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313 Exam 3 Cheat Sheet by

bme     313     psu

Laws

Zeroth Law: If system x = system y & system y = system z, then system x = system z. (Trans­itive)
First Law: Internal energy(ΔU) of an isolated system is constant. No heat lost, only transf­erred.
Second Law: The entropy of any isolated system always increases.
Third Law: The entropy of a system approaches a constant value as the temper­ature approaches absolute zero.
Cyclic Rule: (dP/­d­T)­vd­T/­dV­)(­dV­/P)T= -1

Defini­tions

Adiabatic: No transfer of heat or matter
Diathe­rmal: Heat allowed to transfer, no matter transfer. Can transfer energy in the form of work
Enthal­py(ΔH): Amount of heat content used or released in a system at constant pressure
Irreve­rsible: A process that cannot return both the system and the surrou­ndings to their original condit­ions.

Exam 2

ΔU=m/Ms ΔUcom­b+­mH­2O­/MH2O Cv,m­(H2­O)Δ­T+Δ­TC­cal­ori­meter
ΔHo=­ms­alt­/M­salt ΔHo­sol­uti­on­+m­H2O­/M­H2O Cp,m­(H2­O)Δ­T+Δ­TC­cal­ori­meter
S=k ln(W) W=#of states
Efficiency = 1-|qc­d|­/|q­ab| <1
ΔHo­rt­=ΔH­o­29­8+­∫ΔC­p(T) dT from 298 to T
ΔHcom­bus­tion = ΔUcom­bus­tio­n+­Δ(PV)
For Solids & Liquids: ΔH~= ΔU
Δs=-nR­ln(­P/P­i­)+∫­nC­pm/T dT for Pi to Pf
Δs=nRl­n(V­f­/V­i)­+∫n­Cvm/T dT for Vi to Vf
Isolated System: ΔS=q(1­/T1 - 1/T2)
Isothe­rmal, Ideal: ΔS=nRl­n(V­f­/Vi)
ΔStot­al­=ΔS­+ΔS­su­rro­und­ings
ΔG=nRT Σ xiln­(xi) xi is mole fraction
ΔG = TΔStotal
 

Internal Energy (ΔU)

General
ΔU=q+w
Constant Volume
ΔU=CvΔT = qv
Adiabatic, Reversible
ΔU= w = n(Cpm­-R)ΔT = nCvmΔT
Ideal
ΔU=nC­vmΔT

Enthalpy (ΔH) (State Fxn)

General
ΔH= ΔU+Δ(PV) = ΔU+nRΔT
Constant Pressure
ΔH= CpΔT
Ideal
ΔH = qp
Constant Volume
ΔH= nCpmΔT + VΔP
Even More General
dH= (dH/­d­P)T dT + (dH/­d­T)P dP
Liquids & Solids
(dH/­d­P)T = V(1-Tβ)
Constant Pressure, closed system
ΔH= (Uf+­PV­f)­-(U­i­+P­iVi)
Isobaric
ΔH= n∫Cpm(T) dT = nCpmΔT

Exam 2 Material

Sm(T­)=S­m­(0ok) +∫Cpm/T dT(solid 0-Tf) +ΔHfu­s/Tf + ∫Cpm/T dT(liquid Tf-Tb) +ΔHva­p/Tb ∫Cpm/T dT(gas Tb-T) `
For Ideal Gases: ΔSm=­Rln­(V­f/­V)=­-Rl­n(P­f­/Pi)
ΔG(T)/­T2= ΔG(T)/­T+Δ­H(T­1­)(1­/T­2-­1/T1)
Max Work: Revers­ible, adiabatic, isothermal
Hess's Law: Total Enthalpy change is indepe­ndent of # of steps(­pat­h-i­nde­pen­dent).
ΔA = ΔU-TΔS = ΔH-nRT (Hemholtz)
for ΔGor only include non-pure substa­nces.

Exam 3

ΔGR = ΔG°R+RT ln(QP)
ln(KP) = -ΔG°R/RT
Kx=K­P­(P/­P°)­-ΔV
dA = 𝛾 dσ gamma is surface tension
Work = 8pi𝛾r dr
Force = 8pi𝛾r
h(capi­llary rise/d­epr­ession) = 2𝛾/𝞺gr
𝓾B=𝓾­°+R­Tln­(𝛾[B]) gamma is activity coeffi­cient
ΔGR = ΔG°R­-2.3­03­vRT(pH)
qx = kA(Ts­i-­Ts­o)/L
q''x= -k dT/dx = qx/A
Ėin+­Ė-Ė­out = Ėinte­rnal
q12 = εσA(T­1­4-­T4) - Heat xchange via radiation b/t 2 surfaces
q''s = h(Ts­-T) - Newton's Law of Cooling

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