Question #c9102
1 Answer
There is one, but this applies best at "low or moderate pressure" for pure gases and "for nonpolar liquids or solids" (Physical Chemistry, 6th ed., Levine, pg. 433):
#(epsilon_r - 1)/(epsilon_r + 2) M/rho = (N_A)/(3epsilon_0)(alpha + mu^2/(3k_BT))#
This is the Debye-Langevin equation, where:
#M# is the molar mass.#rho# is the density of the substance.#epsilon_r# is the dielectric constant of the substance. This physically tells you how high the electric flux density can be (and consequently, how long something can hold electric charge).#N_A = 6.0221413 xx 10^23# #"mol"^(-1)# is Avogadro's number.#k_B = 1.38065 xx 10^(-23)# #"J/K"# is the Boltzmann constant.#T# is the temperature in#"K"# .#alpha# is the polarizability of the substance.#mu# is the electric dipole moment of the substance.#epsilon_0 = 8.85418782 xx 10^(-12)# #"C"^2cdot"N"^(-1)cdot"m"^(-2)# is the vacuum permittivity.
We can rewrite the equation as follows:
#stackrel(y)overbrace((epsilon_r - 1)/(epsilon_r + 2) M/rho) = stackrel(m)overbrace((N_Amu^2)/(9epsilon_0k_B))stackrel(x)overbrace(1/T) + stackrel(b)overbrace((N_Aalpha)/(3epsilon_0))#
From this equation, if we plot
That is, we have:
- A slope of
#(N_Amu^2)/(9epsilon_0k_B)# .- A y-intercept of
#(N_Aalpha)/(3epsilon_0)# .
In general, we can say that at constant temperature---with some exceptions, the dielectric constant increases as the electric dipole moment increases.
This is because with an increased electric dipole moment, as the system interacts with the applied electric field, it can store more electric charge (due to the more permanent nature of its dipole), increasing its dielectric constant.
