Chemical Potential (Ch. 5.1)
Relations like dG = SdT + VdP are specifically for individual
substances where the number of moles present is fixed.
In a process, the number of moles of each component may
change, or be redistributed, as the process proceed
,
which
approximates
U
+
P
Vgas
=
U
+
RTngas
So,
U
=
43,900

(8.315)(298.2)(1)
=
41.4
kJ
and
w
=
U

q
=
(8.315)(298.2)(1)
=
2480
J.
S
5.
(a)
0 .9236
0 .7964 = 0.796 J K
Chem 201 Midterm Test Feb. 7, 2011
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0 Time: 70 min. This exam consists of 5 questions.
0 Express your answers in terms of numbers or the given variables; Use R as the
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University of British Columbia
Midterm Examination March 5, 2013
Chemistry 201
Time: 70 min
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Q1) A sample containing two moles of He gas behaves ideally. In state A, it is at (PA, VA, TA).
In the first step of a four step process, it expands isothermally and reversibly to state B, in which
VB = 2 VA. In the second step, it expands adiabatically a
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University of British Columbia
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C205 Lecture Notes  2015
Page 28 of 410
B
Processes by which a system can change:
IRREVERSIBLE Process:
A
The system is not always in equilibrium. The process cannot be reversed
by an infinitesimal change of its properties.
At the end of a cycle, the s
C205 Lecture Notes  2015
Page 37 of 410
Calculation of Expansion Work:
final
In general: dw = Pex dV
w
( P
ex
initial
Pex : the external pressure exerted on the gas,
NOT the internal gas pressure (P Pgas)!
Some Examples:
a) Free Expansion: Pex 0 w 0 U
Problem Set 3
In general, consider all given data good to four significant figures and give your answers to three
figures. For example, use R = 8.315 J K1 mol1 =0.08206 L atm K1 mol1, 1 L atm = 101.3 J, 1 atm
= 101.3 kPa, etc. Note: 1 atm = 760 mm Hg,
Problem Set 5
R = 8.315 J K1 mol1 =0.08206 L atm K1 mol1, 1 L atm = 101.3 J, 1 atm = 101.3 kPa, etc.
_
1.
The decomposition of compound A is studied at a fixed temperature. At time t = 80 min,
only 10% of A is left. What percentage of A has reacted at
ChemicalReaction Equilibrium (Ch 6.16.4)
ChemicalReaction equilibrium is the state in which the
concentrations (or chemical activities) of the reactants and
products have no net change over time.
Usually, it is a dynamic equilibrium in which the rea
Chemical Kinetics (Ch. 21)
Thermodynamics deals with the direction of spontaneous
change for chemical reactions, but says nothing about the
reaction rates.
Chemical kinetics deals with the reaction rates. Therefore, it is
concerned with the details of t
Thermochemistry (Ch. 2.72.9)
Thermodynamics:
(1) In which direction will the reaction go spontaneously?
(2) How much heat will be released?
Changes in a state function, F, do not depend on the details of
the process but only on the initial and final st
Problem Set 5 Answers
1.
Values of [A] for different times are:
t
[A]
0
Co
20
fCo
80 min
0.1 Co
Fraction left at t = 20 min is f, therefore the fraction reacted is 1 ? f
(a) 0 order: C = Co ? k0t, therefore
0.1Co = Co ? k0 80 k0 = 0.9 Co/80
At t = 20 min,
Problem Set 1 Answers
Note: units have been left out in intermediate numerical values below
but should be checked in every calculation.
1.
(a) P
nRT (0.1)(0.08206)( 200)
0.32824 = 0.328 atm
V
5
(0.32824)(101.3) 33.25 = 33.3 kPa
(Use 1 atm = 101.3 kPa)
(b)
5.
Q
0
(1) 1st law:
U
Q
w
U
(2) For ideal gas,
w
P2 (V2 V1 )
CV (T2
nR )
T1
nRP2
P1
nRT2
P2
nRT1
P1
nR T2
P2T1
P1
T1 )
From (1) and (2), we get CV (T2
T2 (CV
P2
T1 )
nR T2
P2T1
P1
CV
Here, P2 2P1 ;
T2 (CV nR ) T1 2nR CV
For ideal gas, CP CV = nR or CP = C
(b)
4.
By considering volume as a function of temperature and pressure, Vm = Vm (T,P), determine a
formula for the slope of a line on a P,T diagram along which the volume is constant (an
isochore), in terms of ( Vm / T)P and ( Vm / P)T. [Ans: the eqn. In
Problem Set 4
In general, consider all given data good to four significant figures and give your answers to
three figures. For example, use R = 8.315 J K1 mol1 =0.08206 L atm K1 mol1, 1 L atm = 101.3
J, 1 atm = 101.3 kPa, etc.
_
1. (a) The reaction i
C205 Lecture Notes  2015
Page 19 of 410
Equipartition Principle (hightemperature limit)
We have just shown that each of the 3 translational degrees of freedom contributes
RT/2 to the translational energy. For an ideal gas, this is the only contribution
C205 Lecture Notes  2015
Page 55 of 410
H =nCP,mT
Considerageneralprocess:
P1 V1 T1
State1(P1,V1,T1) State2(P2,V2,T2)
H1P
isoP
H
Thus,onlytheisoP step
contributestoH.
P1 Vc T2
H2T 0
isoT
P2 V2 T2
The alternative process (1) iso P (2) iso T
P
T
H
H
H
1
2
C205 Lecture Notes  2015
Page 46 of 410
Example: Practice Problem Set #1.Q5
Two moles ideal gas (CV,m = 3R/2) initially at 0 C and 2 atm underwent a full
expansion against a constant external pressure of 1 atm, and did 2 kJ of work.
Calculate U, H, and q
C205 Lecture Notes  2015
Page 181 of 410
BeerLambert Law:
A = log10(T) = log10(I0 / I) = c l
where
= molar extinction coefficient (a constant characteristic
of the sample and an indication of the probability of
the transition)
c = concentration of samp
C205 Lecture Notes  2015
Page 172 of 410
Donnan Membrane Equilibrium General Analysis
Consider a salt AbBa dissolves in a solution:
AbBa = bAa+ + aBb
which establishes an equilibrium across a semipermeable membrane:
Inside
Membrane
Outside
[Cc], [Aa+]in
0. Introduction
physical chemistry establishes and develops the
principles of chemistry
concepts used to explain and interpret observations
on the physical and chemical properties of matter
central theme:
systems
states
processes
topics of physical che
4. Phase Equilibria in Pure Substances
phase: uniform throughout in chemical composition
and physical state
Second Law = dG T,P 0
equilibrium if and only if dG T,P = 0
consider a system with two phases, say and :
G = n + n
assume an amount dn goes from p
1. The Properties of Gases
Notation: (i) I use an overbar to denote molar quantities; the textbook uses a subscript m. Example: molar volume, these notes V , textbook Vm. (ii) I use a
capital P for pressure; the textbook uses p .
simple system to learn th