Physical Cheistry I or Biocheists Che340 Lecture 26 (3/14/11) Yoshitaka Ishii Ch 7.2, 7.4-5, & 7.10 Announceent Exa 2 this Friday. Please be well prepared! HW average 80-85. You will probably have one drop or the hoe work. So don t worry i you did not do very well in one HW. 1
How does in three phases depends on? d(, P) = -S d + dp = -S d 0<S, solid () < S, liquid () < S, gas () Assuing S is constant, (, P) = (0, P) - S Q1. Where is the elting point? Q2. Which line represents solid ()? Q3. How do you explain the phase transition ro solid to liquid using the igure? Cheical pote ential () () or 3 phases () plot when P is increased - How are and b aected by P? (, P + P) ~ {(0, P) + P} S Gas > [Q1] Liquid > Solid Gas > Solid > Liquid 2
7.2 he Pressure-eperature Phase iagra Isobar Increases a) S L G (P higher than triple point Pressure) b) S G (P lower than triple point P) Isother; P Increases c) G L S ( higher than triple point ) L-G coexists above critical point ( c, P Pc) Supercritical Fluid P- Phase iagra ( Solid < Liquid ) Water P- Phase iagra ( Solid > Liquid ) What is the dierence ro the previous graph or solid < liquid? Isother; P Increases c) G S L 3
P7.3) Within what range can you restrict the values o P and i the ollowing inoration is known about CO 2? Use Figure 7.8 to answer this proble. a. As the teperature is increased, the solid is irst converted to the liquid and subsequently to the gaseous state. b. As the pressure on a cylinder containing pure CO 2 is increased ro 65 to 80 at, no interace delineating liquid and gaseous phases is observed. b c. Solid, liquid, and gas 80 phases coexist at equilibriu. a. 5.11 at < P < 73.75 at a 7.4 Providing a heoretical Basis or the P- Phase iagra When two phase and are in equilibriu at a constant P and, (P, ) = (P, ) When (P,) is changed by (dp,d) while keeping the equilibriu, (P, ) +d = (P, )+ d hus, d = d -S d + dp = -S d + dp 0 = (S -S )d - ( - )dp = S d - dp Clapeyron equation dp d S 4
dp usion d 0 or this syste. Q. Which is correct or this syste? a) (sl) > 0 b) (sl) < 0 c) S (sl) > 0 d)s (sl) < 0 dp/d or usion and orization For Fusion dp d usion S 6 5. 510 PaK usion usion 1 Average o S and or Ag, AgCl, Ca, CaCl 2, KCl, a, acl 1 1 22Jol K ~ 6 3 4 10 ol 55 bar K 1 1 (d/dp) usion ~ 0.05K bar -1 egative sign only or H 2 O For aporization dp d S 3 5. 010 PaK 1 510 Average o S and in able 7.3 95Jol ~ 2 10 2 3 ol 2 1 K bar K 1 1 1 (d/dp) ~ 20K bar -1 5
7.5 Using the Clapeyron Equation to Calculate apor Pressure as a Function o For aporization dp d S H H, gas PH R 2 1 H dp P nr Pin Pini in 1 2 d 1 H dp P R ini 1 Clausius-Clapeyron equation 2 d P in H 1 1 ln Pini R in ini HW 8 P7.5) he or pressure o liquid SO 2 is 2232 Pa at 201 K, and H orization = 24.94 kj ol 1. Calculate the noral and standard d boiling points. oes your result or the noral boiling point agree with that in able 7.1? I not, suggest a possible cause. Q. Which equation should we use? P in H 1 1 ln [Q1] Pini R in ini = 201 K P ini = 2232 Pa For noral BP, P in = 1 [Q2] at (convert to Pa) Find in For standard BP, P in = [Q3] 1 bar Find in ini 6
HW 8 P7.6) For water, H orization is 40.65 kj ol 1, and the noral boiling point is 373.15 K. Calculate the boiling point or water on the top o a ountain o height 5500, where the noral baroetric pressure is 380 orr. P in H 1 1 ln P ini R in ini P ini = 1 at & ini = 373.15 K Review Q. How uch is S at noral BP? S = [Q1] H / b Review Q HW 8 P7.7) Use the values or ΔG (ethanol, l) and ΔG (ethanol, g) ro Appendix B to calculate the or pressure o ethanol at 298.15 K. For the transoration C 2 H 5 OH (l) C 2 H 5 OH (g), what is K p? K p = P(ethanol, [Q1] g)/p 0 K p = exp(- G(lg)/R) [Q2] 7
7.5 Using the Clapeyron Equation to Calculate apor Pressure as a Function o For Fusion P Pi usion usion S H d dp d usion usion ~ i i H usion usion i d P H usion in Pi ~ ln ~ usion ini H usion usion ini when << ini 7.10 Conorational ransitions o Biological Polyers G den = H den -S den H den > 0 At lower, is stable S den > 0 At higher, is stable Fraction o Uno olded () At K =1 & = 0.5 8
9 7.10 Conorational ransitions o Biological Polyers C C K C C C C K 1 At K =1 & = 0.5
G den = H den -S den G den 0 = -RlnK H 0 den = R 2 (lnk/) P S 0 den = (H den0 - G den0 )/ C 0 den = (H den0 /) P P7.19) A protein has a elting teperature o = 335 K. At = 315 K, U absorbance deterines that the raction o native protein is = 0.965. At = 345 K, = 0.015. (a) Assuing a two-state odel and assuing also that the enthalpy is constant between = 315 and 345 K, deterine the enthalpy o denaturation. 1-1- 0.965 1- K 315 K 0.036 1-0.015 K 345 K 0.965 ln{k( 2 )} - ln{k( 1 )} = -(ΔH/R)(1/ [Q1] 2-1/ 1 ) 0.015 (b) Also, deterine the entropy o denaturation at = 335 K. Obtain ΔG en (335K) using ΔG en ( 2 )= 2 {ΔG en ( 1 )+ΔH(1/ 2-1/ 1 )} ΔG en (335K) = ΔH [Q2] en (335K) - ΔS en (335K) (c) By SC, the enthalpy o denaturation was deterined to be 251 kj ol 1. Is this denaturation accurately described by the two-state odel? I the answer o (a) ~251 kjol -1, yes 65.67 10
7.6 Surace ension H 2 O olecules axiize the nuber o neighboring olecules or stability A water ors a spherical droplet How does the energy o a droplet depend on the surace area? he work (in Helholz energy A) associated with creating additional surace area d at constant and is da = d, where is the surace tension 7.6 Surace ension (Continued) da = d, (1) where is a surace area. he surace tension (>0) has the units o energy/area or J -2. For a natural process, Helholtz da < 0 d <0 or a natural process For a sphere o a radius r, =4r 2 d = 8r dr (2) (1) & (2) yield da = (8r dr) = (8r ) dr = F sur dr P sur = F sur /4r 2 = (2/r) 11
is dependent on copound! Saple Question here large and sall bubbles o H 2 O. Assue that P outer (~1at) + P sur is in equilibriu with the internal pressure P inner. Which bubble has a higher P inner? P outer + P sur P outer + P sur sall R P inner large r P inner sall Q. Is P inner large > P inner sall correct? P inner = P out + P sur = 1 at + (2/r) 12
How two bubbles use into one. P inner large < P inner sall P7.40 Calculate the dierence in pressure across the liquid-air interace or a water droplet o radius 150 n. P outer + P sur P inner P outer + P sur = P inner P inner -P outer =P sur = 2/r ~ 10 bar Laplace s equation 13
apor Pressure o a roplet (erivation) dp liq dp = d(2/r) = 2d(1/r) [1] Cheical potential o two phase should be sae 0 = d liq d 0=, liqdp liq,dp [2] P(r) and P bulk are P or a droplet By cobining [1, 2] o a radius r and bulk liquid (r = ) (, /, liq 1)dP = 2d(1/r) dp, /, liq dp ~ 2d(1/r) or + dp sur, dp = 2, liq d(1/r) R/P dp =2, liq d(1/r) R P ( r ) P ( r ) dp P P( r ) 2, ln P R bulk liq 2 1/ r, liq 0 d( 1/ r ') 2M 1 / r 1 / r R liq dp liquid Liquid P7.41) Calculate the actor by which the or pressure o a droplet o ethanol o radius 1.00 10 4 at 45.0 C in equilibriu with its or is increased with respect to a very large droplet. Use the tabulated value o the density and the surace tension at 298 K ro Appendix B (table 2.3& table 7.5) or this proble. (Hint: You need to calculate the or pressure o ethanol at this teperature.) P P ) A2 3.697110 ln 3 ln 0( A 1 23.593 11.0043 Calculate P A 325 31.317 3 0 () Pa 4 hen, P use 6.01 10 Pa K P ( r ) 2, ln P( r ) R liq 2M 1 / r 1 / r R liq 14