Recap….

1. Recap…. Distributions and averaging The ideal gas law Thermal & Kinetic Lecture 4 Distributions, free expansion, the ideal gas law LECTURE 4 OVERVIEW

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

3. 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

4. Distributions ? So, to determine <x>, the appropriate integral is...?

5. 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

6. 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)

7. 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)dvx 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

8. 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,) ?

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

10. 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:

11. 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>

12. 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:

13. 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))

14. 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)?

15. 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…..”

16. Free expansion As dU=0 for a free expansion, if dT =0 (i.e. no temperature change takes place) then it follows that: 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: In words: “U – the total energy of the gas – does not depend on P” Writing U as U(T,P), the following relationship holds: The internal energy of an ideal gas is a function of T only.