# equations physical chemistry

vanderwaals interaction

v = -1/r^6(cuu + cuu* +cdisp)

multiply by na to get per mole

multiply by na to get per mole

average kinetic energy of translation of a particle

Etrans=3/2KT

energy of translation per mole

Etrans,m = 3/2K×nA×T

k×nA=R gas constant

Etrans,m=3/2RT

k×nA=R gas constant

Etrans,m=3/2RT

dispersion interaction

v=-3/2 polarizability vol1 x vol2 × (ionization potential 1 ×potential 2/I1 +I2) ×1/r^6

gas

epsilon<

liquid

etrans~~epsilon

solid

epsilon>>etrans

charge charge interaction

v = Q1 × Q2 / 4piepsilon0 × r

charge dipole interaction colinear

v = -u1 Q2/ 4piepsilon r^2

u* induced dipole interaction

u* = 4pi epsilon0 × polarizability volume × E (electric field)

charge dipole interactions at an angle

v = -u1 Q2 / 4pi epsilon0 × R^2 × costheta

dipole dipole interactions colinear

v = -u1u2/2pie0 r^3

addition of dipole moments

u = (u1^2 + u2^2 + 2u1u1costheta)^0.5

Pythagoras 3d dipole moments

u =(ux^2 + uy^2 +uz^2)^0.5

Keesom interactions

rotating ermanent dipoles

C = 2u1^2 u2^2 / 3(4pie0)^2 KTr^6

C = 2u1^2 u2^2 / 3(4pie0)^2 KTr^6

if v << kt

keesom interaction holds

dipole induced dipole interaction

V = -μ1^polarizability vol2^2/4pie0 r^6

effect of changing conditions for fixed amount of gas

p1 v1/ t1 = p2 v2/ t2

root mean square speed for n molecules

c = ((s1^2 + s2^2 + … sn^2)/N)^1/2

Phase Rule

F = C – P + 2

F= number of degrees of freedom

P = number of phases at equilibrium with one another

F= number of degrees of freedom

P = number of phases at equilibrium with one another

dipole dipole at an angle

v = μ1×μ2(1-3cos^2theta)/4pie0r^3

ideality assumptions for van der waals equation of gases

1. molecular size is negligible

2. no intermolecular interactions

only true at low pressures or high temperatures

2. no intermolecular interactions

only true at low pressures or high temperatures

corrections to vdw equation for gases

pressure reduced by intermolecular forces

p -> p + a(n/v)^2

excluded volume as molecules cannot occupy the same space

v -> v – nb

p -> p + a(n/v)^2

excluded volume as molecules cannot occupy the same space

v -> v – nb

a

correction factor to account for intermolecular attractions

b

correction factor to account for finite size of molecule

Categories: Physical Chemistry