenient representation is to think of the moving molecule as having diameter 2d, all other molecules being points 13  Imagine a typical molecule of equivalent diameter 2d moving with speed v through a gas of equivalent point particles Figure 5 : A molecule with an equivalent diameter 2d traveling with speed v sweep out a cylinder of base area d2 and length vt in a time t 14  In time t our molecule sweeps out a cylinder of cross - sectional area = πd2 of length = vt    The number of particles contain in this cylinder =  d 2 v t   n Where n  number of molecules per unit volume  Since our molecule (equivalent diameter 2d moving with speed v) and the point particles do exerted forces on each other, this is also the number of collisions experienced by the molecule in time t  Number of collisions in time t =  d  v t   2 n 15  The distance moved by our molecule to have such collisions = v t  The mean free path  is the average distance traveled by molecules between collisions vt      d 2 v t   n 1    d2 n  This equation is based on the picture of a molecule hitting stationery targets. Actually the molecule hits moving targets 16  Let moving targets are right angle to our molecule as shown V (A, B) = V (A, E) + V (E, B) V (A, B)  v relative  v2  v2  2v 17  The mean free path vt      d 2 v relative t   n vt    d   2  2 vt n 1   2  d2 n 18  Average rate of collision  From kinetic theory 1 PV m N C2 3 3P V 3n R T 3R T 3R T C2     mN m NAn NAm M 3R T C 2  v rms  M v rms  The average rate collision rate   19  Example The molecule diameters of different kinds of gas molecules can be found experimentally by measuring the rates at which different gases diffuse into each other. For nitrogen, d = 3.15 x 1010 m has been reported. What are the mean free path and the average rate of collision for nitrogen at room temperature ( T = 300 K) and at atmospheric pressure? 20 ANSWER 21 22 23 End of the Chapter 02 24