introduction to first law the state of system changes when heat is transferred to or from the system...
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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:
TEssentially, 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:
ΔE =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 regularitypreserved...."
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 = h
• h = Planck’s constant = 6.6261 x 10-34 Js
• = frequency in s-1
– Blackbody radiation: E m-2 s-1 = T4
• 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 + Nh• Fluorescent bulb? Bioluminescence? (chemical, not blackbody)
Thermodynamics and Photosynthesis
H2O, CO2
System
Inputs
Biomass, O2, HeatOutputs
Light energy (h)mineralsphosphate
This is a fully open system.
Thermodynamics and Photosynthesis
H2O, CO2
System
Inputs
Biomass, O2, HeatOutputs
hmineralsphosphate
We assume the major conversion process is:
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 + HeatEfficiency= Chemical Energy/Photon Energy
Thermodynamics and Photosynthesis
So to calculate the efficiency of production of 1 mole of O2:
1 mole of photons= 1 einsteinPhoton Energy= (8-9 einsteins)(6*1023 photons/mol) h
h= planck’s constant= 6*10-34 Js= c/=(3*108 m/s)/(680*10-9 m)
Photon Energy= 1400-1570 kJChemical Energy= 485 kJ/mol * 1 mol O2
%Efficiency= 485/1570 * 100 = 31%
Path Dependence and Independence
State 1
State 2
q and w are path dependent variables. A state change that occurs over the blue path has qblue, wblue.
Over the red path we get qred, wred.
All we know from the 1st Law is qblue+wblue= qred+wred
State Variables are NOT Path Dependent
State 2
State 1
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.
ΔE =E2 −E1 = dE1
2
∫
dE=0∫
dE
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
This is valid for solids liquids and gases, and for pure materials and for mixtures.
ΔE = CvdTT1
T2
∫
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 - PdVdH= dq - VdP
So for constant volume pressure processes: H= qP
Reversible or Irreversible State Change
State A
State B
Ethermal
= +
Heat
Emech.
= 0
Ethermal
= 0
Emech.= +
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.
Reversibility of Paths: PV work
State A
State B
Ethermal
0
Emech.
= 0
Ethermal
= -
Emech.
= +
Heat
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.
Work from Reversible vs. Irreversible Processes
State 1
State 2
State 1
State 2
V V
P P
Ideal Isotherm
V1 V1 V2
Constant pressure expansion Isothermal reversible expansion
V2
w=PopdV
If the opposing pressure is 0,w = 0 thus ΔE =q
P =nRTV
w=− PdVV1
V2
∫ =−nRTV
V1
V2
∫ dV
w=−nRT lnV2( )−ln V1( )( ) =−nRTlnV2
V1
⎛
⎝ ⎜
⎞
⎠ ⎟
Pop =Psys
Phase Changes
Physical changes
fusion (melting) = solid to liquidfreezing = liquid to solidvaporization = liquid to gascondensation = gas to liquidsublimation = solid to gas
Heat Capacity and Phase Changes
During a phase change, the heat capacity changes discontinuously:
T2TfusionT1 T3Tvap
CP
Solid Liquid Gas
InteractionsInteractions+translation
translation
C=Ctrans+ Crot+ Cvib+ Celec+...
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)
H2O (l), T1= 35 °C, P=1 atm
H2O (l), T2= 100 °C, P=1 atm
H2O (g), T1= 35 °C, P=1 atm
H2O (g), T2= 100 °C, P=1 atm
H(35 °C)
H(100 °C)
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.
TA TB
Energetics of Phase Transitions
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 reactionsMaking 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:
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)
ΔH = Hproducts
products∑ − Hreactants
reactants∑
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/molC(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 Reaction is Exothermic H>0 Reaction is Endothermic H=0 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.