KINETIC THEORY

How does the kinetic model develop a root mean square velocity? What is root mean square?• Half the class average the numbers below, square the average,

then take the square root of the squared averages.• Half the class square the numbers below, average the squares

then take the square root.• 2.1, 3.5, 4.0, 5.1, 6.5, 7.0

KINETIC THEORY ASSUMPTIONS

Large number of molecules of identical mass, m, which behave as point particles

Molecules move randomly and obey Newton’s Laws of motion

Molecules are, far apart on average & are small compared to their relative separations.

When molecules collide or hit walls they bounce elastically And no time is spent in collisions

PRESSURE & TEMPERATURE IN A GAS

Pressure Pressure is due to the molecules

colliding with the container walls. P = = in Pascal 1 Pa = Temperature Temperature is a measure of the

average Kinetic Energy of molecules in a substance.

MOLES, MOLECULES & BOLTZMAN

mole A mol, n = molecules Avagadro’s number, NA is the # of molecules

in a mole: 6.022x1023 molecules mol-1

The number of molecules, N = n NA

Boltzman’s Constant Boltzman’s Constant: k = = 1.38x10-23 J K-1

KINETIC MODEL OF GAS

When molecules bounce off walls Δmv due to change in direction.

There must be a force on molecules from wall (Newton’s II Law).

There must be an equal and opposite force on wall from molecules (Newton’s III Law).

Each time there is a collision between molecules & wall, a force is exerted on wall.

Average of all microscopic forces on the wall over time means there is effectively a constant force on the wall.

AVERAGE ENERGY

½mv2 = Ekavg=kT Where T is in Kelvins and k =

1.38x10-23 J/K

vrms is the root mean square velocity. Root mean square is the overall

distribution of speeds. (Not the Average!!!)

vrms = =

REMEMBER

Density =

LUNGS EXAMPLE

A person’s lung can hold about 6.0 L of air at body temperature of 37°C. Given air is 21% O2, how many oxygen molecules are in their lungs? PV = NkT so N = and T = 310K N = N = 1.4x1023 molecules # O2 = (0.21) N = (0.21) 1.4x1023 # O2 molecules= 2.9x1022 O2

AVG KE OF MOLECULES

What is the average KE of oxygen molecules in the air? Assume the air is at a temperature of 21°C. 21°C = 294 K KEavg = kT = 1.38 x10-23 (294 K) KEavg = 6.09x10-21J

ROOT MEAN SQUARE VELOCITY

Since KE = ½ mv2 = kT v = mass = (Molecular Mass, M) / NA

So v = = =

INTERNAL ENERGIES

Internal energy, U, is the sum of all its potential and kinetic energies. In an ideal gas, because there are no interactions between molecules, other than perfectly elastic collisions, there is no PE.

Internal energy of monoatomic ideal gas: U= or U = if using moles

BASKETBALL’S INTERNAL ENERGY

What is the internal energy of a basketball at 290 K that holds 0.95 moles of air molecules? U = 3/2 nRT U = 3/2 (0.95) 8.31 (290K) U = 3400 J

SOURCES

http://chemwiki.ucdavis.edu/@api/deki/files/8676/root_mean_square_speed_3.JPG?size=bestfit&width=337&height=312&revision=1

www.schoolphysics.co.uk