!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! EXAM REVIEW !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
The exam will have two sections. One section that is multiple choice, largely based on
definitions given below in the Chapter Reviews. You will not have an eqn. sheet
provided on this part of the exam. You will hand this in and then there will be a second
part which is a problem solving part. During the problem solving part, you will be
provided with equations that you may need to solve a particular problem. There will be
more equations than you need. You may be asked to derive an eqn. in one or more
instance. You will be expected to know basic equations like the ideal gas law.
Remember if you set up the problem correctly you will get major credit, so you may want
to set up the problems without using your calculator and then go back at the end and
punch in the numbers.
The review sheets follow. They are words and concepts that should be very familiar to
you. Again, I will take the multiple choice questions from these sheets. Also the problem
solving will be problems based on concepts taken from problems assigned for homework,
given on a quiz, and listed on the review sheets. Look over the sheets and ask questions
about them on Wed at the review. Note that I have included the Constants and eqns.
sheet at the bottom which will be included with your exam.
Review for Chapters 1,2,3, and part of 4
Definitions: (The meanings of these words and phrases should be very familiar to
you!)
System
State of the System
Processes
Surroundings
Closed System
Absolute Temp. Scale
Open System
Adiabatic
Isolated System
Reversible Process
Irreversible Process
Kinetic Theory of Gases
Isothermal
Isobaric
Isochoric
mean free path
Macroscopic System
Microscopic System
Thermodynamic State
State Functions or State Variables
Equilibrium State
Kinetic Theory of Gas
Ideal Gas Law
Dalton’s Law of Partial Pressure
Boyles Law
Charles Law
SI Units
P, T Phase Diagram
Dependent Variable
Independent Variable
Equation of State
Extensive Variable
Intensive Variable
Phase Transitions
Gibbs Phase Rule
Component
Heat of Melting (Fusion)
Freezing Point
Heat of Vaporization
Enthalpy of Sublimation
Isotherms on PV diagram
Adiabat on PV diagram
Kinetic Energy
1st Law of Thermodynamics
Probability Distribution
Maxwell-Boltzmann Distribution
Van der Waals Eqn of State
Work
Heat
Hess’ Law
Effusion
Grahams Law of Effusion
Constant Pressure Heat Capacity
Endothermic
Thermochemical Eqn.
Combustion
Molar Internal Energy
Root Mean square speed
H = U+ PV
Ideal Monatomic Gas Cvm = 3/2R
Path Dependent
Expression for dH for Ideal Gas
Molar Internal Energy
Hess' Law
H= U + PV
Cp = Cv + nR
2nd Laws of Thermodynamics
Heat Engine
Potential Energy
Internal Energy
Momentum, Pressure
Average Speed of a Gas (how can you get it)
Virial Eqn.
Compressibility Factor
External Pressure
Adiabatic
Diffusion
Bomb Calorimeter
Constant Volume Heat Capacity
Exothermic
Heat of Formation, Enthalpy of Formation
Average Bond Energies
Collision Frequency
Kirchoff's Law
Ideal Monatomic Gas Cpm = 5/2 R
Isenthalpic
Closed System
Expression for dU for Ideal Gas
Enthalpy
Enthalpy of Formation
dU = q + w
Cpm = Cvm + R
Carnot Heat Engine
Calculations: You should be familiar with the following types of Calculations
Using the Ideal Gas Law, or a form of it such as Boyles or Charles’ Law to Solve
for P,V, T, or n, Use of nonideal eqns. such as virial eqn. or compressibility factor
Use of Grahams law of effusion, calculating average or rms speed, mean free path
Expressing U in terms of Heat and Work, Use of the 1st Law of Thermodynamics
Use and/or derivation of Expressions for Internal Energy, Work, Heat, Enthalpy
of different types of processes such as reversible, isothermal, adiabatic, isobaric,
isochoric from the basic definitions
Use of the Gibbs Phase Rule
Dependence of Enthalpy on Temperature
Use of Heat Capacity to determine enthalpy or internal energy changes
Calculation of H and U for a chemical reaction from heats of formation, heats
of combustions, bond enthalpies, or a more general use of Hess’s Law
Calculation of H or U at an elevated temperature (Kirchoff's Law)
Expressions for dH and dU for an ideal gas
Converting between H and U
Finding H for phase changes or heating a material that not only is heated, but
goes through phase changes
Use of the concept of a State Function in calculations
From Chp 4
General statements that define natural processes talk about the observed efficiencies in
converting heat into work, the direction of heat flow, and the fact that the disorder of the
universe seems to be increasing. These can be formulated into a mathematical statement
of spontaneity which involves the entropy. In any irreversible process the entropy of the
universe increases. Because entropy is a state variable, the entropy change of the system
can be calculated using a reversible process having the same initial and final states as the
irreversible process. In any reversible process the entropy of the universe remains
constant and therefore this constitutes the minimal or maximal work case.
The Carnot efficiency tells about the maximum efficiency realizable for a process
which converts heat to work. The expression for the Second Law of Thermodynamics
Stot > 0 for an irreversible process is strictly applicable to the system and the
surroundings.
REMEBER that if you do your calculation using SI units, your answer will also be
in SI units. Units are your friend.
Constants and Equations
PV = nRT,
Vi/Ti = Vf/Tf,
Z1 = 1 / 2 d2 cav(N/V)
PiVi = PfVf,
Etrans = n 3/2 RT
(P+(an2 /V2 ) (V-nb)=nRT
Pi/Ti = Pf/ Tf
f(c) = 4c2 (m/2kT)3 / 2 exp(-mc2 /2kT)
Z1 = 1 / 2 d2 cav(PNav/RT) Z11 = (2)½ /2 d2 cav(PNav/RT)  = cav/z1 c = (3RT/Mm)1 / 2
c =(2RT/Mm)1 / 2 c = (8RT/(m))1 / 2
F = ma
F = -kx
PE=mgh
KE = ½ mv2
U = Q + W, W = -Pext V, W=  F dl W= - P dV, W = -nRT ln(Vf/Vi) F= C-P+2
W = nRT ln(Pf/Pi) Q = n C  Q = m C  dU = Cv dT,
dH = Cp dT, H=U +PV,
Qp,  U = Qv, c = Cvm /R, Tf/Ti = (Vi/Vf)1/c (PiVi)g = (PfVf)g,
H = U + ngRT
g = Cpm/Cvm,
H = U + (PV) Hrxno(T2) = Hrxno(T1) + T rCpo dT
rxno = prod n Hfo - react n Hfo  S = nR ln(Vf/Vi)  S=-nR ln(Pf/Pi)  = 1- Tc/Th
Cpm = R + Cvm K = oC + 273.15
1atm = 101325 Pa
1 atm = 14.7 psi 1atm = 101325 Pa 1 bar = 105 Pa
g = 9.81 m/s2
R= 0.08206(Latm)/(mol K)
1L = 0.001 m3
R=8.314J/mol K
g=9.81m/s2
1cm3 = 1mL