ADIABATIC PROCESS Q = 0

Transcription

1 THE KINETIC THEORY OF GASES Mono-atomic Fig.1 1

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3 3 Average kinetic energy of a single particle Fig.2 INTERNAL ENERGY U and EQUATION OF STATE For a mono-atomic gas, we will assume that the total energy U is equal to the number of atoms multiplied by the average kinetic energy of each. (We are disregarding any possibility of excitation or motion inside the atoms itself. The kinetic energy of the center of mass motion is all the energy there is in the atom.) U = N <½ m v2> (2) The kinetic theory of gases thus gives as an equation of state for the gas PV = (2/3) U To gain some generality, we express the expression above as PV = (γ - 1) U (γ = 5/3 for the case of a mono-atomic gas) For non mono-atomic gases γ is different than 5/3. (3)

4 Once the equation of state of a system is known, different thermodynamic processes can be studies in detail. Here let s consider an adiabatic process ADIABATIC PROCESS Q = 0 As an application of the expression above for the equation of state, let s consider an adiabatic process, one in which the gas container is thermally well isolated, i.e. no heat-transfer occurs. No heat-transfer is added or removed to the gas, Q = 0 As the piston is pressed, work is done on the gas and the internal energy increases. We would like to predict the changes in the gas volume as a consequence of changes in pressure. Q = 0 Insulation Pressure P F External force Fig. 3 Gas isolated from heat transfer. Using the first law of thermodynamics, du = Q - W, we obtain U = - W When the volume changes by dv, the work done by the gas W is PdV. Therefore, U = - (P dv ) (4) On the other hand, according to the equation of state PV= ( -1)U, which gives, du = d [PV / ( -1) ] = [ 1/ ( -1) ] [P dv + V dp ] (5)

5 From (4) and (5), - (P dv ) = [1/ ( -1)] [P dv + V dp ] - ( -1) (P dv ) = P dv + V dp - (P dv ) = VdP Dividing by PV - (dv) /V = (dp) / P - lnv + const = ln P const = ln V + ln P = ln (P V This implies, P V = Const (adiabatic process) (6) ( = 5/3 for the case of a mono-atomic gas.) Q = 0 Insulation Pressure P F P adiabatic process V Fig. 4 Gas undergoing an adiabatic process. V

6 What happens at THERMAL EQUILIBRIUM conditions Case: Two different gases in the same container Consider a gas mixture of two types of molecules, 1 and 2, If molecules of type-1 were standing still, they would be speed by their collision with the faster moving molecules of type-2. If the type-1 molecules were to pick up too much speed, then, again through collisions, they would pass the energy back to the molecules of type-1. The question is to find the rule that determines the relative speed of the two types of molecules. Insulation m 1 m 2 Fig. 5 A mixture of molecules of different masses. As demonstrated in the appendix to these notes, the rule is, < ½ m 1 v 1 2 > = < ½ m2 v 2 2 > (7) (the heavier atoms move slower than the lighter ones.) Case: Two different gases in different containers Figure 6 shows two different gases, initially occupying equal volumes. A movable piston separate the gases. What is the relative speed of the molecules once a thermal equilibrium is reached?

7 Insulation 1 2 Movable piston Fig. 6 Initially, we could argue that the final equilibrium state is reached when the pressure at both side is the same. According to expression (1), P = (2/3) (N/V) <½ mv 2 >; thus a condition of MECHANICAL EQUILIBRIUM, would imply, (2/3) (N 1 /V 1 ) <½ m 1 v 2 1 > = (2/3) (N 2 /V 2 ) <½ m 2 v 2 2 > It turns out, as argued in the Feynman lectures Vol-1, chapter 39), this condition is not sufficient. Something else must happen. The condition that must be satisfied is, <½ m 1 v 2 1 > = <½ m 2 v 2 2 > In the two cases mentioned above, the mean kinetic energy of the molecules turns out to be a property of the thermal equilibrium condition, and not of the particular gas.!!! That is, the notion that we had about the gas reaching a common temperature after letting them to interact for awhile mean that the gas reaches a situation in which the average kinetic energy per molecule is the same. Definition of temperature We have found that the average kinetic energy is a property of the temperature and not of the particular gases used. This

8 suggests using the average kinetic energy itself as the temperature. Indeed, the simplest approach could be to call the average energy itself the temperature. The higher the average kinetic energy the higher the temperature. This would constitute our absolute temperature scale. [This possibility is not at all disconnected form the discussion of defining an absolute temperature scale within the context of discussing the concept of entropy, as will be seen in the next chapter.] Unfortunately, the scale of temperature has been chosen differently. A constant conversion factor k is introduced between the average kinetic energy of a molecule < ½ mv 2 > and the, thus defined, absolute temperature scale T. < ½ mv 2 > 3/2 k T Definition of (8) absolute temperature The unit of the absolute temperature T is denominated the Kelvin (K) k = 1.38 x J/K The Ideal Gas Law Now we can put the new definition of temperature into Eq (1). The expression PV = (2/3) U = (2/3) N < ½ m v 2 > becomes, PV = N k T (9) This result indicates, equal volumes of different gases, at the same pressure and temperature, have the same number of molecules, because of Newton s laws. In terms of moles n = N/N o (where N o 6.02 x ), expression (9) become, PV = n N o k T. Making N o k = Jmol -1 K -1 =R, one obtains,

9 PV = n R T (10) Question: Is expression (9) still valid when considering diatomic molecules? First, let s put expression (9) again in terms of the average kinetic energy per molecule, PV = (2/3) U = (2/3) N < ½ m v 2 > (mono-atomic gas) Notice that, during its derivation we were invoking the collision of mono-atomic molecules of mass m that and translation motion of speed v against a wall. If the molecules were, for example, diatomic, what would its internal motion (i.e. vibrational oscillations between the two atoms) affect its translational motion? Apparently none, since all what matter is the net impact against the walls of the container (what the molecule does in between collisions does not affect the pressure the molecules exert on the walls. That pressure results only from the collisions of the molecules against the walls.) Since the translational motion of the molecules is characterized by the velocity of its center of mass v CM, if we repeat the derivation of the kinetic theory of the gases, we will obtain, PV = (2/3) N < ½ M v CM 2 > (di-atomic gas) (11) where M is the mass of the di-atomic gas.)

10 An extra step (argued in the Feynman Lectures on Physics, Vol-1, Chapter 39) lead to the following result. < ½ M v CM 2 > = (3/2) kt (12) Hence, from (12) and (11), PV = N k T (gas composed of (13) di-atomic molecules) Internal energy of a gas composed of di-atomic molecules m 1 m 2 We have learned that the In this case, the atoms are Condition of thermal joined equilibrium implies <½ m 1 v 2 1 > = <½ m 2 v 2 2 > What is the condition of thermal equilibrium in this case? Answer: <½ m 1 v 1 2 > = <½ m 2 v 2 2 > That is, whether or not are forces involved, the average kinetic energy of an atom is the same (it depends only on the temperature.)

11 Summary (diatomic molecules) One consideration <½ m 1 v 1 2 > = <½ m 2 v 2 2 > = (3/2) kt Total kinetic energy = <½ m 1 v 1 2 > + <½ m 2 v 2 2 > = 3 kt Alternative consideration r CM K CM = < ½ M v CM 2 > = (3/2) kt Total kinetic energy = K CM + K internal-motion Since the total kinetic energy = 3kT then the above two results imply, K internal-motion = (3/2) kt Internal motion

12 K internal-motion = 2 ( ½ k T) + 1 ( ½ k T) = (3/2) kt

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