Thermal & Kinetic Philip Moriarty School of Physics & Astronomy

1. Recap…. Distributions and averaging The ideal gas law Thermal & Kinetic Philip Moriarty School of Physics & Astronomy B403, x 15156, philip.moriarty@nottingham LECTURE 3 OVERVIEW

2. Last time…. Heat capacity, latent heats. Vapours and vapour pressures. The riddle of the drinking duck.

3. Magnitude of Avogadro’s number illustrates scale of problem. With current best supercomputers: follow a few 1000 atoms for – at best – a few ns. ? It’s not the individual units which are of interest – it’s the collective, statistical behaviour of the system. Therefore probability theory very important. Consider the ‘ball-and-spring’ model of a solid we discussed earlier. Is there 0 energy at 0K? We’re going to focus on classical descriptions – however, QM ultimately underlies interatomic and intermolecular interactions. Towards kinetic theory: some points to bear in mind Not a good strategy to attempt to explain properties of matter by simply tracking detailed behaviour of every individual unit.

4. Distributions ? Can you write down an expression that gives the total number of people? Dh → dh. Get smooth (quasi-continuous) curve Dh NB See also: “Dimensional analysis and distributions”, Modelling lecture notes 1st year laboratory errors manual (PLEASE read this!!) Histogram of heights N(h) h N(h)dh: no. of people with heights between h and h+dh

5. f4 f3 f2 ? f1 What’s the probability of finding this continuous system in a specified, individual state? In other cases, system may be capable of existing in any one of a continuous range of states. Introduction to the Mathematics of Averaging First, consider a system which can only exist in certain discrete states: fi(i=1,2,….) The probability of occurrence of these states is Pi (i = 1,2…..) If the function Q has the values Qi when the system is in the states i = 1,2, …. then the average value of Q (denoted <Q>) is given by: ANS: 0

6. If dx is sufficiently small, the probability of finding the system in a state between x and x+dx is given by f(x) dx. We now consider a quantity Q(x). The average value of Q is: 1.3 In the limit dx → dx, we have the integral (evaluated over all values of x): 1.4 1.5 The function f(x) dx is a probability and hence: Introduction to the Mathematics of Averaging Let’s now suppose that the state of the system is represented by a coordinate, x. Although the probability that the system is associated with a value that’s precisely x is 0, the probability that the value lies within x and x+dx is finite.

7. The ideal gas law: Molecular velocities Molecules are featureless points, occupy negligible volume. Total number of molecules (N) is very large. ? Molecules follow Newton’s laws of motion. What’s the average velocity of a molecule? Molecules move independently making elastic collisions. No potential energy of interaction (no bonding). Gas confined in a large rectangular box (wall area: A, side length: a) Assumptions This hypothetical gas is termed an ideal gas. Monatomic gas such as He or Ne at low pressures shows approximately ideal gas behaviour. ANS: 0, but average speed is NOT 0

8. The ideal gas law: Molecular velocities We could assume molecules all move parallel to either the x, y- or z-axis and all have the same speed (this is a big assumption) Instead, let’s do the derivation rather more rigorously by taking into account the distribution of molecular velocities (distributions are at the core of thermal and kinetic physics.) Molecules are specularly reflected at walls. Momentum change of individual molecule on hitting wall: 2mvx. Gas molecules have a distribution of speeds (different vector lengths in figure to left)

9. The ideal gas law: Momentum changes The fraction of all molecules which have x components of velocity in the range vxto (vx+ dvx) = f(vx) dvx. Here f(vx)dx represents a probability Total number of molecules having velocity in this range is: Nf(vx) dx Hence, total change of momentum per second due to molecules within the given velocity range is: To determine the rate of change of momentum due to collisions, we need to know frequency of collisions. Only those molecules within a distance vx of the wall will collide with it within 1 second. This will be a fraction vx/a of the total number of molecules in the box. vx a

10. The ideal gas law: Velocity distributions By symmetry, f(vx)=f(-vx). Hence integrand is symmetrical about vx=0. Now, integrate to get total change of momentum due to molecules of all velocities hitting a wall: ? Why are the integral limits (0,) ?

11. The value of the integral in Eqn. 2.7 in the notes is: • 0 • <vx> • <vx2> • None of these

12. The ideal gas law: Pressure and Volume Value of integral is mean value of vx2 = <vx2>. The expression for the total change of momentum per second then becomes:

13. If the total change of momentum per second is mN<vx2>/a, then which of the following is true?: • P = mN<vx2>/a • P = mN<vx2> • PV = mN<vx2> • P/V = <vx2>

14. The ideal gas law: Pressure and Volume <vx2> = <vy2> = <vz2> = 1/3 <vx2 + vy2 + vz2> = 1/3<v2> <v2> is the mean square speed of the molecules. Hence:

15. 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))

16. Free expansion: Joule’s experiment U(r) Hence, no work is done during the adiabatic free expansion of an ideal gas. r 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. U(r) r X If the gas is ideal there are no intermolecular interactions. Joule’s (and our!) question: is U ≡ U(T) or U ≡ U(T,P) or U ≡ U(T,V)?

17. 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 “Free 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…..”