General Physical Chemistry I

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1 General Physical Chemistry I Lecture 5 Aleksey Kocherzhenko February 10, 2015"

2 Last time "

Diffusion, effusion, and molecular collisions" Diffusion" Effusion"

collisions:" Mean free path, : average distance a molecule travels

of collisions made by one molecule per s" 1 Time of flight, z : average

3 Diffusion, effusion, and molecular collisions" Diffusion" Effusion" Graham s law: " " " 1 2 r / M " (@ fixed T and p)" Molecular collisions:" Mean free path, : average distance a molecule travels between collisions" Thomas Graham" Collision frequency, z : average # of collisions made by one molecule per s" 1 Time of flight, z : average time a molecule spends between two collisions" vrms = z 1 = z Mean free path:" RT p = pna 2 l

4 Real gasses" d Perfect gas: " no intermolecular interactions; total energy = kinetic energy! Perfect gas isotherms" p / 1 V Real gas: non-negligible! intermolecular interactions! Repulsive potential (positive)" Attractive potential (negative)" Similar to perfect gas at high T Real gas isotherms" Equilibrium distance" Compression with no change in pressure"

5 Real gasses"

6 The critical temperature" CO 2" (varies greatly)! p c V c Ø A perfect gas can never be liquefied (no interactions to hold molecules together); a real temperatures below some value T c (the critical temperature) can! or above T c : no distinction between gas and liquid; single phase fills entire volume!

7 We define the compression factor:! The compression factor" Molar volume of real gas" Z = V m Vm 0 Molar volume of perfect gas (under same conditions)" From perfect gas law:!vm 0 = RT p We can then rewrite the compression factor as:" Z = pv m RT or, multiplying by the amount of gas in the numerator and denominator:" Z = pv m RT = pv RT For a perfect gas, "Z =1 Deviation of the compression factor from 1 reflects deviation of real gas behavior from perfect gas behavior:" Z<1 attractive interactions dominant " Z>1 repulsive interactions dominant "

Heike Kamerlingh Onnes The virial equation of state" Kamerlingh Onnes proposed an empirical equation of state for real gasses by adding extra terms to the perfect gas law!

8 Heike Kamerlingh Onnes The virial equation of state" Kamerlingh Onnes proposed an empirical equation of state for real gasses by adding extra terms to the perfect gas law! ) p = RT V pv RT = Z ) p = RT V Z At fixed," perform Taylor expansion of of inverse molar volume,"v m 1 apple 1+B (T ) V + C (T ) V Z T in terms = /V Virial equation of state" 2 nd virial coefficient" 3 rd virial coefficient" Ø Virial coefficients depend on intermolecular interactions and can be selected to reproduce the experimentally observed relation between,, and! Ø The 1 st virial coefficient is dominant (except for highly compressed gasses)! Ø Virial coefficients depend on ; at T B where, known as Boyle s temperature, the real gas behaves (almost) like a perfect gas! T B (T B )=0 p V T

The van der Waals equation of state" The virial equation of state can approximate the behavior of real gasses, but gives limited physical insight!

V 0 = V Accessible volume" Total volume" b m Excluded volume" If molecules are assumed to be hard spheres à their centers cannot be closer than :" 2r = d Volume around a molecule

9 The van der Waals equation of state" The virial equation of state can approximate the behavior of real gasses, but gives limited physical insight! Ø Van der Waals derived an approximate equation of state for real gasses based on simple physical considerations! ü Molecules are NOT infinitely small:! V 0 = V Accessible volume" Total volume" b m Excluded volume" If molecules are assumed to be hard spheres à their centers cannot be closer than :" 2r = d Volume around a molecule that is not accessible to other molecules" b m = 4 3 (2r)3 2 N A =4 Johannes van der Waals Multiply by Avogadro s # to convert to molar excluded volume" 4 3 r3 N A Divide by 2 to avoid double-counting (exclusion of volume requires 2 molecules)"

10 The van der Waals equation of state" ü Molecules interact with each other! Intermolecular interactions are relatively short-range à molecules that are far from walls experience, on average, the same interaction with molecules on all sides " Pressure for a gas of interacting molecules" p = RT V 0 a V Proportionality constant" Molecules close to walls interact only with molecules within the volume of the gas (we neglect interaction with walls); on average, this interaction is attractive" Thus, intermolecular interactions reduce the pressure that molecules exert on walls " We expect the reduction in pressure to depend on the # of molecules per unit volume; this dependence turns out to be quadratic:" Pressure for a gas of non-interacting 2 hard-sphere molecules"

V 0 = V The van der Waals equation of state" We have found that:" p = RT V 0 a V b m 2 ) Reduces to perfect gas equation of state @ high T and low p Van der Waals equation of state" apple

11 V 0 = V The van der Waals equation of state" We have found that:" p = RT V 0 a V b m 2 ) Reduces to perfect gas equation of high T and low p Van der Waals equation of state" apple p + a V 2 (V bm )= RT Van der Waals parameters (empirical):" a>0, large value when intermolecular interactions are strong" a V 2 p b m > 0, large value when molecular size is large" b m V

12 Van der Waals vs. experimental isotherms " Van der Waals isotherms" Experimental isotherms" CO 2" Van der Waals loops : unrealistic reduction in pressure as volume is reduced" Ø Van der Waals loops are replaced by horizontal straight lines that divide the loops into areas of equal size " Ø The resulting isotherms then closely resemble experimental isotherms"

13 Critical point from van der Waals equation" dp dv = Critical point" RT (V b m ) 2 + 2a 2 V 3 Ø The critical point is an inflection point for the critical isotherm" d 2 p d 2 V Tc c,v c ) Ø The tangent to the critical isotherm at the critical point is zero" dp dv Tc c,v c ) Ø From the van der Waals equation: " 2 p = RT V b m d 2 p dv 2 = a V Ø Taking the 1 st and 2 nd derivatives of this expression, we find (check this!):" 2 RT 6a 2 (V b m ) 3 V 4

14 Critical point from van der Waals equation" Ø The critical point is an inflection point for the critical isotherm:" d 2 p d 2 V Tc c,v c ) we found:" d 2 p dv 2 = 2 RT 6a 2 (V b m ) 3 V 4 Ø The tangent to the critical isotherm at the critical point is zero:" dp dv Tc c,v c ) we found:" dp dv = RT (V b m ) 2 + 2a 2 V 3 ) 2 RT c 6a 2 (V c b m ) 3 (V c ) 4 =0 ) RT c (V c b m ) 2 + 2a 2 (V c ) 3 =0 ) V c 2RT c 2 (V c b m ) 3 = 6a 2 Vc 4 (condition the critical point)" is the critical molar volume: hence, the total volume at the critical point is " (Divide l.h.s. and r.h.s. by 2)" ) RT c (V c b m ) 2 = 2a Vc 3 V c

15 Critical point from van der Waals equation" RT c (V c b m ) 3 = 3a V 4 c RT c (V c b m ) 2 = 2a V 3 c ) ) 1 (V c b m ) 1 (V c b m ) RT c (V c b m ) 2 = 3a Vc 4 2a V 3 c = 3a V 4 c Multiply both sides by" 1 3 V 4 c (V c b m ) ) 2 3 V c = V c b m or" b m = 1 3 V c ) V c =3b m ) Express critical temperature:" T c = 2a (V c b m ) 2 RV 3 c) T c = 2a (3b m b m ) 2 27Rb 3 m or:" T c = 8a 27Rb m We now just need to find p c... "

16 Critical point from van der Waals equation" We know that" ) R 2b m V c =3b m ) p c = a = T c = p = 8a 27Rb m RT V b m ) p = RT 1 a V m b m V m But we have found that" p c = a 27b 2 m ) a V 2 and" V = V m 2 or, at the critical point:" p c = = 4a 27b 2 m 8a 27R b m a 9b 2 m RT 2 c 1 a V c b m V c 1 = a 27b 2 m 3b m 2 We have expressed the critical temperature, pressure, and molar volume in terms of van der Waals parameters a and "b m Ø For a van der Waals gas, critical molar volume = 3 times the excluded molar volume "

slow down molecules, increase average separation between them" Ø Letting a gas expand without heating results in the gas cooling

17 Liquefaction of gasses" Ø Achieved by cooling a gas beyond its boiling point at the pressure of the experiment" Ø T #, v rms #, slow molecules can capture each other by intermolecular forces" Getting molecules from smaller to larger separation requires energy" Ø To slow down molecules, increase average separation between them" Ø Letting a gas expand without heating results in the gas cooling (Joule-Thomson effect)" Ø Cooling by repeated expansion cycles is commonly used to liquefy gasses " William Thomson, 1 st Baron Kelvin" James Joule"

it " Ø The process is repeated several times, with the gas cooling down some more on each iteration" Ø The Linde apparatus allows

18 Liquefaction of gasses" Linde apparatus" Carl von Linde" Ø Gas is forced through a tube to a throttle, where it expands and cools down" Ø It is cycled back to the compressor through an outer tube, exchanging heat with gas moving towards the throttle and cooling it " Ø The process is repeated several times, with the gas cooling down some more on each iteration" Ø The Linde apparatus allows liquefying all gasses (N 2, O 2, H 2, He) and separating the constituents of air (since each gas liquefies at a different temperature)"

Chapter 1. The Properties of Gases Fall Semester Physical Chemistry 1 (CHM2201)

Chapter 1. The Properties of Gases 2011 Fall Semester Physical Chemistry 1 (CHM2201) Contents The Perfect Gas 1.1 The states of gases 1.2 The gas laws Real Gases 1.3 Molecular interactions 1.4 The van