Transcript No Slide Title

Introduction to First Law
The state of system changes when heat is transferred to or from the
system or work is done.
E2-E1= q + w (with the proper sign conventions)
But it was not obvious early on that this “Law” held.
Joule in around 1843 did the definitive experiment:
T
Essentially, the amount of heat
generated in the vessel by the
movement of the paddle is
exactly the potential energy lost
by the weight.
The First Law
• The total energy of the system plus surroundings is
conserved.
• Energy is neither created nor destroyed.
– Energy can be transferred
• Heat (q)
• Work (w)
– Change in energy is equal to the sum of heat and work:
E  E2  E1  q  w
In a public lecture, Joule rejoiced in this understanding: "...the phenomena of
nature, whether mechanical, chemical or vital, consist almost entirely in a
continual conversion of attraction through space [PE], living force [KE], and
heat into one another. Thus it is that order is maintained in the universe-nothing
is deranged, nothing ever lost, but the entire machinery, complicated as it is,
works smoothly and harmoniously. ...every thing may appear complicated and
involved in the apparent confusion and intricacy of an almost endless variety of
causes, effects, conversions, and arrangements, yet is the most perfect regularity
preserved...."
First Law Thought Experiment
• On a clear night, the temperature can drop very quickly.
Where does the heat go?
– Radiation energy in one photon of light: E = hn
•
•
h = Planck’s constant = 6.6261 x 10-34 Js
n = frequency in s-1
– Blackbody radiation: E m-2 s-1 = sT4
• s = Stefan-Boltzmann constant = 5.67 x 10-8 J m-2 s-1 K-4
• Pay attention to units! Energy per unit area(m2) per unit time (s)
– Blackbody radiation is a daily experience
•
•
•
•
“Red” hot objects look red due to blackbody radiation
Thermos silver in color to reflect blackbody radiation (better insulation)
Light bulb (incandescent): w (electrical) = EIt = q + Nhn
Fluorescent bulb? Bioluminescence? (chemical, not blackbody)
Thermodynamics and Photosynthesis
Inputs
Outputs
H2O,
CO2
Biomass, O2,
Heat
Light energy (hn)
minerals
phosphate
System
This is a fully open system.
Thermodynamics and Photosynthesis
Inputs
Outputs
H2O,
CO2
hn
minerals
phosphate
Biomass, O2,
Heat
We assume the major conversion
process is:
System
H2O + CO2
O2 + (CH2O)
The enthalpy change for the process is H°= 485 Joules/mol
(from H° tables in back of text, for example)
In order to get this process to go we need: light, and a catalytic
system. This takes place in the chloroplast of the plant.
Thermodynamics and Photosynthesis
We assume the major conversion process is:
H2O + CO2
O2 + (CH2O)
H°= 485 Joules/mol
It has been shown that it takes about 8-9 photons to make one O2.
The first law tells us that the total energy input must equal total energy
output.
Photon Energy = Chemical Energy + Heat
Efficiency= Chemical Energy/Photon Energy
Thermodynamics and Photosynthesis
So to calculate the efficiency of production of 1 mole of O2:
1 mole of photons= 1 einstein
Photon Energy= (8-9 einsteins)(6*1023 photons/mol) hn
h= planck’s constant= 6*10-34 Js
n= c/=(3*108 m/s)/(680*10-9 m)
Photon Energy= 1400-1570 kJ
Chemical Energy= 485 kJ/mol * 1 mol O2
%Efficiency= 485/1570 * 100 = 31%
Path Dependence and Independence
q and w are path dependent variables.
State 1
A state change that occurs over the blue path
has qblue, wblue.
Over the red path we get qred, wred.
State 2 All we know from the 1st Law is
qblue+wblue= qred+wred
State Variables are NOT Path Dependent
State 1
2
dE

E  E2  E1  dE
1
 dE  0
State 2
dE is called an exact differential since it may be directly integrated
to produce the variable.
Heat and Work are inexact differentials. It is not, in general possible
to write down equations like that for energy.
Though for adiabatic paths: dw appears like an exact differential.
And for isochoric paths: dq appears like an exact differential.
Measuring Energy
So what is energy as a function of temperature (holding volume constant)?
E= q + w
If we consider only pressure volume work then at constant volume…
E= q
And since we know that CV= (dq/dT)V
T2
E 
 C dT
v
T1
This is valid for solids liquids and gases, and for pure materials and for mixtures.
Measuring Enthalpy
Remember H E + PV
Since both E and PV are state properties, H must be a state
property. It is therefore an exact differential!
Thus basic calculus applies:
dH= dE + d(PV)= dE + PdV + V dP
Since for reversible pressure-volume systems,
dE= dq + dw= dq - PdV
dH= dq - VdP
So for constant volume pressure processes:
H= qP
Reversible or Irreversible State Change
Reversible: State is changed by differential amounts along a path. At
any moment a small change in the opposing force will alter the
direction of the state change.
Irreversible: “All at once”-- the method of the change is such that it
is not possible to reverse the direction.
Heat
State A
Ethermal
=+
State B
Emech.
=0
Ethermal
=0
Emech.= +
Reversibility of Paths: PV work
Here in the irreversible case you would have to do significant
mechanical work to restore the initial state.
In the reversible case, a small differential change in the weight
can cause a reversal.
State A
Heat
Ethermal
0
Ethermal
=-
State B
Emech.
=0
Emech.
=+
Work from Reversible vs. Irreversible Processes
State 1
State 1
Ideal Isotherm
P
P
P
State 2
V1
V
V2
Constant pressure expansion
State 2
V1
Isothermal reversible expansion

w   PdV  
V1
If the opposing pressure is 0,
w = 0 thus E  q
V2
V
V2
w  Pop dV
nRT
V
V2

V1
Pop  Psys
nRT
dV
V
V 
w  nRT ln V2   ln V1   nRT ln  2 
V1 
Phase Changes
Physical changes
fusion (melting)
freezing
vaporization
condensation
sublimation
= solid
= liquid
= liquid
= gas
= solid
to
to
to
to
to
liquid
solid
gas
liquid
gas
Heat Capacity and Phase Changes
During a phase change, the heat capacity changes discontinuously:
Liquid
Solid
CP
Interactions
Gas
Interactions+translation
translation
T1
Tfusion
T2
Tvap
C=Ctrans+ Crot+ Cvib+ Celec+...
T3
Energetics of Phase Transitions
Consider vaporization of water at particular temperature
There are many paths to take. Let’s consider primarily reversible paths, at constant
pressure. (Basic lab conditions).
For a constant pressure process:
wP= -P V= -P (Vphase2-Vphase1)
We also know
H= qp
Since we are talking about constant P, and only allowing PV work
E= H - P(V)
Energetics of Phase Transitions
H2O (l), T1= 35 °C,
P=1 atm
H(35 °C)
TA
H2O (l), T2= 100 °C,
P=1 atm
H2O (g), T1= 35 °C,
P=1 atm
TB
H(100 °C)
H2O (g), T2= 100 °C,
P=1 atm
Assuming CP constant over this range:
H(35 °C)= CP(l) TA + H(100 °C) + CP(g) TB
= H(100 °C) + (CP(g) - CP(l))(- TA)
Since -TA= TB
H(35 °C)= H(100 °C) +  CP (-65 °C)
If we calculate this from tabular value we get 2443 kJ/kg.
This is how sweating cools you.
Phase Changes and Volume Considerations
Freezing a liquid to a solid:
E = Hfreezing - P V
(H E + PV)
Here you simply can’t ignore V of either phase. But difference between E and
H will be small.
Vaporizing a liquid to a gas:
E = Hvaporization + (PV)liquid-(PV)gas
 Hvaporization - nRT
Here the volume of liquid can be approximated as so much smaller than the
volume of gas produced that it can be ignored.
E and H will be very different!
Chemical Changes
Chemical changes
Overall chemical reactions
Making and breaking of molecular bonds
We are now going to consider a chemical changes:
nAA + nB B +….  nCC + nD D +….
We will mostly be considering changes at constant P so that the
enthalpy of the reaction is equal to the heat evolved.
Two Basic Rules of Reaction Enthalpy
We generally speak about reaction enthalpy because most chemical
process occur at constant pressure, thus, the heat generated by the
reaction is a direct measure of the enthalpy.
Rule 1: The enthalpy of a particular reaction at standard temperature
and pressure is given by:
H 
H
products
products 
H
react ants
react ants
Rule 2: The enthalpy of a particular overall reaction can be derived
by summing the enthalpy of a set of subreactions
(Hess’s law of heat summation)
Example: Rule 1
Rule 1:
C(graphite) + O2 (g)  CO2 (g)
H= HCO2 -HC(graphite)-HO2= -393.51 kJ/mol
Note: HC(graphite)=HO2=0
Note also that this is at Standard Temperature and Pressure:
H= H°298K
Example: Rule 2
Rule 2: Making Diamonds from Pencils
We want the enthalpy for
C(graphite)  C(diamond) H=?
But we have:
C(graphite) + O2 (g)  CO2 (g)
H°298K= -393.51 kJ/mol
C(diamond) + O2 (g)  CO2 (g)
H°298K= -395.40 kJ/mol
Subtraction gives the correct overall reaction. So
H= (HCO2 -HC(graphite)-HO2)-(HCO2 -HC(diamond)-HO2)
= HC(graphite)- HC(diamond)
= -393.51+395.40= 1.89kJ/mol
Vocabulary
When
H<0
H>0
H=0
Reaction is Exothermic
Reaction is Endothermic
Reaction is Thermoneutral
Generally, the more exothermic the reaction the more likely a reaction
will occur spontaneously… but there are other things to consider.
We will have to consider whether or not the molecules are likely to be
in a configuration that allows the reactions to occur for one.