Miscellaeous Notes The ed is ear do t get behid. All Excuses must be take to 233 Loomis before 4:15, Moday, April 30. The PHYS 213 fial exam times are * 8-10 AM, Moday, May 7 * 8-10 AM, Tuesday, May 8 ad * 1:30-3:30 PM, Friday, May 11. The deadlie for chagig your fial exam time is 10pm, Moday, April 30. Homework 6 is due Tuesday, May 1 at 8 am. (NO late turi). Course Survey = 2 bous poits (soo to accessible i SmartPhysics) Lecture 18, p 1
Lecture 18 Review & Examples Lecture 18, p 2
Chemical Potetial with Potetial Eergy The potetial eergy per particle, PE, makes a additioal cotributio N PE to the free eergy: F = U - TS. So, the chemical potetial (m = df/dn) gais a additioal cotributio: d(n PE)/dN = PE. It s the eergy that oe particle adds to the system. Examples: Atom i gravity: m kt l PE = N/V particle desity T m kt l T mgh Molecule with bidig eergy, D: m kt l T D Lecture 18, p 3
Simple example: The Law of Atmospheres (1) Questio: How do the gas desity ad pressure p of the gas vary with height from the earth s (or ay plaet s) surface? Compare regio 2 at height h with regio 1 at height 0. (equal volumes) For every state i regio 1 there s a correspodig state i regio 2 with mgh more eergy. Assume that the temperature is i equilibrium (T 1 =T 2 ). h Pressure: p(h) p 2, 2 h p 1, 1 0 0 Earth s surface p(0) Write the ideal gas law like this: p = kt. ( N / V) Lecture 18, p 4
The Law of Atmospheres (2) The two regios ca exchage particles (molecules), so imagie that they are coected by a arrow tube. The rest of the atmosphere is the thermal reservoir. p 2 N 2 Equilibrium: m 1 m 2 Chemical potetial (ideal gas): Solutio: kt l 1 Q kt l 2 1 2 1 m kt m 1 2 l 1 Q kt l 2 Q mgh p 1 N 1 Ideal gas law: p/ = kt (T 1 = T 2 i equilibrium) h Lecture 18, p 5
The Law of Atmospheres (2) The two regios ca exchage particles (molecules), so imagie that they are coected by a arrow tube. The rest of the atmosphere is the thermal reservoir. p 2 N 2 Equilibrium: m 1 m 2 Chemical potetial (ideal gas): Solutio: 1 2 kt l kt l mgh Q kt l 2 1 2 1 mgh e Q mgh / kt m kt m 1 2 l 1 Q kt l 2 Q mgh p 1 N 1 Ideal gas law: p/ = kt (T 1 = T 2 i equilibrium) p p 2 1 e mgh / kt (a rigorous derivatio of our earlier result.) h Lecture 18, p 6
ACT 2: Chemical Potetial of Diatomic Gas How does icludig the rotatioal eergy ad etropy of H 2 chage its m, compared to what you d have if it were moatomic? I other words, what is the effect o F per molecule? A) No chage B) Decreases m C) Icreases m D) Not eough ifo Lecture 18, p 7
Solutio How does icludig the rotatioal eergy ad etropy of H 2 chage its m, compared to what you d have if it were moatomic? I other words, what is the effect o F per molecule? A) No chage B) Decreases m C) Icreases m D) Not eough ifo Molecular rotatio icreases U. Icreases F. It also icreases S (more microstates). Decreases F. Which effect is larger? The additio of rotatio modes must decrease F!!! (proof o ext slide) If F were ot lowered whe the molecules rotated, they would ot rotate! What does this do to equilibrium? m H2 is ow less tha 2m H, so the reactio will proceed i the directio of icreasig H2, util equality is restored. m kt l T D Lecture 18, p 8
Cotributio of Iteral Modes to m Proof that iteral modes lower F: For ideal gases, the iteral cotributios to F are idepedet of the other cotributios. F it /N is their cotributio to m=df/dn C v 7/2 Nk Vibratio 5/2 Nk Rotatio F it = U it - TS it 3/2 Nk Traslatio We ca calculate F it by rememberig that the iteral modes cotribute to C v (T): C ( ) T T V(it) Uit C V(it) ( )d Sit d 0 0 10K 100K 1000K T it it it V(it) 0 T T F U TS 1 C ( )d 0 This is always true, because C V is always positive. So the thermal excitatio of the iteral modes of a molecule always lowers its m, ad thus icreases its equilibrium cocetratio. Lecture 18, p 9
Work from Free Eergy Due to No-equilibrium m Mechaical Work (isothermal expasio of N particles at temp T) 1 2 Quasi-static expasio (a exteral force o the pisto prevets it from beig free expasio): vacuum Thermal Bath, T W by = -DF = -DU + TDS = NkT l(v f /V i ) Costat T DU = 0 (ideal gas) Look familiar? Same old physics, differet laguage. Ca we use these cocepts to do somethig ew??? Lecture 18, p 10
ACT 1 Cosider this situatio. Two differet ideal gases, at equal pressures. If oe starts o the left, ad oe o the right, each will expad to fill the volume. N G N P Thermal Bath, T I his situatio, how much work, if ay, could be extracted? A) 0 B) (N G N P ) kt C) (N G +N P ) kt*l(2) Lecture 18, p 11
Solutio Cosider this situatio. Two differet ideal gases, at equal pressures. If oe starts o the left, ad oe o the right, each will expad to fill the volume. N G N P Thermal Bath, T I his situatio, how much work, if ay, could be extracted? A) 0 B) (N G N P ) kt C) (N G +N P ) kt*l(2) Each of the compoets idividually loses free eergy i the expasio (after the expasio, there s obviously o loger the possibility of expadig ad doig work!); therefore, each compoet idividually could do work up to DF = N kt l2. Ca we actually get work out of this system? Lecture 18, p 12
How to Get the Workout Semipermeable membrae that oly lets gree molecules through. N G N P Thermal Bath, T Equilibrium: N G N P Thermal Bath, T Both gases will have expaded to fill the etire volume. The purple molecules will have doe work, W = N p ktl2, o the membrae (quasistatic expasio). The free eergy of the gree molecules is lost (free expasio). Questio: Ca we get the gree gas to do work as well? Lecture 18, p 13
Fuel Cells Hydroge gas. You pay for this. H 2 Oxyge gas (air). No cost. O 2 Aode Cathode Platium catalyst H 2 2H2(p + e - ) Semipermeable membrae Oly protos ca pass. p + O 2 +4p + 4e 2H 2 0 + eergy Electric circuit e - Load (e.g., motor) H 2 O Water. The waste product. The fuel cell requires cocetratio gradiets i two places: At the platium catalyst. Keep the cocetratio of p + ad e - below equilibrium value, so the reactio will cotiue. Proto diffusio out of the aode (through the membrae) does this. Betwee the cathode ad aode. Keep the proto (ad electro) cocetratio low o the cathode side, so diffusio will cotiue. Water productio does this. Lecture 18, p 14
Fuel Cells (2) The semipermeable membrae forces the electros to take a differet path to the low cocetratio regio. This path, through the load, is where we get useful work from the fuel cell. Ulike typical heat egies, which first bur fuel to make heat, there is o coversio to heat. o Carot efficiecy limit. The platium catalyst speeds the H 2 2(p + e - ) reactio by icreasig the H 2 cocetratio locally (at the surface). We ll see ext week that Fuel cells are used i specialty applicatios. However: 2 2 2 K Top executives from Geeral Motors Corp. ad Toyota Motor Corp. Tuesday expressed doubts about the viability of hydroge fuel cells for mass-market productio i the ear term ad suggested their compaies are ow bettig that electric cars will prove to be a better way to reduce fuel cosumptio ad cut tailpipe emissios o a large scale. (WSJ, 3/5/2008) p H e Lecture 18, p 15
ACT/Discussio: Covertig Fuel to Work The work output comes from the excess free eergy (above the equilibrium value) of the startig chemicals (the fuel). I a atmosphere cotaiig O 2, H 2 is fuel. There s lot s of H 2 i the H 2 O i the oceas. Do we therefore have plety of fuel? A) yes B) o C) maybe Lecture 18, p 16
Solutio The work output comes from the excess free eergy (above the equilibrium value) of the startig chemicals (the fuel). I a atmosphere cotaiig O 2, H 2 is fuel. There s lot s of H 2 i the H 2 O i the oceas. Do we therefore have plety of fuel? A) yes B) o C) maybe Not from this source. The hydroge i the water is i thermal ad chemical equilibrium, so its free eergy is miimum. If we have aother eergy source (e.g., solar), we ca use it to dissociate the H 2 O, ad the carry the H 2 aroud. This might be useful, but the H 2 is ot the fuel; it s just the eergy trasport medium. Lecture 18, p 17
Exercise: Gas Purificatio Suppose that some gas (helium) is preset at five parts per millio i the atmosphere, at 300K. You wat a 99% pure sample of it at 1 atm. About how much work must be doe per mole to get that? (igore chages of F due to the other costituets.) a) What s the differece i m betwee the iside ad outside of the sample tak (cotaiig the pure helium)? Which has bigger m? b) How much DF per mole? Lecture 18, p 18
Solutio Suppose that some gas (helium) is preset at five parts per millio i the atmosphere, at 300K. You wat a 99% pure sample of it at 1 atm. About how much work must be doe per mole to get that? (igore chages of F due to the other costituets.) a) What s the differece i m betwee the iside ad outside of the sample tak (cotaiig the pure helium)? Which has bigger m? m, T kt l T T i 21 5 20 m l 4 10 J l2 10 i mout kt 5 10 J out b) How much DF per mole? (5x10-20 J) (610 23 ) = 310 4 J i out 0.99 210 6 5 10 That s about 0.01 kwh, would cost about $0.001 (cost of electric power). Ufortuately real life extractio methods are much less efficiet. 5 Lecture 18, p 19
Exercise: Gas Purificatio (2) Suppose we wat to take advatage of semipermeable techology. Ca we avoid havig to do ay work to obtai our helium? Cosider this apparatus: Before: After: vacuum Earth s atmosphere He Earth s atmosphere Semipermeable membrae. Oly He ca pass through. Let s igore the cost of makig the iitial vacuum-filled box. A) What is the equilibrium pressure of the helium i the box? B) How much work is eeded to produce 1 mole of He at 1 atm? Lecture 18, p 20
Solutio Suppose we wat to take advatage of semipermeable techology. Ca we avoid havig to do ay work to obtai our helium? Cosider this apparatus: Before: After: vacuum Earth s atmosphere He Earth s atmosphere Semipermeable membrae. Oly He ca pass through. Let s igore the cost of makig the iitial vacuum-filled box. A) What is the equilibrium pressure of the helium i the box? Igore the o-helium compoets of the atmosphere. I equilibrium, the helium pressure must be the same iside ad out: p = 510-6 atm. B) How much work is eeded to produce 1 mole of He at 1 atm? You ll eed to start with a vacuum box that is 210 5 times larger tha you wat, the compress it to get 1 atm. If you do the itegral (I wo t) you ll get the same 310 4 J/mole as before. Lecture 18, p 21