Constants: 1 L atm = 101.325 J 1 atm = 1.01325x105 Pa = 1.0132

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Constants:
R  8.314
J
L atm
 0.0821
mol K
mol K
1 L atm = 101.325 J
1 atm = 1.01325x105 Pa = 1.0132x105 kg m-1 s-1
kB = 1.38x10-23 J K-1
1 torr = 133.3224 Pa
Ideal Gas: PV = nRT | (1) Gas particles are small – negligible size/volume (2) Gas particles exert no forces on each other (3) Particles in constant
motion; elastic collisions with walls produce pressure => applies ONLY at low [ ], low P, high T
Van der Waal’s adjustments: [ P  a ( n ) 2 ][V  nb]  nRT ; a = attractions b/w gaseous molecules – varies much b/c diff molecules have diff IMFs; b =
V
extended volume gas molecules occupy – varies little b/c volume of diff gases varies little.
Partial Pressures: P  P  P  P  ...  (n  n  n  ...) RT  n RT ; P  n A P
tot
A
B
C
A
B
C
tot
A
tot
V
V
ntot
Diffusion/Effusion: spreading one substance through another/escape of one substance through hole; rate 
Collisions: Z  ( 1 )( N ) 8RT A (rate of effusion)
w
4 V
M
1
Speed, Temp, Mass: N m(u ) 2  RT ; u  3RT ; u 
A
rms
rms
mp
2
M
1 r1
; 
M r2
M 2 ; Molecule-Wall
M1
2 RT ;
8RT
uavg 
M
M
Intermolecular Forces: Ionic V  1 ; Ion-dipole V  1 (orientation dependent); Dipole-dipole V  1 (randomly), V  1 (in liquid); Ion-induced
r
r2
r3
r6
dipole V  1 ; Dipole-induced dipole V  1 ; Induced dipole-induced dipole V  1
r4
r6
r6
Thermo Terms:
System – portion confined by boundaries
Thermal Equilibrium – macroscopic condition of a system characterized
Surroundings – part of universe outside system
by constancy of pressure, temperature
Heat – means by which energy is transferred; hot -> cold
Exothermic – heat given off (/\H < 0)
Work – product of force over displacement
Endothermic – heat absorbed (/\H > 0)
Temperature – degree of hotness/coldness based on motion of molecules
Exergonic – release of E in form of work
Isothermal – process at constant T
Endergonic – absorption of E in form of work
Adiabatic – process in thermally isolated system
State Functions: Temperature, enthalpy, heat capacity, pressure, entropy, Gibbs’ energy
3 Laws of Thermo: (1) U  q  w (conservation of E); (2) Suniv  0 (towards disorder); (3) S > 0 except for perfect crystals at 0 K
Common formulae: w   Pext V ; q  CT ; U  q  w  ncv T (ideal ) ; U univ  0 ; H  nc p T (ideal ) ; U univ  0 ; c p  cv  R (ideal )
Bond Enthalpies: H bond  H broken  H formed  H react  H prod
Constant Pressure - H  U  PV  q p  U  (n g ) RT ; PV  n g RT (ie. at STP); NOTE: negative work = work done by system on
surroundings
Heat capacities: Monoatomic c  3 R ; Linear c  5 R ; Non-linear cv  3R ; [based on rotational, translation energies, vib ~ kBT contributes little
v
v
2
2
V2
Isothermal: w   PdV   nRT ln V2 (reversible); w   q ; U  0 (b/c /\T=0)

V1
V1
Adiabatic: q = 0; U  w
Entropy: S  kB ln W , W = # available microstates (like probability – flipping coins)…S > 0 always (W > 1); S  qrev  nR ln V2 (constant T);
T
V1
S 
2
G
nc
nc
T (constant V);
T
S   p dT  nc p ln 2 (constant P); S univ  
T T v dT  ncv ln T12
T
T
T1
T1
1
T2
T
As state function: Carnot result for reversible heat engine showed: qh  qi  0 => qh is state function b/c value does not change in cycle
Th Ti
Th
=> dqrev is state function
 T
Gibbs Energy: G  H  TS (constant T, P); G  0 (spontaneous/product favored); G  0 (non-spontaneous/reactant favored);
G  0 (equilibrium)
dG  dw' (reversible, constant T, P – Gibbs’ E is max work possible, not including PV work); dG  VdP (constant T);
dG   SdT ; G   ST (constant P – linear!)
Other:
V  A
| z1 z2 | N Ae 2 ;
40 d
Svap  88  5JK 1mol 1 = Trouton’s rule – entropy of vap is similar for all liquids b/c /\Hvap and T change proportionally except H-bonders b/c they
have fewer allowed configurations
Processes:
a) Isothermal, reversible
V
2
U  0 ; H  0 ; w   P dV  nRT ln V2  find Volumes with PV=nRT; q = -w
V V
V1
1
b) Isothermal, irreversible
U  0 ; H  0 ; w   Pext V ; q = -w
c) Adiabatic, reversible
q = 0; U  C T  3 RT ; H  C T  5 RT ; w  U
v
p
2
2
d) Adiabatic, irreversible
q = 0; U  Cv T  w ; H  C p T
Don’t forget Hess’ Law (supercooled H2O problem)
ump
uavg
urms
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