Physical Chemistry Test Questions – Flashcards
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Intensive variable
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independent of amount of substance. Examples: density
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Extensive variable
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depends on mass, volume, etc. The ratio of two extensive variables is an intensive variable. Examples: volume, internal energy,
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Equilibrium
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The state of the system is independent of time.
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Short-term equilibrium
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Steady state: System appears to not change with time but actually does slowly.
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Ideal gas
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Molecules do not interact, collisions are elastic, molecules fill container entirely, molecules are infinitely small and far apart. Obeys PV=nRT Real gases approximate at high T and low P.
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Real gas
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Approximate ideal gases at high T and low P. Follow (P+ [a/Vbar]^2(Vbar-b) = RT where a term is the correction factor for pressure and b is the correction factor for volume. a is measure of strength of molecular interactions and b is the volume of the molecules.
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Zeroth law of thermodynamics
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If A is in thermal equilibrium with B which is in thermal equilibrium with C, A is in thermal equilibrium with C. (Transitive property of thermodynamics)
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Thermal equilibrium
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Defined by temperature. Equilibrium is reached when two systems have the same temperature.
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Boyle's law
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PV = constant (isothermal) PV^gamma = constant (adiabatic) P1V1^gamme = P2V2^gamma (adiabatic) P1V1=P2V2 (isothermal)
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Charles' Law
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Volume is proportional to temperature at constant pressure.
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Avogadro's law
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Volume is proportional to the amount of moles of a substance if temperature and pressure are constant.
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Compressibility
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Ideal gas: Z= PVbar/RT =1. Real gases follow the virial equation: Z= PVbar/RT = 1 + B/Vbar + C/Vbar^2 + ...Van Der Waals gas: Z = (Vbar/Vbar-b)- (a/VbarRT)
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Cubic expansion coefficient
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(alpha) = (1/Vbar)(dVbar/dT)P,n
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Isothermal compressibility
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(kappa) = -(1/Vbar)(dVbar/dP)T kappa is always positive.
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Value of alpha and kappa
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a/k = (dP/dT)Vbar which is hard to measure.
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Work
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dw= -Pext dV (On a system when a finite change in volume) dw= PdV (if reversible) w is positive if work done ON the system w is negative if work done BY the system
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Heat
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(q) positive when energy enters the system negative when energy leaves the system.
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Critical point of isotherm
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(dP/dV)T = Tc= 0 Van Der Waals equation fails at this point.
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Internal energy
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U - associated with constant volume processes Delta U = w +q For adiabatic: Delta U = w For no work done (constant volume): Delta U = q Delta U is negative if the system loses energy in heat and work. Delta U is positive if the system gains energy in heat and work
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Heat capacity
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dq/dT at constant volume OR pressure. Cv= dq/dT = (dU/dT)v (At constant volume) Cp = dq/dT = (dH/dT)p (At constant pressure) Cp-Cv = R for an ideal gas Cp/Cv = gamma = 5/3 for an ideal gas
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Enthalpy
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H = U+PV - associated with constant pressure processes. dq = dH dH = CpdT
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Cv=? (monatomic ideal gas)
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Cv = 3/2R (U)T = 3/2 RT
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Cp =? (monatomic ideal gas)
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Cp=5/2R (H)T = U + RT = 3/2RT + RT = 5/2 RT
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Joule experiment
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No work was done because Pext = 0. No measured change in temperature, so dU = 0. Which meant (dU/dV)T =0 (No Change in internal energy with change in volume at constant temperature) This is true for an ideal gas but not a real gas. There is a change in temp, but it is very small.
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q at constant volume
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qV= delta U
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q at constant pressure
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qP = delta H
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Joule thomson expasion
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insulated pipe, q=0, w=P1Vbar1 - P2Vbar2 = Delta U Joule-Thomson coefficient: muJT = (dT/dP)H for a real gas muJT= 0 for an ideal gas
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Stoichiometry numbers
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Positive for products, negative for reactants
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Exothermic
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Heat flows out of the system with this type of reaction to restore the system to the original temperature. q is negative
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Endothermic
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Heat flows in from the surroundings with this type of reaction to restore the system to the original temperature. q is positive.
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Enthalpy of formation
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DeltafH = change in enthalpy in which one mole of the substance in its standard state at a given temperature is formed from its elements. (DeltafH = 0 for an element in its standard state)
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Cp reactant - Cp product
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Delta r Cp.
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Calculate Delta H at some temp other than 298 K
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deltarH (298) + integral from T1 to T2 of deltarCp
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Calclate deltarH for a Reaction at 0K
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1. Warm up to 298 (H298-HT) 2. Reaction at 298 (delta H298) multiply by moles 3. cool down to 0K (HT-H298)
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solvation
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solute interacts strongly with the solvent.Heat is given off. If water is the solvent, this is called hydration. Decrease in energy.
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overcoming the attraction of solute and solvent molecules
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heat is absorbed to overcome the attraction between solute and solvent.
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entropy
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dq/T= dS If reversible reactions were possible, delta S of the universe for that system would equal 0, but since all reactions irreversible, delta S of the universe is greater than zero. Spontaneous processes increase S and waste energy. delta S universe is NEVER less than zero.
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Two things that drive processes
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Desire to get to higher entropy and desire to get to lower energy
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Delta S for a reversible change in an ideal gas
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Two steps: 1) Isochoric (constant volume) 2) Isothermal (change in volume) Delta S = integral Cv/Tdt + nRlnV2/V1 (also works for irreversible)
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