The Ideal Gas: From Kinetic Theory and Boltzmann's Law 

Ideal Gas Law from Kinetic TheoryFrom the above study, we found that the average kinetic energy is proportional to the temperature, according to where the angular brackets denote the average value. From this, we can calculate the average momentum imparted to a wall of the container (suppose it is a cubic box with side lengths L) surrounding the ideal gas. Specifically, an average of onethird of this kinetic energy will be directed at a wall in a given direction, say perpendicular to the xaxis, (since v^{2} = v_{x}^{2} + v_{y}^{2} + v_{z}^{2}). Thus we have < mv_{x}^{2}> = k_{B} T. The force exerted by one particle collision is given by the momentum impulse: where in the last equality, is the change in xmomentum of the particle, and is the time between colisions by that particle. Taking average values, we find that = 2 <mv_{x}> (the factor 2 occurs because the momentum changes from mv_{x} to mv_{x} upon elastically bouncing off the wall), and = 2 L / <v_{x}> (this second factor of 2 occurs because the particle needs to make a round trip across the box length L). This gives the average force imparted by a collision to be F_{x }= <mv_{x}^{2}> / L, and so the pressure due to all N particles, on the wall of area A, is Since the gas constant is just R = N_{A} k_{B} , where N_{A} is Avogadro's number, we have the ideal gas law P V = n R T where n is the number of moles of gas particles. Java applets showing the kinetic origin of the ideal gas lawRun the following two applets simulating gas particles in a container: For this 1dimensional simulation, how does moving the piston convert external work to an increase in the gas kinetic energy (temperature)? In this simulation, we have precise control over the piston motion. Is it possible to reduce the volume of the chamber without heating up the gas? Conversely, can we expand the volume of the chamber without cooling down the gas? Try to do this. Does this violate the ideal gas law? Does this violate the first law of thermodynamics? (Hint: consider how much work is being done, and whether the motion is adiabatic.) This is for you to play around with. What experiment will you try? Follow the given directions, and answer the questions about the relationship between N, P, v, and V. Write your answers, experiments tried, observations and ideas in your lab notebook or on a lab writeup. 



Boltzmann's LawThe MB distribution is a specific case of a more general law that states that the distribution of states with different energies is given by which is also called the Boltzmann distribution. Since above for the ideal gas, this reduces to the MB distribution in that case. This distribution can be derived from the idea that the ratio of probabilities between two states of energy E_{1} and E_{2}, P(E_{1})/P(E_{2}) = function of (E_{1}  E_{2}) only = f(E_{1}  E_{2}). (See Chapter 17.1 in McQuarrie.) This last assumption simply states that only relative energies matterthe absolute zero of energy can have no physical meaning. From this, we can also have a simple interpretation of the fundamental law of statistical mechanics, namely Boltzmann's formula where is the number of states corresponding to a given energy. The connection can be made by using the thermodynamic expression for the free energy A = E  TS, which generalizes energy to include thermal effects. Now this shows that TS plays the role of an energy, or more precisely, a free energy. Using this in P(E), we find P(state) = exp((ETS)/k_{B}T) = exp(E/k_{B}T) exp(ln) = exp(E/k_{B}T) In other words, when we have states corresponding to a given energy, the probability is multiplied by that number. Entropy of an Ideal GasClassically, calculation of the entropy of the ideal gas is much simpler than using a full quantum description. When a gas particle which was originally in volume V_{0} is reduced to a volume V, the number of possible states (positions) it can be in is reduced by a factor of (V/V_{0}). If we have N particles, this factor must occur N times, we find the change in entropy in going from V_{0} to V is Since the pressure is given by P =  (dA/dV) = (T dS/dV) = T k_{B} N (1/V), we recover the ideal gas law! : P V = N k_{b} T Notice that the idea of pressure, which mechanically is a force per area, arises without any mention of force, momentum, kinetic energy, or velocity! It arises solely in consideration of possible positions of the particles. Do you find it puzzling that a calculation of the entropy can give an expression for the pressure which is the same as that arising from a detailed analysis of the force imparted on container walls due to particle collisions? Question: In the above, we assumed that the particle volumes are negligible. At higher densities, however, the molecular or particle volume, b, can be a noticeable fraction of the total volume. One approximate way to take this into account is to say that the total available volume at any time is reduced by N b. Use this to arrive at a modified expression for the change in entropy, . Furthermore, at higher densities, the energetic interactions between particles also become significant. We can approximate the total interaction energy by assuming it is proportional to the number of particleparticle contacts. This in turn is proportional to the density squared, and is proportional to the total volume: E = a V (N/V)^{2} = a N^{2} / V where a is the van der Waals interaction parameter. Now using the relations P =  (dA/dV)_{N,T} , A = E  TS, and the expressions above for E and , calculate an expression for the pressure. Does this look familiar? (As above, we rely on d()/dV = dS/dV. Why is this true?) 