General Physical Chemistry I Lecture 5 Aleksey Kocherzhenko February 10, 2015
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General Physical Chemistry I
Lecture 5
Aleksey Kocherzhenko
February 10, 2015"
Last time…"
l
Diffusion, effusion, and molecular collisions"
Diffusion" Effusion"
Mean free path:"� =RT
pNA�p2
Graham’s law:" " "
"(@ fixed T and p)"
r / M� 12
Thomas Graham"Mean free path, :average distance a molecule travels between collisions"
�
Time of flight, :average time a molecule spends between two collisions"
z�1
Collision frequency, :average # of collisions made by one molecule per s"
z
Molecular collisions:"
vrms =�
z�1= �z
Real gasses"
Real gas: non-negligible!intermolecular interactions!
Perfect gas: "no intermolecular interactions;total energy = kinetic energy!
� � d
Attractive potential (negative)"Repulsive potential (positive)"
Equilibriumdistance"
Perfect gas isotherms"
Real gas isotherms"
Similar to perfect gas at high T
Compression with no change in pressure"
p / 1
V
Real gasses"
The critical temperature"
Ø A perfect gas can never be liquefied (no interactions to hold molecules together);a real gas @ temperatures below some value (the critical temperature) can!Tc
(varies greatly)!CO2"
Vc
pc
Ø @ or above : no distinction between gas and liquid; single phase fills entire volume!
Tc
The compression factor"
Deviation of the compression factor from 1 reflects deviation of real gas behavior from perfect gas behavior:"
We define the compression factor:!
Z =Vm
V 0m
Molar volume of real gas"
Molar volume of perfect gas (under same conditions)"
From perfect gas law:!V 0m =
RT
pWe can then rewrite the compression factor as:"
Z =pVm
RT
Z < 1 – attractive interactions dominant "
Z > 1 – repulsive interactions dominant "
or, multiplying by the amount of gas in the numerator and denominator:"
Z =pVm⌫
⌫RT=
pV
⌫RT
For a perfect gas, "Z = 1
Virial equation of state"
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!
Heike Kamerlingh Onnes
2nd virial coefficient" 3rd virial coefficient"
pV
⌫RT= Z ) p =
⌫RT
VZ
At fixed ,"perform Taylor expansion of in terms of inverse molar volume,"V �1
m = ⌫/V
TZ
) p =⌫RT
V
1 +B (T )
⌫
V+ C (T )
⇣ ⌫
V
⌘2+ ...
�
Ø The 1st virial coefficient is dominant (except for highly compressed gasses)!
Ø Virial coefficients depend on ; at where , known as Boyle’s temperature, the real gas behaves (almost) like a perfect gas!
T B (TB) = 0TB
Ø Virial coefficients depend on intermolecular interactions and can be selected to reproduce the experimentally observed relation between , , and !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!
Johannesvan der Waals
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 � bm⌫
Total volume"Accessible volume" 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"
Divide by 2 to avoid double-counting (exclusion of volume requires 2 molecules)"
Multiply by Avogadro’s # to convert to molar excluded volume"
bm =43⇡ (2r)3
2NA = 4
✓4
3⇡r3
◆NA
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 "
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:"
Proportionality constant"
Pressure for a gas of interacting molecules" p =
⌫RT
V 0 � a⇣ ⌫
V
⌘2
Pressure for a gas of non-interacting hard-sphere molecules"
The van der Waals equation of state"
V 0 = V � bm⌫
We have found that:"
p =⌫RT
V 0 � a⇣ ⌫
V
⌘2
)p+ a
⇣ ⌫
V
⌘2�(V � bm⌫) = ⌫RT
Van der Waals equation of state"
, large value when intermolecular interactions are strong"
a > 0
, large value when molecular size is large"bm > 0
Van der Waals parameters(empirical):"
a⇣ ⌫
V
⌘2⌧ p
bm⌫ ⌧ V
Reduces to perfect gas equation of state @ high T and low p
Van der Waals isotherms"
Van der Waals vs. experimental isotherms "
CO2"
Experimental isotherms"
“Van der Waals loops”: unrealistic reduction in pressure as volume is reduced"
Ø The resulting isotherms then closely resemble experimental isotherms"
Ø “Van der Waals loops” are replaced by horizontal straight lines that divide the loops into areas of equal size "
Critical point from van der Waals equation"Critical point"
p =⌫RT
V � bm⌫� a
⇣ ⌫
V
⌘2
Ø From the van der Waals equation: "
dp
dV= � ⌫RT
(V � bm⌫)2 +
2a⌫2
V 3
d2p
dV 2=
2⌫RT
(V � bm⌫)3 � 6a⌫2
V 4
Ø Taking the 1st and 2nd derivatives of this expression, we find (check this!):"
Ø The critical point is an inflection point for the critical isotherm"
d2p
d2V
����Tc
= 0 @(pc, Vc)
Ø The tangent to the critical isotherm at the critical point is zero"
dp
dV
����Tc
= 0 @(pc, Vc)
Critical point from van der Waals equation"Ø The critical point is an inflection point
for the critical isotherm:"d2p
d2V
����Tc
= 0 @(pc, Vc)
Ø The tangent to the critical isotherm at the critical point is zero:"
dp
dV
����Tc
= 0 @(pc, Vc)
(condition holds @ the critical point)"
d2p
dV 2=
2⌫RT
(V � bm⌫)3 � 6a⌫2
V 4
we found:"dp
dV= � ⌫RT
(V � bm⌫)2 +
2a⌫2
V 3
we found:"
2⌫RTc
(Vc ⌫ � bm⌫)3 � 6a⌫2
(Vc ⌫)4= 0
)� ⌫RTc
(Vc ⌫ � bm⌫)2 +
2a⌫2
(Vc ⌫)3= 0
)
) 2RTc
⌫2 (Vc � bm)3 =
6a
⌫2V 4c
) RTc
⌫ (Vc � bm)2 =
2a
⌫V 3c
is the critical molar volume: hence, the total volume at the critical point is "Vc Vc⌫
(Divide l.h.s. and r.h.s. by 2)"
1
(Vc � bm)
2a
V 3c
=3a
V 4c
)
Critical point from van der Waals equation"RTc
(Vc � bm)3 =
3a
V 4c
1
(Vc � bm)
RTc
(Vc � bm)2 =
3a
V 4c
)
RTc
(Vc � bm)2 =
2a
V 3c
Multiply both sides by" 13V
4c (Vc � bm)
) 2
3Vc = Vc � bm or" bm =
1
3Vc
) Vc = 3bm
)Express critical temperature:"
Tc =2a (Vc � bm)
2
R V 3c
)
Tc =2a (3bm � bm)
2
27Rb3m
or:" Tc =8a
27Rbm
We now just need to find ... "pc
))
pc =R
2bm· 8a
27R bm� a
✓1
3bm
◆2
Critical point from van der Waals equation"
Tc =8a
27Rbm
p =⌫RT
V � bm⌫� a
⇣ ⌫
V
⌘2We know that" and" V = ⌫Vm
) p =RT
Vm � bm� a
✓1
Vm
◆2
or, at the critical point:" pc =
RTc
Vc � bm� a
✓1
Vc
◆2
But we have found that"
Vc = 3bm =
=4a
27b2m� a
9b2m=
a
27b2m
)
pc =a
27b2m We have expressed the critical temperature, pressure, and molar volume in terms of van der Waals parameters and "bma
Ø For a van der Waals gas, critical molar volume = 3 times the excluded molar volume "
Liquefaction of gasses"Ø Achieved by cooling a gas beyond its boiling point at the pressure of the experiment"
Ø , slow molecules can capture each other by intermolecular forces"T # , vrms #
Ø To slow down molecules, increase average separation between them"
Getting molecules from smaller to larger separation requires energy"
Ø Letting a gas expand without heating results in the gas cooling (Joule-Thomson effect)"
James Joule"William Thomson, 1st Baron Kelvin"
Ø Cooling by repeated expansion cycles is commonly used to liquefy gasses "
Liquefaction of gasses"Linde apparatus"
Ø Gas is forced through a tube to a throttle, where it expands and cools down"
Carl von Linde"
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 (N2, O2, H2, He) and separating the constituents of air (since each gas liquefies at a different temperature)"
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