Elasticity

Question 1

Young’s modulus equation

Answer 1

E= σ/ε = FL/Ax
E: young modulus
σ: stress
ε: strain

Question 2

when does Young’s modulus apply?

Answer 2

when the object is being stretched by a force (under tension). Only for solids.

Question 3

Shear modulus equation

Answer 3

G = FL/Ax = F/Aθ
G: shear modulus
θ: angle between the height and slanted edge (imagine parallelepiped)

second equation uses small angle approximation

Question 4

when does the Shear modulus apply?

Answer 4

when a force F is being applied to opposite edges of an object. (which is the same as a torque). only for solids.

Question 5

Bulk modulus equation

Answer 5

K = -V dP/dV
K: Bulk modulus
V: volume
P: pressure

Question 6

derive the Bulk modulus equation for an isothermal and an adiabatic gas

Answer 6

isothermal:
K= -V dP/dV (sub in the differential of P=NKT/V)
K = NKT/V = P

Adiabatic:
K= -V dP/dV (sub in the differential of P=CV^-γ
K= V [ γC / V^(γ+1) ]
K= γC /V^γ = γP

Question 7

how do you derive the Clausius Clapeyron equation

Answer 7

at thermal equilibrium. dG = 0 (Gibbs free energy)
dG = Vdp - SdT
hence:
VₓdP - SₓdT = VᵧdP - SᵧdT
dp(Vₓ-Vᵧ) = dT(Sₓ-Sᵧ)
dp/dt = (Sₓ-Sᵧ)/(Vₓ-Vᵧ)
dp/dt = ΔS/ΔV

(can sub in that ΔS = ΔH/T if needed)

Question 8

Explain what each term in the Clausius-Clapeyron equation is

Answer 8

dP/dT = ΔH/TΔV

RHS:
ΔH: the change in enthalpy for a given event
ΔV: the volume change that results from that event
T: the temperature the event happens at

LHS:
the ratio of the change in pressure and temperature if the conditions under which that event occurred at, changed.

(example, if something were to melt at 100K but you wanted to know the temperate it would melt at at 2atm instead of 1atm, use dT= ΔT and calc for it. sub in T=100K and dP=ΔP=(2-1)atm)

everything is EITHER per unit mass or per unit moles