Thermal & Kinetic Lecture 4 Free expansion, the ideal gas law LECTURE 4 OVERVIEW Recap…. Free expansion: the Joule effect The ideal gas law Introduction to Boltzmann factors

Last time…. Distributions and averaging. Ideal gases - approximations. Towards the ideal gas law…..

Free Expansion: Joule’s experiment <v2> and thus total molecular kinetic energy ↑ with increasing T Is <v2> also a function of P or V? Joule designed an experiment in 1845 to address this question. Take an ideal monatomic gas where all the ‘molecules’ are single atoms, e.g. He, Hg vapour, Ar …. In this case there are no internal motions of the gas molecule (eg vibrations, rotations). For a monatomic gas the kinetic energy ((½)m<v2>) is the total energy, U. Now consider allowing gas to expand into a vacuum adiabatically NB No heat energy added or removed: adiabatic (From the Greek: a (not) + dia (through) + bainein (to go))

Free expansion: Joule’s experiment U(r) U(r) r If the gas is ideal there are no intermolecular interactions. Hence, no work is done during the adiabatic free expansion of an ideal gas. There is also no heat energy flowing into or out of the gas. Therefore, the total energy of an ideal gas remains unchanged during a free expansion. Joule’s (and our!) question: is U ≡ U(T) or U ≡ U(T,P) or U ≡ U(T,V)?

Free expansion: Joule’s experiment ! It is easy to get confused here. In this case we’re concerned with the free expansion of the gas – there are no pistons or other mechanisms whereby the system could do work or work could be done on the system. In later lectures we’ll see examples of other systems where adiabatic processes involve work. Erratum: In the first paragraph under “Adiabatic expansion: the Joule effect” in the notes, the last sentence should read “An adiabatic process occurs when the system is completely thermally isolated from its surroundings…..”

The internal energy of an ideal gas is a function of T only. Free expansion Writing U as U(T,P), the following relationship holds: As dU=0 for a free expansion, if dT =0 (i.e. no temperature change takes place) then it follows that: In words: “U – the total energy of the gas – does not depend on P” Joule did not detect a temperature change during the free expansion of the gas used in his experiment and hence he used this as proof of the following statement: The internal energy of an ideal gas is a function of T only.

The ideal gas law: Boyle’s law As stated before, PV = (1/3) Nm<v2> and U=N(½ m<v2>)  PV = (2/3)U Taking the result of Joule’s experiment into consideration, U is a function of T only for an ideal gas*. So: PV = f(T), or PV = constant for a given temperature. (Boyle’s law) Since the temperature T is proportional to the average kinetic energy we can write: <Ek> T  ½ m<v2> = cT (where c is a constant)  ½ m<v2>=(3/2)kT (k is Boltzmann’s constant) *Note that Joule’s experiment was done with air and not an ideal gas. This is an important complication to which we’ll return but need not concern us now.

The ideal gas law: Boyle’s law So, there’s a constant conversion factor 3k/2 between the mean kinetic energy of a molecule and a unit of temperature (which we call a Kelvin). BUT…… “Why have you decided to choose the constant of proportionality as (3/2)k?” Bear with me……! Accepting that <Ek>=(3/2)kT, we can rewrite PV=(1/3)Nm<v2> as: PV=NkT Consider 1 mole of gas: N = NA  PV=RT (R is the gas constant (R=NAk)) NB Don’t confuse N (total no. of molecules) with n (no. of moles)

When is a gas an ideal gas? ! The ideal gas law is a good approximation to the behaviour of real gases at low pressures.

….intermission…. To recap: we have derived the ideal gas law from a microscopic picture of the gas. “Fundamentally, we assert that the gross properties of matter should be explainable in terms of the motion of its parts”, RP Feynman, Lectures on Physics I, p. 40-1 ….but we need to go further. Thus far we’ve considered the average speed of the molecules but know little about the details of: the distribution of velocities, or the positions/ arrangements of the atoms. We also need to determine just why we say that ½ m<v2> = 3/2 kT. Considering the distributions rather than the average quantities will provide us with a much better insight into just what temperature represents……..

Let’s make a deal…… http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html An American game show in the seventies involved the following scheme: A contestant was given a choice of three doors, behind one of which there is a prize (car, speedboat, $$$ etc…). Behind the other two doors are ‘gag’ prizes (chicken, donkey…etc..). After the contestant chooses a door, the host reveals an empty door amongst the two left and asks the contestant if they would like to change their choice to the other door. ? If the contestant decides to change, are their chances of winning a prize: greater, less, or, (c) exactly the same? ANS: a

Boltzmann factors “Available energy is the main object at stake in the struggle for existence and the evolution of the world.” Ludwig Boltzmann “At thermal equilibrium all microscopic constituents of a system have the same average energy” (G & P, p. 421) …….now let’s consider the distribution of energy in the system. For a system in thermal equilibrium at a certain temperature, the components are distributed over available energy states to give a total internal energy U. …but what is the probability of finding a particle in a given energy state? The probability of finding a component of the system (eg. an atom) in an energy state e is proportional to the Boltzmann factor: (NB: T is in Kelvin)