Introduction to First Law The state of system changes when heat is transferred to or from the system or work is done. E 2 -E 1 = q + w (with the proper.

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.