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									Some Basics

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.



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 -50 -50 -50 -50

0.333 2.00 1.00 0.50 0.333

91.59 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



from this Charles showed that temp is proportional to volume

ie. and so

VT V  1/P 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

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??


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?

2 Wrev   V 1 pdV V

pV =



V2  nRT dV / V  nRT ln ( ) V1 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

define and

dqrev 0 T

dS = dq/T

S2  S1  S 




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



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: # combinations 1 4 6 4 1

4 heads, 0 tails 3 heads, 1 tail 2 heads, 2 tails 1 head, 3 tails 0 heads, 4 tails

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 )

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