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Some Basics
PV= nRT
Energy units joules/ calories
Mole, and Avogadro’s#,
=============================
Boyle in 1662 observed that as the pressure is increased on
a gas its volume decrease
proportionally to the weight
or pressure . A plot of the data
gave a hyperbola.
pressure
volume
At a const. temp. twwwwa433he pressure x volume gave
a constant; so P proportional to 1/V
Or V x P = const.
Data for Agon gas
Temp
Volume
o
C
Liters
100
2.00
100
1.00
100
0.50
P atm
15.28
30.52
60.99
Pres. x Vol
100
0.333
91.59
-50
-50
-50
-50
2.00
1.00
0.50
0.333
8.99
17.65
34.10
49.50
Charles law (1789) and Absolute zero- The volume
of a gas was measured as its temperature was
decreased. The volume decreased linearly with
temperature. If the centigrade scale is used, than the
temperature point where the volume is equal to zero
is –273.16oC
volume
temp
from this Charles showed that temp is proportional to volume
ie.
VT
and
V  1/P
so
V  T/P, and …..?
Dalton in 1801 showed that if uncreative gases were
mixed, their volumes were additive. This means that the
total pressure exerted by all of the gases is equal to the
sum of the individual pressures that each gas exerts, ie.
its partial pressure
P total = P1 + P2 + P3
============================
Gay-Lussac’s in 1809 observes that when equal volumes of
hydrogen and chlorine gas are reacted in UV light and gave
the same total volume of HCL gas a product.
We might expect that
10 cc Hydrogen + 10 cc Chlorine  20 cc HCl
If a molecule of hydrogen contains one atom, and
chlorine one atom, than in this reaction scheme than ½
of volume of HCl would be produced compared to the
starting volumes of H and Cl
H + Cl  HCl
This does not work because we have
If we assume there are two atoms /molecule of H
and Cl
Then
H2 + Cl2
 2HCl
This suggests the very important notion that H2 and Cl2 are
molecules of these gases. Gay-Lussac’s state their law as:
at a given pressure and temperature, gases combine in
simple proportions by volume, and the volume of any
gaseous product bears a whole-number ratio to that of any
gaseous reactant
This implies that there is a simple relationship between
volume and the number of molecules that occupy that
volume and Avogadro in 1811 was the 1st to propose that
equal volumes of gases at the same temperature and
pressure contain equal numbers of molecules.
If we know the weight of a given volume of gas, we can
determine its molecular weight. For example one liter of
hydrogen weighs 0.09 grams, and one liter of oxygen
weighs 1.43 grams. Then the weight of hydrogen
compared to oxygen is 0.09/1.43 or 0.063 time as heavy
as each oxygen molecule. If the molecular weight by
convention of oxygen is taken as 32, than the molecular
weight of hydrogen is 2.016.
At zero oC and one atmosphere 32.0 grams of
oxygen (one mole) occupies 22.4 liters, and
ideally 2.016 grams of hydrogen occupies 22.4
liters too. This is called the molar volume.
Zero oC and one atmosphere is called STP.; so in your
own words what is a mole of oxygen?
Now let’s go on to Avogadro’s number, or the
number of molecules in one mole and see how it
can actually be determined.
When Radium { 22688Ra }decomposes it emits an alpha
particle or a helium nucleus
226
88Ra
2 alpah particles { 21He+ } + radon {22286Ra}
the alpha particles can be counted with a Geiger
counter, so we know how many are forming
alpha particles quickly pick up electrons and form
stable helium
`
2
+
1He
+ electon  helium { 21He}
the volume of helium that is formed can be measured
In an experiment (my Freshman Chem. book) 1.82 x1017
alpha particles are measured and they occupy a space of
0.00734 ml at 19oC. We already know that one mole of an
ideal gas occupies 22.4 liters of space at STP or zero oC and
one atmosphere.
So first compute the volume occupied by one mole at 19oC
in liters
For your homework calculate Avogadro’s # from these
data and compare it to the accepted value.
=====================================
Basic Physics Force = m a
vel = distance/time = s/t = v
acceleration = v/t = s/t2 = a
pressure = Force/area
Regression analysis??
y=mx+b
Some Thermodynamics:
The First Law
U2 - U1 = q - w
work
change in
internal energy
of an object
heat
object
reservoir
b
U = q1-w1
U = q2-w2
For example one gram of H2O at 25oC is
evaporated and condensed; the condensed gram
of water at 25oC will have the same internal
energy as it did previously.
If only pV work is done and the pressure of the
system is constant
wrev =  pdV
What is the work of a reversible expansion of a
mole of an ideal gas at 0oC from 2.24 to 22.4
liters?
Wrev   VV 12 pdV
pV =
Wrev  nRT
nRT

V2
dV / V  nRT ln (
V1
V2
)
V1
Wrev = 1mole x1.987 cal K-1 mole-1
x 273 K x 2.303 log (22.4/2.24)
Wrev = 1.25 Kcal mole-1
Internal energy, heat and work
What is the energy required to vaporize water at
100oC???
when one mole of water is vaporized at 100oC the work is
w = p V = RT = 1.987 cal K-1 mole-1 x 373.15K
w= 741.4 cal mole-1
The energy or heat required to vaporize water at 100oC
requires energy to separate the liquid molecules;
that is 529.7 cal g-1
q = 18.02 g mole-1 x 539 cal g-1 = 9725 cal mole-1 ;
For a mole of water, the internal energy U = q - w
U= 9725 cal mole-1 - 741 cal mole-1
U = 8984 cal mole-1
Enthalpy
U = q -  pV) at constant pressure
q= (U2 + pV2) - (U1+ pV1)
We define U + pV as the enthalpy, H
q = H2-H1 = H
or the heat adsorbed in a process at constant
pressure
Heat Capacity
Heat Capacity, C = ratio of heat absorbed/mole to the
temperature change = q/T
At constant pressure
q = U+pV = H
Cp = dH/dT
i.e. the calories of heat
adsorbed/mole by a substance/oC
so
H= Cp(T2-T1)
At constant volume
U = q - pV
U = q
Cv = dU/dT
What is the relationship between Cp and Cv?
The Second Law
Lord Kelvin (1824-1907): “It is impossible by a
cyclic process to take heat from a reservoir
and convert it into work without at the same
time transferring heat from a hot to a cold
reservoir”
Clausius: “It is impossible to transfer heat
from a cold to a hot reservoir without at the
same time converting a certain amount of the
work to heat”
i.e. work can only be obtained from a
system when it is not at equilibrium
It can be shown (see any p-chem book, page
262 Baird) that the max. efficiency of a
sequence of isothermal and adiabatic process
is
eff = (TH-TL)/TH = (qH + qL)/qH
rearranging
qH qL

0
TH TL

qi
0
Ti

dqrev
0
T
define
dS = dq/T
and
S2  S1  S 

2
1
dqrev
T
at absolute zero the entropy is assumed to be
zero
Consider 1 mole of H20 (l)---> H20 g at 100oC
SH20= dq/T = 1/Tdq = 1/T

n 1
Hdn
n 0
= dHvap/T = 9,720 cal/373K= + 26cal/degK mole
ssurroundings= a negative 26cal/degK

S total = zero
When spontaneous processes occur
there is an increase in entropy
When the net change in entropy is zero the system is
at equilibrium. If the calculated entropy is negative the
process will go spontaneously in reverse.
S = dq/T
1. What is it about a gas that makes it have more
entropy when it is expanded, then when it is
compressed or in the liquid state?
Let’s say that in the reaction of
A ---> B
B has more entropy than A
2. What is it about B that gives it more entropy?
1st consider a box with a 4 pennies; if we place them
with heads up and then shake the box, we get:
4 heads, 0 tails
3 heads, 1 tail
2 heads, 2 tails
1 head, 3 tails
0 heads, 4 tails
# combinations
1
4
6
4
1
we might consider this to be the normal state, or
equilibrium state, because there are more
combinations to “go to”
Free Energy
The concept of free energy comes from the
need to simultaneously deal with the
enthalpy energy and entropy of a system
G = H -TS
G = U+PV - TS
dG= dU + PdV + VdP -TdS -SdT
dH = dU +PdV
at const temp and pressure
G= H -TS
What is the free energy for the process of converting
1mole of water at 100oC and one atm. to steam at one
atm.
H= H vap
Svap = 1/T  dq = Hvap/T
TS = Hvap
G= Hvap - TS
G= Hvap - Hvap= 0
Equilibrium Constants
dG= dU +VdP + pdV -TdS -SdT
for a reversible process
TdS = dq
dU -dq+dw = 0
so dG= +VdP -SdT
at const. temp
(G/P)T = V; and if const. temp is stated all the
time
dG/dP= V
dG =nRT dP/P
G2 -G1 = nRT ln(P2/P1)
At standard state
G = Go + nRT ln(P)
G = Go + nRT ln(P)
for a reaction
A + B--> C + D
for A
we have GA =GoA +RT lnPA
it is the free energy of the products minus the
reactants that is of interest
G =Gprod - Greact
for reactants A and B
GAB = GoA + GoB+ RT lnPA + RT lnPB
For A + B--> C + D
G  Go  RT ln
(PC ) (PD )
(PA )(PB )
if the reaction goes to completion
G= zero
Go  RT ln
(PC )(PD )
(PA )(PB )