1. 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 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.

2. 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:

3. 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...."

4. 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 =  T 4 s = Stefan-Boltzmann constant = 5.67 x 10 -8 J m -2 s -1 K -4 Pay attention to units! Energy per unit area(m 2 ) 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)

5. Thermodynamics and Photosynthesis H 2 O, CO 2 System Inputs Biomass, O 2, Heat Outputs Light energy (h ) minerals phosphate This is a fully open system.

6. Thermodynamics and Photosynthesis H 2 O, CO 2 System Inputs Biomass, O 2, Heat Outputs h minerals phosphate We assume the major conversion process is: H 2 O + CO 2 O 2 + (CH 2 O) 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.

7. Thermodynamics and Photosynthesis We assume the major conversion process is: H 2 O + CO 2 O 2 + (CH 2 O)  H°= 485 Joules/mol It has been shown that it takes about 8-9 photons to make one O 2. 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

8. Thermodynamics and Photosynthesis So to calculate the efficiency of production of 1 mole of O 2 : 1 mole of photons= 1 einstein Photon Energy= (8-9 einsteins)(6*10 23 photons/mol) h h= planck’s constant= 6*10 -34 Js = c/ =(3*10 8 m/s)/(680*10 -9 m) Photon Energy= 1400-1570 kJ Chemical Energy= 485 kJ/mol * 1 mol O 2 %Efficiency= 485/1570 * 100 = 31%

9. 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 q blue, w blue. Over the red path we get q red, w red. All we know from the 1st Law is q blue +w blue = q red +w red

10. 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.

11. 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 C V = (dq/dT) V This is valid for solids liquids and gases, and for pure materials and for mixtures.

12. 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= q P

13. Reversible or Irreversible State Change State A State B  E thermal = + Heat  E mech. = 0  E thermal = 0  E mech. = + 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.

14. Reversibility of Paths: PV work State A State B  E thermal  0  E mech. = 0  E thermal = -  E mech. = + 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.

15. Work from Reversible vs. Irreversible Processes State 1 State 2 State 1 State 2 V V P P Ideal Isotherm V1 V2 Constant pressure expansion Isothermal reversible expansion V2 If the opposing pressure is 0, w = 0 thus

16. Phase Changes Physical changes fusion (melting)= solid toliquid freezing= liquid to solid vaporization= liquidto gas condensation= gasto liquid sublimation= solidto gas

17. Heat Capacity and Phase Changes During a phase change, the heat capacity changes discontinuously: T2T2 T fusion T1T1 T3T3 T vap CPCP Solid LiquidGas Interactions Interactions+translation translation C=C trans + C rot + C vib + C elec +...

18. 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: w P = -P  V= -P (V phase2 -V phase1 ) We also know  H= q p Since we are talking about constant P, and only allowing PV work  E=  H - P(  V)

19. H 2 O (l), T 1 = 35 °C, P=1 atm H 2 O (l), T 2 = 100 °C, P=1 atm H 2 O (g), T 1 = 35 °C, P=1 atm H 2 O (g), T 2 = 100 °C, P=1 atm  H(35 °C)  H(100 °C) Assuming C P constant over this range:  H(35 °C)= C P (l)  T A +  H(100 °C) + C P (g)  T B =  H(100 °C) + (C P (g) - C P (l))(-  T A ) Since -  T A =  T B  H(35 °C)=  H(100 °C) +  C P (-65 °C) If we calculate this from tabular value we get 2443 kJ/kg. This is how sweating cools you. TATA TBTB Energetics of Phase Transitions

20. Phase Changes and Volume Considerations Freezing a liquid to a solid:  E =  H freezing - 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 =  H vaporization + (PV) liquid -(PV) gas   H vaporization - 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!

21. Chemical Changes Chemical changes Overall chemical reactions Making and breaking of molecular bonds We are now going to consider a chemical changes: n A A + n B B +….  n C C + n D D +…. We will mostly be considering changes at constant P so that the enthalpy of the reaction is equal to the heat evolved.

22. 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)

23. Example: Rule 1 Rule 1: C(graphite) + O 2 (g)  CO 2 (g)  H= H CO2 -H C(graphite) -H O2 = -393.51 kJ/mol Note: H C(graphite) =H O2 =0 Note also that this is at Standard Temperature and Pressure:  H=  H° 298K

24. Example: Rule 2 Rule 2: Making Diamonds from Pencils We want the enthalpy forC (graphite)  C (diamond)  H=? But we have: C (graphite) + O 2 (g)  CO 2 (g)  H° 298K = -393.51 kJ/mol C (diamond) + O 2 (g)  CO 2 (g)  H° 298K = -395.40 kJ/mol Subtraction gives the correct overall reaction. So  H= (H CO2 -H C(graphite) -H O2 )-(H CO2 -H C(diamond) -H O2 ) = H C(graphite) - H C(diamond) = -393.51+395.40= 1.89kJ/mol

25. Vocabulary When  H<0Reaction is Exothermic  H>0Reaction is Endothermic  H=0Reaction 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.