Stat_Phys_4th Chapter.pdf

Statistical Physics Mr. P.R.S.Tissera Department of Physical Sciences & Technology Faculty of Applied Sciences Sabaragamuwa University of Sri Lanka Macroscopic Property of a Gas and the Ideal Gas Law Figure 1 : Gas confined to...

Statistical Physics Mr. P.R.S.Tissera Department of Physical Sciences & Technology Faculty of Applied Sciences Sabaragamuwa University of Sri Lanka Macroscopic Property of a Gas and the Ideal Gas Law Figure 1 : Gas confined to a cylinder fitted with a movable piston 2  Suppose the gas to be in contact with an idealized device called a thermal reservoir, which we can regard as a body maintained at a temperature T, such that the temperature of the reservoir does not change when our gas cylinder comes into thermal equilibrium with it  If we wish to change the pressure p, we add or subtract weights on the piston  The space above the piston is assumed to be evacuated, so that there is no air pressure pushing down on the piston  The volume V can be changed simply by changing the position of the piston  The amount of gas might be changed by allowing gas to enter the chamber, thereby changing the number of molecules N 3  Dependence of V on N  Keep the temperature and the pressure constant  Allow gas to enter or leave the chamber  Measure the resulting volume V by observing height of the piston  Assume that we know the mass of each molecule and the total mass of gas that is present in the cylinder, thus we can determine N, the total number of molecules Figure 2 : Curve of V vs N 4  The data points appear to follow a straight line Conclusion : There is a direct proportion between V and N  Further more, by replacing the gas in the cylinder with an equal number of molecules of a different gas at the same pressure and temperature, the new gas occupies the same volume Conclusion: The volume occupied by a gas at a particular pressure and temperature is independent of the type of the gas or the size or mass of its molecules; the volume depends only on the number of molecules V C N p, T constant  This equation is some times known as Avogadro’s Law 5  Dependence of V on p  Keep the number of particles N and the temperature T constant  Change the pressure p  Measure the resulting volume V Figure 3 : Curve of V vs P 6 Conclusion:  The volume V occupied by the gas appears to depend inversely on the pressure p, with the temperature and the number of particles held constant p  C 1 1 N, T constant V  This equation is called Boyle’s Law which states that if the number of particles and the temperature is kept constant, then the pressure of the gas is inversely proportional to its volume 7  Dependence of V on T  Keep pressure p and number of particles N constant  Vary the temperature T  Measure the resulting volume V Figure 4 : Curve of V vs T 8 Conclusion:  The volume V occupied by the gas varies linearly with the temperature T, when the pressure and the number of molecules are held constant V  C11 T p, N constant  This equation is called Charles’s Law or GayLussac’s Law which states that if the number of particles and the pressure is kept constant, then the temperature of the gas is directly proportional to its volume 9  Equation of State  Combine three equations in Avogadro’s, Boyle’s and Charles’s Laws into a single equation that includes all three of the observed relationships, as follows: pV  k k - Boltzmann constant = 1.38066 x 1023 J NT K1  It can be shown that using the above equation, we can obtain Avogadro’s, Boyle’s and Charles’s Laws kT k N V    N  C N p, T constant  V    T  C11 T p, N constant   p   p  p  k N T  N, T constant  1 1  C1 V V 10  pV  k pV  k NT NT N The number of moles, n  NA p V  n NA k T pV  nR T Where; R  N A k  This equation is called the Ideal Gas Law   R  6.023 x 1023 mol 1 1.38066x 1023 J K 1   8.3145 J K 1 mol1 11  Example An insulated cylinder fitted with a piston contains oxygen at a temperature of 20C and a pressure of 15 atm in a volume of 22 liters. The piston is lowered, decreasing the volume of the gas to 16 liters, and simultaneously the temperature is raised to 25C. Assuming oxygen to behave like an ideal gas under these conditions, what is the final pressure of the gas? 12 Answer 13  The Ideal Gas Model  A model is a simplified version of the system that permits calculations to be made but still yields physical insight  A model might begin with a set of simplifying assumptions that permit the system to be analyzed using an existing set of laws  Because the model is a simplification of nature, the final result is generally not a true or complete description of nature, but if we have been cleaver at forming the model, the final result may prove to be a very good approximation of the behavior of the system 14  A gas confined in a container is an example of a complex system that is difficult to analyze using Newton’s law  The molecules can collide inelastically, and the energy of the collision can be absorbed by the molecules as internal energy in a variety of ways  Keeping track of these processes for all the molecules would be a project of hopeless complexity  We simplify this problem by inventing a model that describes the microscopic properties of the real gas  This model, which we call the ideal gas model, proves to be entirely consistent with the concept of the ideal gas that we developed experimentally 15 Assumptions made in ideal gas model 1. A gas consists of particles, called molecules - If the gas is an element or a compound and is in a stable state, we consider all its molecules to be identical 2. The molecules are in random motion and obey Newton’s law of motion 3. The total number of molecules is large 4. The volume of the molecules is a negligibly small fraction of the volume occupied by the gas 5. No appreciable force act on the molecules except during a collision 6. Collisions are elastic and of negligible duration 16 Kinetic Calculation of the Pressure Figure 5 : A cubical box of edge L containing an ideal gas. A molecule of the gas is shown moving with velocity v toward side A1 17  The change in the particle’s momentum Final momentum  Initial momentum =  mvx  ( mvx ) =  2 mvx  Because the total momentum is conserved in the collision, the momentum imparted to A1 is 2 mvx  Suppose that this particle reaches A2 without striking any other particle on the way  The time required to cross the cube = L vx  At A2, it again has its x component of velocity reversed and return to A1. Assuming no collisions with other molecules, 2L  The time taken by the round trip = vx 18  The average impulsive force exerted by this molecule on A1 2 m vx m v 2x Fx   2L L    vx   The total force by all molecules on A1 m v 2x1  v 2x2  v 2x3  v 2x4 ................  F  L  The pressure on A1 m v 2x1  v 2x2  v 2x3  v 2x4 ................  p  L3 19   If N is the total number of particles in the container, m N  m Then the density of gas    L3 N L3  The pressure on A1  v 2x1  v 2x2  v 2x3  v 2x4 ................  p    v 2x N  For any particle, v 2  v 2x  v 2y  v 2z  We have many particles and also they are moving entirely at random motion 1 2 Therefore, it can be concluded that v  v  v 2 x 2 y 2 z  v 3 1  The pressure on A1 p   v2 3 20   From this equation, we can calculate the root mean square speed 3P v rms  v 2   EXAMPLE: Calculate the root mean square speed of hydrogen molecules at 0.00C and 1.00 atm pressure, assuming hydrogen to be an ideal gas. Under these conditions, hydrogen has a density  of 8.99 x 102 kg m3. ANSWER: v rms  3 P   3 1.01 x 105 Pa   1838 ms 1  8.99 x 10  2 Kg m 3 21 Table 2 : Some Molecular Speeds at Room Temperature (300 K) Molar mass M vrms (m/s) Translational Kinetic Gas (g/mol) Energy per Mole (J/mol) Hydrogen 2.0 1920 3720 Helium 4.0 1370 3750 Water vapour 18.0 645 3740 Nitrogen 28.0 517 3740 Oxygen 32.0 483 3730 Carbon dioxide 44.0 412 3730 Sulfur dioxide 64.1 342 3750 22 Kinetic Interpretation of the Temperature 1 p   v2 3 1 pV   V v2 3 nM But   V 1  pV  n M v2 3  The total translational kinetic energy of the gas 1 2  m v 12  v 22  v 32  v 24  v 52 .............   1 2  m N v2  23 But mN = nM The total translational kinetic energy of the gas  1 2  n M v2  1 2 1 2 pV  n M v2   nM v  3 3 2  2 1 2 n R T   nM v  3 2   Thus the average translational kinetic energy per mole is 1 3 M v2  R T 2 2  We may consider the above equation as a connection between a macroscopic property; temperature, and a microscopic property; the kinetic energy of a molecule 24  The average translational kinetic energy of a molecular is 1 3 R 3 m v  2 T  k T 2 2 NA 2  The root mean square speed 3k T v rms  v 2  m  Thus, the ratio of the root mean square speed of molecules of two different gases at a particular temperature v rms 1 m2  v rms 2 m1 25  The ratio of the number of molecules of the two gases that pass through the porous walls in a short time interval is defined as the separation factor   The separation factor v rms1 m2 M2     v rms 2 m1 M1  The diffusion process through porous walls is one method used to separate the atoms of an element by mass into its different isotopes 26 EXAMPLE 27 ANSWER 28 End of the Chapter 04 29