Surface tension

Question 1

the equation for the surface energy

Answer 1

E = 1/4 * n * N * ε
ε: binding energy
n: nearest neighbours
N: surface atoms per unit area
E: surface energy per unit area

(1/4 comes from ½ε as ε is shared. and the number of nearest neighbours is ½n at the surface)

Question 2

what is ɣ and what does it’s sign tell you

Answer 2

ɣ is the surface free energy per unit area

at a liquid-liquid interface:
if ɣ is positive the fluids wont mix
if ɣ is negative the fluids will mix

liquid-solid interface:
if ɣ is positive the fluid will maximise its surface area of contact with the solid (called wetting)

Question 3

what’s the contact angle

Answer 3

the angle between the container wall and the meniscus. (for a convex meniscus it is >90°)

Question 4

derive the equation to minimise the surface free energy of the meniscus (young’s equation)

Answer 4

draw a tube cross-section with the meniscus of the fluid inside. simplify the meniscus to a right-angled triangle of height h and length l.

surface free energy of the meniscus:
E = (ɣₛₗ -ɣₛ₉)h + ɣₗ₉*(H-l) (H: hypoteneus length)

minimised for ∂E/∂h = 0
∂E/∂h = (ɣₛₗ -ɣₛ₉) + ɣₗ₉(h/H) (h/H = cos(θ)
hence: ɣₗ₉*cos(θ) = (ɣₛ₉ -ɣₛₗ)
^^ is known as the Young equation

Question 5

derive the capillary action equation

Answer 5

capillary action: (how much the fluid rises in a thin tube)
if a thin tube is placed in a fluid then the height above the fluid’s surface reached by the fluid inside the tube is h.

Δ surface free energy= 2πRh(ɣₛₗ -ɣₛ₉)
Δ G.P.E = πR²hρ (<– the mass) * g * ½h (<– centre of mass is at ½h)

E(h) = 2πRh(ɣₛₗ -ɣₛ₉) + ½πR²h²ρg
∂E/∂h = 2πR(ɣₛₗ -ɣₛ₉) + πR²hρg = 0
hence: h = -2(ɣₛₗ -ɣₛ₉) / ρgR
=2ɣₗ₉cos(θ)/ρgR (from youngs equation)

Question 6

show how surface tension and free energy are the same

Answer 6

draw a wire frame with a string going across it. a film of liquid on one side. the side length is L.

work done to move the string (to expand the liquid film)
dW = Fdx

surface area increase:
dA = 2Ldx (2 is cause the film has a front and back)

Surface energy increase:
dE = ɣdA = 2ɣLdx

dW = dE
2ɣLdx = Fdx
F/L = 2ɣ

hence, the force per unit length (F/L) and the energy per unit area (ɣ) are the same

Question 7

variance formula

Answer 7

σ²= < f²> - < f>²

Question 8

how would you show that the probability distribution of velocities can be written in a particular form

Answer 8

use the normalisation:
∫ p(v) dV = 1 (limits are 0 to infinity)
p(v) is the probability distribution
dV is the volume element of the velocity distribution

dV = 4πv²dv

Question 9

how would you re-derive the ideal gas equation using probability distribution of the particles velocity

Answer 9

for a cubic space the number density n of gas particles:

particle collisions with any face = nvA*dt (assumes all particles travel towards the wall)
the momentum transferred per particle= 2mv

average momentum transfer = dp = 2mnAdt<v²>
dp = 2mnAdt ∫ v² p(v) dv (since <v²> = ∫ v² p(v) dv)

F= dp/dt and P= F/A
hence P = 2mn*∫ v² p(v) dv

plug in an expression for p(v) (the probability distribution) then solve for PV.

this is the method to derive the Boltzmann factor, b. (b will be given in the p(v) expression)