More on probability?

Question 1

< v> equation

Answer 1

sqrt (8KT/mπ)
K: Boltzmann constant

Question 2

mean free path equation and meaning

Answer 2

mean free path (λ) :
the average distance a particle travels before colliding with another

λ = 1/nσ√2
n: number density
σ: collision cross-section

Question 3

relative velocity formula

Answer 3

for 2 particles in an ideal gas
vᵣ =√2 * < v>
if they have the same mass

Question 4

derive the mean free path and mean collision time formulae

Answer 4

for 2 particles of radius r that would just barely touch:
σ = π(2r)² (collision cross-section)

volume swept out = σvᵣdt
probability the 2nd particle is in that volume = nσvᵣdt

-dPₛ = Pₛ(t) * nσvᵣdt
dPₛ/Pₛ = nσvᵣdt
hence: Pₛ = e^(-nσvᵣt) = e^(-t/T) (T: tau symbol)
∴T = 1/nσvᵣ = 1/nσ< v>√2

since λ = T*< v>
then λ = 1/nσ√2

Write n (number density) as N/V

Question 5

assumption used for particles motion

Answer 5

when taking a cube, 1/6 of the particles in that volume will be moving toward any given face

Question 6

derive the viscosity of a fluid

Answer 6

Draw 2 lines, L is the distance between them. the top one is moving with speed u= u₀. Label x as parallel to the lines and z as perpendicular
Draw arrows (along x) between the plates, larger near the top.

stress caused by the transfer of momentum to the gas:
F = - η ∂u/∂z

Take a section of area A and height λ
number of particles moving up or down: 1/3 * n * < v>
momentum gained per particle: mλ ∂u/∂z
∴ F = -1/3 * < v>nmλ ∂u/∂z

hence η = 1/3 * < v>nmλ

Question 7

heat flux equation

Answer 7

q = -K * ∇T
T: temperature (as a function of position)
K: thermal conductivity
q is underlined

Question 8

derive the thermal conductivity equation

Answer 8

take a plate of area z₀. draw a box of height λ above and below that. If the energy is a function of height E(z) then:

average energy flow per unit area, per unit time:
q = 1/6 *n< v><E(z-λ)> - 1/6 n< v><E(z+λ)>
(-1/6 for the particles moving down)

Since E= CᵥdT:
<E(z-λ)> - <E(z+λ)> = 2λ∂E/∂z = 2λCᵥ∂T/∂z

∴ q = -1/3 * n< v>λCᵥ* ∂T/∂z
since q = -K*∇T
K = 1/3 * n< v>λCᵥ

Question 9

Flux of molecules equation

Answer 9

J = -D *∇n
D: diffusion coefficient
n: number density

Question 10

derive the diffusion coefficient equation

Answer 10

Draw a cube and split it into an upper and lower section. both of height λ. label the height of the middle as z₀.

the mean number of molecules that can cross an area:
1/6 * < v> n(z₀-λ)
hence J = 1/6 * < v> n(z₀-λ) - 1/6 * < v> n(z₀+λ)
= -1/6 * <v> [ -n(z₀-λ) + n(z₀+λ)]</v>

sub in that: n(z₀+λ) - n(z₀-λ) = 2λ ∂n/∂z

J = -1/3 *λ< v> * ∂n/∂z

since J = -D*∇n
D = 1/3 *λ< v>