Study Guide for Physical Chemistry

 

Many of formulae will be given on the exam.  However, the basic facts given below should be familiar to you.  This is a foundation that you will be able to use to interpret many chemical problems.

 

Kinetic Theory of Gases

The kinetic theory of gases is based on the idea that molecules have kinetic energy 1/2 mv2

where m is the mass and v is the velocity.  The Boltzmann factor, exp{ -DE/kBT }gives the probability of the ratio of the population of two energy separated by DE.  Applying this factor to kinetic energy of molecules leads to the kinetic theory of gases.  In other words the velocity distribution function is exp{ -mv2/2kBT } in one dimension. 

 For an ideal gas the internal energy is U = 3/2nRT

R is the gas constant and it has units of energy/mole-K.

kB is Boltzmann's constant and it has units of energy/K.

R is related to Boltzmann's constant by R = NAkB where NA is Avagadro's number.

 

First law: DU = w + q  (integral),  dU = dw + dq (differential)

U is the internal energy , DU is the internal energy change

w is the work done , q is the heat transferred

U is a state function, while w and q are path functions

w = - PdV , PV = nRT , w = -(nRT/V)dV , w = -nRT ln(V2/V1) for a change in volume DV = V2 - V1

The internal energy change is calculated using DU = CvDT or dU = Cv dT (differential form) where Cv is the heat capacity, Cv = (U/T)v.  We can also define a heat capacity at constant pressure, CP = (H/T)P.

For constant pressure processes we define the enthalpy, H = U + PV = U + nRT using the ideal gas law.

For a chemical reaction the relationship is DH = DU + DnRT where n is the number of moles of gas created or consumed in the reaction.  The temperature dependence of the enthalpy is dH = Cp dT

 

Second law: Clausius inequality is a definition of entropy: dS ³ dq/T.  The equal sign applies for a reversible process, and dS > dq/T applies for an irreversible process.  One significance of entropy is that it defines the direction of spontaneous change for processes occuring at constant temperature, volume or pressure.

S = R ln(V2/V1) constant temperature (isothermal)

S = CV ln(T2/T1) constant volume (ischoric)

S = CP ln(T2/T1) constant pressure (isobaric)

The thermodynamic temperature scale is defined by qcold/qhot = Tcold/Thot.

Thermodynamic efficient is h = 1 - Tcold/Thot.  This is maximum theoretical efficiency of a heat engine.

The entropy also has statistical definition S = kBlnW, where W is the statistical weight.  W is the number of configurations available to a system at constant energy.  This was the original definition of Boltzmann and is useful because you reason about chemically important entropy changes in polymers, proteins and solids using this type of model.  The classic example is the CO molecule in crystalline form.  Each CO can exist in one of two possible configurations, CO or OC.  Thus, for a crystal of N CO molecules the number of configurations is 2N.  Thus, the statistical entropy of a CO crystal is S = N kB ln 2.  For exactly one mole of CO molecules S = NAkB ln 2 = R ln 2.

The entropy of mixing is DS = -R(x1lnx1 + x2lnx2)

 

Third Law: The entropy of all perfect crystals is zero at absolute zero.  If a crystal is not perfect such as CO discussed above where two configurations are possible, then there is a residual entropy at absolute zero of temperature.  This permits calculation of the third law entropy, i.e. the entropy relative to absolute zero.

As one approaches T = 0 K, the heat capacity also approaches zero.  Thus, it is theoretically impossible to actually reach 0 K.  This also makes calculation of the entropy difficult near zero K.  However, Debye worked out a theory that says that the heat capacity depends on the cube of temperature near zero Kelvin. 

S = SDebye + CP ln (Ta/TDebye) + DHa/Ta + CP ln (Tb/Ta) + DHb/Tb + …

For phase transition a, b … etc.

 

Free Energy Functions:  To predict the direction of spontaneous change at constant temperature and volume we define the Helmholtz free energy A, as DA = DU - TDS or dA = dU - TdS and Gibb's free energy G, as DG = DH - TDS or dG = dH - TdS.

dG < 0 and dA < 0 for spontaneous processes at constant P,T and V,T, respectively.

dG = 0 and dA = 0 at equilibrium.  This includes phase equilibria.

 

 

 

The Master Equation and Maxwell Relations:

dU = TdS - PdV                                                                    

dH = TdS + VdP                                       H = U + PV

dA = - SdT - PdV                                                A = U - TS

dG = - SdT + VdP                                     G = H -TS , G = A + PV

Since these are all state functions they are exact differentials. 

T = (U/S)V  , P = -(U/V)S  à (2U/SV) = (2U/VS) = (T/V)S = -(P/S)V

T = (H/S)P  , V = (H/P)S  à (2H/SP) = (2H/PS) = (T/P)S = (V/S)P

S = -(A/T)V  , P = -(A/V)T à (2A/SV) = (2A/VS) = (S/V)T = (P/T)V

S = -(G/T)P  , V = -(G/P)T à (2G/SP) = (2G/PS) = (S/P)T = -(V/T)P

The Max well relations come from the cross derivatives.

 

Phase Transitions

The Clapeyron equation that defines coexistence curves in condensed phases is derived from

dP/dT = dS/dV (note that this is the third Maxwell relation)

dP/dT = DS/DV = DH/TDV  à dP = DH/DV(dT/T)  à P2 - P1 = DH/DVln(T2/T1)

For transitions between liquid and vapor (vaporization) and solid and vapor (sublimation) the change in molar volume is so large that we neglect the volume of the condensed phase.  Thus, we use the ideal gas law: DV » Vgas = RT/P.  Note that I assuming that entropy, enthalpy and volume are molar quantities here.

 dP/dT  = PDH/RT2  à dP/P = DH/R(dT/T2)  à ln(P2/P1) = -DH/R (1/T2 - 1/T1).

This is the Clausius-Clayperon equation.

 

Solutions

Raoult's Law: Pj = xjPj*

Henry's Law: Pj = xjkH,j

Dalton's Law: Pj = yjPtotal in the vapor

Construct a two component phase diagram for A and B, PA = xAPA* , PB = xBPB*,  xA+ xB = 1.

P = PA* + xA( PB* - PA*) defines the line between the liquid and the two phase region.

yA = xAPA*/P =  xAPA*/[PA* + xA( PB* - PA*)]

The lever rule states that na/nb = (yA-z)/(z-xA) where a is liquid and b is gas.

Activity aj = gjxj in terms of mole fraction.  The activity coefficient is gj.

 

Colligative properties

Chemical potential for a single component m = Gm where Gm is the molar free energy.

For multiple components mj= mjo + RT ln xj (ideal) or mj= mjo + RT ln aj  (non-ideal).

Using these expressions we derive the freezing point depression and boiling point elevation. 

Osmotic pressure PV = nRT or P = cRT

 

Partition functions

Q = qN/N! for indistinguishable particles

q = qtransqrotqvibqelec

You should know that the energy levels for each of these come from solutions of the Schrodinger equation for translation, rotation, vibration and electronic levels, respectively.

It is important to understand that the partition functions of the kinematic variables depend on temperature dependent population of levels.  Translational levels are nearly continuous. Rotational energy level spacings are less than thermal energy, and the number of accessible levels is typically in the tens or hundreds.  Vibrational levels have a spacing somewhat greater than thermal energy or on the order of thermal energy at 300 K.

It is good to know that thermal energy is about kBT » 200 cm-1 at 300 K (since kB = 0.687 cm-1/K).

Alternatively RT = 2.4 kJ/mol at 300 K since R = 8.314 J/mol-K.