equations physical chemistry – Flashcards

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question
vanderwaals interaction
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v = -1/r^6(cuu + cuu* +cdisp) multiply by na to get per mole
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average kinetic energy of translation of a particle
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Etrans=3/2KT
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energy of translation per mole
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Etrans,m = 3/2K×nA×T k×nA=R gas constant Etrans,m=3/2RT
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dispersion interaction
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v=-3/2 polarizability vol1 x vol2 × (ionization potential 1 ×potential 2/I1 +I2) ×1/r^6
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gas
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epsilon<<etrans
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liquid
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etrans~~epsilon
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solid
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epsilon>>etrans
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charge charge interaction
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v = Q1 × Q2 / 4piepsilon0 × r
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charge dipole interaction colinear
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v = -u1 Q2/ 4piepsilon r^2
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u* induced dipole interaction
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u* = 4pi epsilon0 × polarizability volume × E (electric field)
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charge dipole interactions at an angle
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v = -u1 Q2 / 4pi epsilon0 × R^2 × costheta
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dipole dipole interactions colinear
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v = -u1u2/2pie0 r^3
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addition of dipole moments
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u = (u1^2 + u2^2 + 2u1u1costheta)^0.5
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Pythagoras 3d dipole moments
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u =(ux^2 + uy^2 +uz^2)^0.5
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Keesom interactions
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rotating ermanent dipoles C = 2u1^2 u2^2 / 3(4pie0)^2 KTr^6
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if v << kt
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keesom interaction holds
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dipole induced dipole interaction
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V = -μ1^polarizability vol2^2/4pie0 r^6
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effect of changing conditions for fixed amount of gas
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p1 v1/ t1 = p2 v2/ t2
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root mean square speed for n molecules
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c = ((s1^2 + s2^2 + ... sn^2)/N)^1/2
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Phase Rule
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F = C - P + 2 F= number of degrees of freedom P = number of phases at equilibrium with one another
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dipole dipole at an angle
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v = μ1×μ2(1-3cos^2theta)/4pie0r^3
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ideality assumptions for van der waals equation of gases
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1. molecular size is negligible 2. no intermolecular interactions only true at low pressures or high temperatures
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corrections to vdw equation for gases
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pressure reduced by intermolecular forces p -> p + a(n/v)^2 excluded volume as molecules cannot occupy the same space v -> v - nb
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a
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correction factor to account for intermolecular attractions
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b
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correction factor to account for finite size of molecule
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