WEBVTT
Kind: captions
Language: en
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welcome in todays class we will revise and
summarize what all we have learnt about the
00:00:23.359 --> 00:00:29.610
statistical thermodynamics of ideal gases
where as you have seen that we are mainly
00:00:29.610 --> 00:00:38.850
focusing on gases which are comprised of atoms
such as helium or argon or diatomic molecules
00:00:38.850 --> 00:00:46.899
such as carbon monoxide nitric oxide hcl
or hbr which can be classified as heteronuclear
00:00:46.899 --> 00:00:54.760
diatomic molecules or gases like hydrogen
deuterium oxygen etcetera which can be clubbed
00:00:54.760 --> 00:01:01.449
under the common name of homonuclear diatomic
molecules and we have already seen that the
00:01:01.449 --> 00:01:10.570
microscopic information regarding any of these
systems is is actually collected and
00:01:10.570 --> 00:01:16.400
contained in what is known as the canonical
partition function
00:01:16.400 --> 00:01:23.490
so what is this canonical partition function
if you maintain the ideal gas at a given temperature
00:01:23.490 --> 00:01:31.280
volume and with a constant number of particles
either atoms or molecules in that case if
00:01:31.280 --> 00:01:38.210
you start from the hamiltonian of the system
you can write down all informations regarding
00:01:38.210 --> 00:01:46.090
the underlying accessible microscopic states
along with their associated probabilities
00:01:46.090 --> 00:01:53.240
in the form of this canonical partition function
and as shown here this is what the expression
00:01:53.240 --> 00:02:00.550
for this canonical partition function is so
as you see in addition to depending on this
00:02:00.550 --> 00:02:05.920
capital n the number of particles it depends
on the small q that is the single particle
00:02:05.920 --> 00:02:12.980
partition function and we know that for monatomic
ideal gases small q will derive contributions
00:02:12.980 --> 00:02:18.950
from the overall translational motion of the
atom and the underlying electronic and nuclear
00:02:18.950 --> 00:02:25.430
structure for a monatomic ideal gas whereas
if you have a diatomic ideal gas then two
00:02:25.430 --> 00:02:31.810
additional terms will come into the picture
which are due to the rotation of the molecule
00:02:31.810 --> 00:02:39.040
about an axis and also the vibration of the
chemical bond that is connecting the two
00:02:39.040 --> 00:02:44.190
atoms that are constituting the molecule of
the gas
00:02:44.190 --> 00:02:50.650
and we have discussed in detail like how we
can calculate each of these quantities given
00:02:50.650 --> 00:02:56.670
the situation and in general we know that
if we know the thermodynamic quantities
00:02:56.670 --> 00:03:04.940
like v t and the number of particles in that
case we are able to find out q trans q electronic
00:03:04.940 --> 00:03:12.680
q nuclear provided you have information regarding
the microscopic structure in terms of the
00:03:12.680 --> 00:03:20.160
the underlying energy states of the electronic
and nuclear nature and also the transition
00:03:20.160 --> 00:03:26.820
energy between the first and the the ground
and the first excited electronic state
00:03:26.820 --> 00:03:33.709
with this background in mind then we can
we have already talked about how to obtain
00:03:33.709 --> 00:03:41.500
the thermodynamics from the canonical partition
function and as you know that all the thermodynamic
00:03:41.500 --> 00:03:48.569
properties that are of interest in our case
can be derived if you know the dependence
00:03:48.569 --> 00:03:57.400
of lnq on properties like temperature volume
or the number of particles so with this background
00:03:57.400 --> 00:04:05.040
in mind what we have also pointed out that
the underlying structure plays an important
00:04:05.040 --> 00:04:11.870
role in overall thermodynamics that is because
because that is because the presence of this
00:04:11.870 --> 00:04:16.609
underlying structure gives add additional
corrections
00:04:16.609 --> 00:04:24.190
so the corrections that add up to the dominant
term in each of these thermodynamic properties
00:04:24.190 --> 00:04:30.750
and we have seen that in the case of monatomic
ideal gas simply because we can partition
00:04:30.750 --> 00:04:37.690
the overall single particle canonical partition
function into the contribution from the center
00:04:37.690 --> 00:04:44.540
of mass that is structureless and the internal
the structure all the thermodynamic properties
00:04:44.540 --> 00:04:52.199
that we could derive they also had explicit
contributions arising from these three factors
00:04:52.199 --> 00:04:57.020
which added together to give us the net thermodynamic
property
00:04:57.020 --> 00:05:03.570
now if we look at what happens in the case
of the monatomic ideal gases in the case of
00:05:03.570 --> 00:05:12.120
the diatomic ideal gases what we find is
in this case because of the presence of these
00:05:12.120 --> 00:05:19.150
two extra terms that you see here there are
two extra terms which are appearing in all
00:05:19.150 --> 00:05:24.970
the thermodynamic properties which can be
attributed to the effect of discrete energy
00:05:24.970 --> 00:05:30.970
levels present at this present for the
rotational and the vibrational degrees of
00:05:30.970 --> 00:05:38.090
freedom for the diatomic case and as a result
we would say that well the presence of these
00:05:38.090 --> 00:05:45.449
two extra terms for example q rot gives the
its corresponding extra terms in each of the
00:05:45.449 --> 00:05:51.160
thermodynamic properties that we have shown
here similarly the presence of q vib leads
00:05:51.160 --> 00:05:57.930
to this vibrational contribution to each of
the thermodynamic properties that we see
00:05:57.930 --> 00:06:05.699
now we have seen all that and now let us try
and understand what happens if i want to derive
00:06:05.699 --> 00:06:12.759
the ideal gas equation starting from the canonical
partition function now we have already seen
00:06:12.759 --> 00:06:20.139
in the last lecture that the pressure is related
to the derivative of lnq with respect to volume
00:06:20.139 --> 00:06:26.130
keeping temperature and the number of particles
constant now if i think about what the way
00:06:26.130 --> 00:06:34.110
the structure of small q on which this capital
q depends we find that in both the cases the
00:06:34.110 --> 00:06:43.880
explicit volume dependence of the of the small
q term comes through this translational
00:06:43.880 --> 00:06:50.710
partition function so in the case of the monatomic
ideal gas this corresponds to the national
00:06:50.710 --> 00:06:58.130
motion of the single atom in the given box
and here it corresponds to the translational
00:06:58.130 --> 00:07:03.630
motion of the molecule as a whole represented
in terms of the translational motion of the
00:07:03.630 --> 00:07:10.460
center of mass in the confining volume under
the given condition of temperature
00:07:10.460 --> 00:07:17.400
now what we find here is its possible then
to write down what the q trans is irrespective
00:07:17.400 --> 00:07:23.530
of whether we are dealing with a monatomic
ideal gas or a diatomic ideal gas and this
00:07:23.530 --> 00:07:32.340
is given in terms of the volume of the gas
divided by the the cube of the thermal debroglie
00:07:32.340 --> 00:07:37.100
wavelength lambda where we have already seen
that the thermal debroglie wavelength lambda
00:07:37.100 --> 00:07:43.770
depends on temperature and the mass of each
particle and therefore the way the monatomic
00:07:43.770 --> 00:07:51.640
and the diatomic gases will differ from each
other will be in terms of this the quantity
00:07:51.640 --> 00:07:52.640
m
00:07:52.640 --> 00:08:04.110
now if we use this then taking a natural logarithm
of q trans and then estimating the derivative
00:08:04.110 --> 00:08:12.419
directly gave us the pressure in terms of
nkbt by v we have already done this derivation
00:08:12.419 --> 00:08:21.069
so what i wanted to emphasize over here is
that the pressure that the here the when i
00:08:21.069 --> 00:08:28.509
derive the pressure the result that i get
this is the same as far as the monatomic ideal
00:08:28.509 --> 00:08:35.060
gas or the diatomic ideal gas is concerned
and that is because both of them have identical
00:08:35.060 --> 00:08:42.190
terms giving rise to the corresponding
to the translational motion in the volume
00:08:42.190 --> 00:08:50.460
v and therefore i would expect that the pressure
term that i see here would be nkbt by v in
00:08:50.460 --> 00:08:56.100
both the cases irrespective of whether i am
looking at the translational motion of an
00:08:56.100 --> 00:09:01.000
atom or the translational motion of a diatomic
molecule
00:09:01.000 --> 00:09:07.430
similarly we have also checked if the theoretical
development that i have been presenting here
00:09:07.430 --> 00:09:15.200
for molecular thermodynamics of ideal gases
can reproduce the known form of chemical potential
00:09:15.200 --> 00:09:21.430
of an ideal gas where we know that if i want
to predict what the chemical potential of
00:09:21.430 --> 00:09:29.980
the ideal gas is going to be at a given temperature
and pressure this is dependent on what the
00:09:29.980 --> 00:09:37.860
chemical potential of the gas is at a temperature
t under standard conditions of one bar pressure
00:09:37.860 --> 00:09:45.560
plus an additional term that depends on the
pressure at which i am going to observe the
00:09:45.560 --> 00:09:53.000
chemical potential of the ideal gas now
we have already carried out this derivation
00:09:53.000 --> 00:09:58.570
and indeed we have been able to show that
starting from the definition of chemical potential
00:09:58.570 --> 00:10:06.510
in terms of l n capital q then it is possible
to find out what exactly the form of chemical
00:10:06.510 --> 00:10:14.120
potential would be in terms of the parameters
like mass of each particle temperature then
00:10:14.120 --> 00:10:16.690
q electronic u nuclear etcetera
00:10:16.690 --> 00:10:26.700
now here what once again i would like to highlight
is that this expression has given me one set
00:10:26.700 --> 00:10:33.470
of terms that is independent of pressure and
dependent on temperature only and it also
00:10:33.470 --> 00:10:40.480
depends on the kind of system that we are
working on so very easily i can say that this
00:10:40.480 --> 00:10:47.700
must be the chemical potential at one bar
pressure at a given temperature t so when
00:10:47.700 --> 00:10:53.680
i have one bar pressure by definition this
small p is p by p naught and p naught where
00:10:53.680 --> 00:10:59.760
p naught is the standard pressure of one by
one bar therefore i put this equal to zero
00:10:59.760 --> 00:11:08.680
so this must be the mu naught ok now what
i have here is this additional term kbtlnp
00:11:08.680 --> 00:11:16.760
which reproduces correctly the form that i
expected from our prior knowledge of thermodynamics
00:11:16.760 --> 00:11:25.740
we also looked at the predictions of the internal
energy for an ideal gas and compared the predictions
00:11:25.740 --> 00:11:32.430
of our theory that we are presenting here
with what we know from for example equal
00:11:32.430 --> 00:11:39.390
partition theorem or joules law and here what
we found is that for a monatomic ideal gas
00:11:39.390 --> 00:11:45.940
like helium if you know the underlying microscopic
structure and the energy values associated
00:11:45.940 --> 00:11:55.630
its found that not two high temperatures the
internal energy of the system is essentially
00:11:55.630 --> 00:12:02.550
corresponding to the translational energy
and at low temperatures the correction terms
00:12:02.550 --> 00:12:09.190
appearing because of the underlying electronic
structure is negligibly small but if you go
00:12:09.190 --> 00:12:16.490
to very high temperatures then what you will
find is you will find that the contributions
00:12:16.490 --> 00:12:21.940
from this correction terms appearing because
of the underlying macroscopic structure of
00:12:21.940 --> 00:12:24.360
the gas is becoming significant
00:12:24.360 --> 00:12:30.029
so as you understand that if you go for
temperatures like thousand five hundred kelvin
00:12:30.029 --> 00:12:36.880
or two thousand kelvin this number is going
to become larger and larger and will start
00:12:36.880 --> 00:12:43.470
competing at very large temperatures with
the translational contribution so basically
00:12:43.470 --> 00:12:51.610
the message here is as follows if we are at
a few hundreds of kelvin in that case the
00:12:51.610 --> 00:12:59.100
internal energy of the gas is dominated by
the translational energy translational contribution
00:12:59.100 --> 00:13:06.019
to the internal energy and the translational
contribution as we have seen here this is
00:13:06.019 --> 00:13:13.950
dependent on kt right and therefore and there
is no other dependents in here like volume
00:13:13.950 --> 00:13:20.519
doesnt appear therefore u is a function
of temperature and temperature only
00:13:20.519 --> 00:13:25.920
it is only at very high temperature that we
start seeing an additional correction term
00:13:25.920 --> 00:13:32.400
in the form of u electronic but at a normal
temperatures a few hundred of kelvin these
00:13:32.400 --> 00:13:40.390
correction terms are not important and therefore
whatever experiments that we do are good
00:13:40.390 --> 00:13:47.680
enough to it will be explained with the equipartition
theorem with this background in mind let us
00:13:47.680 --> 00:13:58.300
now go ahead and try and talk about this property
the entropy of an ideal gas the way we will
00:13:58.300 --> 00:14:07.990
calculate this quantity is at this point we
know what u is what f is in terms of lnq from
00:14:07.990 --> 00:14:12.240
lnq so s is equal to u minus f divided
by t
00:14:12.240 --> 00:14:19.209
a few lectures back i asked you to work
this out and this would be the expression
00:14:19.209 --> 00:14:28.200
for entropy as calculated starting from the
current model of the canonical partition
00:14:28.200 --> 00:14:36.089
function for this monatomic ideal gas and
here what we find is this s electronic this
00:14:36.089 --> 00:14:43.420
is a term which is a correction to the big
term that you see here that depends on the
00:14:43.420 --> 00:14:48.722
underlying electronic structure of the atom
as you can see very well the all the terms
00:14:48.722 --> 00:14:56.160
that are appearing here they are connected
to q trans that is they are connected to the
00:14:56.160 --> 00:15:04.320
contribution to entropy from the translational
motion of the atom in the box and this is
00:15:04.320 --> 00:15:11.019
the contribution from the underlying electronic
structure and this equation is known as the
00:15:11.019 --> 00:15:17.950
sackur tetrode equation and in todays class
we are going to use this equation a lot and
00:15:17.950 --> 00:15:21.910
so it is very important that we understand
what are the different quantities that are
00:15:21.910 --> 00:15:23.170
involved
00:15:23.170 --> 00:15:31.279
let me write down explicitly what s electronic
is what you must note over here is that
00:15:31.279 --> 00:15:39.089
that in this particular expression i have
s electronic is proportional to nkb because
00:15:39.089 --> 00:15:47.240
this appears as a pre factor for both the
terms that you see here ok now then i have
00:15:47.240 --> 00:15:54.079
two terms all of which depends on parameters
that are associated with the electronic structure
00:15:54.079 --> 00:16:01.040
of the atom present in the system and
of course its temperature dependence comes
00:16:01.040 --> 00:16:08.240
through the appearance of beta now once i
understand what s electronic is i find that
00:16:08.240 --> 00:16:14.149
in order to be able to use the sackur tetrode
equation i would like to simplify it a little
00:16:14.149 --> 00:16:18.040
bit and this is what i shall be doing
00:16:18.040 --> 00:16:26.630
as i understand that instead of writing out
specifically the total expression for the
00:16:26.630 --> 00:16:35.079
for the entropy i can very easily write down
the entropy as v by lambda cubed into e to
00:16:35.079 --> 00:16:44.910
the power of five by two divided by n this
plus s electronic as i have shown here so
00:16:44.910 --> 00:16:51.079
what is lambda lambda square by definition
is equal to h square divided by two pi m kt
00:16:51.079 --> 00:16:58.610
and in the slide that i was showing to you
i had this net expression put in here instead
00:16:58.610 --> 00:17:08.980
of lambda cube now if i go back and try to
see that well i have v and i have n so what
00:17:08.980 --> 00:17:17.040
happens if i introduce this quantity small
n which is capital n divided by v
00:17:17.040 --> 00:17:33.190
if that is so then i can rewrite the expression
of s as nkb ln e to the power of five by two
00:17:33.190 --> 00:17:51.110
plus nkb ln one by lambda cube into one divided
by n by v if that is so then s turns out to
00:17:51.110 --> 00:18:10.350
be nothing but five by two nkb minus nkb ln
small n into lambda cube so what i have done
00:18:10.350 --> 00:18:17.460
is i understand that whatever i see here appears
in the denominator so when i take it up a
00:18:17.460 --> 00:18:22.820
negative sign comes over here because i have
a logarithm here and therefore i can very
00:18:22.820 --> 00:18:32.710
easily say that capital s in this case and
of course you have this plus s electronic
00:18:32.710 --> 00:18:49.179
so that is nkb multiplied by five by two minus
ln n lambda cube plus s electronic right
00:18:49.179 --> 00:18:59.320
so as you see then an alternative and simpler
form of the sackur tetrode equation involves
00:18:59.320 --> 00:19:07.750
this number density n and the thermal debroglie
wavelength lambda we are going to use this
00:19:07.750 --> 00:19:15.970
equation now to do some of the following checks
so first we are going to check for the accuracy
00:19:15.970 --> 00:19:23.769
of the theoretical prediction of entropy of
an ideal gas first by examining whether the
00:19:23.769 --> 00:19:33.299
entropy that we are calculating here preserves
that extensivity of entropy as denoted by
00:19:33.299 --> 00:19:41.629
the basic formulation of thermodynamics we
are also going to check if the current formalism
00:19:41.629 --> 00:19:48.229
given in terms of the sackur tetrode equation
can correctly predict what would happen when
00:19:48.229 --> 00:19:59.119
i mix two ideal gases irreversibly or reversibly
and finally it is possible to estimate the
00:19:59.119 --> 00:20:05.760
absolute value of entropy using the sackur
tetrode equation if you know which thermodynamic
00:20:05.760 --> 00:20:13.200
state your system is present in that case
you can use these theoretical estimates and
00:20:13.200 --> 00:20:19.550
compare them with experimental results which
will tell you whether the theoretical prediction
00:20:19.550 --> 00:20:26.249
that you are making using the formalism of
molecular thermodynamics is correct or not
00:20:26.249 --> 00:20:32.830
so let us take each of these one by one so
the first thing that we check is the extensive
00:20:32.830 --> 00:20:39.299
nature of calculated entropy so the first
thing that i do is in order to do this all
00:20:39.299 --> 00:20:47.989
the time we understand that if there is a
property x which is a an extensive property
00:20:47.989 --> 00:20:58.470
in that case it must be a function of capital
n in such a way that if i replace n by n prime
00:20:58.470 --> 00:21:09.429
and n prime is equal to two n then x will
be replaced by x prime such that x prime would
00:21:09.429 --> 00:21:18.029
be equal to two x so that is the basic definition
of an extensive property and here we know
00:21:18.029 --> 00:21:26.250
from basic thermodynamics that entropy is
an extensive property and we have used a molecular
00:21:26.250 --> 00:21:30.679
thermodynamics to find out some expression
for entropy
00:21:30.679 --> 00:21:37.899
now we are going to test if the expression
that we have obtained for entropy in terms
00:21:37.899 --> 00:21:46.870
of the sackur tetrode equation correctly tells
me that if i replace n by two n or in other
00:21:46.870 --> 00:21:53.940
words if i replace n by n prime such that
n prime is equal to two n if i have done everything
00:21:53.940 --> 00:22:00.499
correctly then if the sackur tetrode equation
is correct then the entropy should become
00:22:00.499 --> 00:22:06.989
twice of its original value so in order to
do this what we do is if this is the expression
00:22:06.989 --> 00:22:14.909
for s then what we do is we write down the
expression for s prime where in case while
00:22:14.909 --> 00:22:23.359
calculating s prime i am replacing n by n
prime obviously the number density small n
00:22:23.359 --> 00:22:31.190
is represented by small n prime and s electronic
this is also becoming s prime electronic
00:22:31.190 --> 00:22:41.629
now if you would remember that s electronic
is a term that is nkb into one particular
00:22:41.629 --> 00:22:49.239
sum plus another particular sum and these
two do not depend on n so in s electronic
00:22:49.239 --> 00:22:59.749
if i replace n by n prime so s prime electronic
will now become n prime kb into the same x
00:22:59.749 --> 00:23:09.320
one plus x two correct so if that happens
then what i can do is i can now go ahead and
00:23:09.320 --> 00:23:17.980
replace n prime by two n and check what is
small n prime and s prime electrical s
00:23:17.980 --> 00:23:26.619
the electronic s i realize over here that
when i float n equal to n prime in or n
00:23:26.619 --> 00:23:34.210
prime equal to two n the other extensive property
that i have in this expression is volume then
00:23:34.210 --> 00:23:40.320
volume goes from a value of v to v prime where
v prime is equal to two v
00:23:40.320 --> 00:23:47.629
now let us check what happens to the ratio
n by v as i understand that n by v is equal
00:23:47.629 --> 00:23:54.230
to n prime by v prime under this condition
and therefore i should be able to say that
00:23:54.230 --> 00:24:01.830
well the number density is not an extensive
function it remains the same even if i replace
00:24:01.830 --> 00:24:09.919
n by twice its value now if i put this back
so i understand that here n prime is two n
00:24:09.919 --> 00:24:17.639
capital n prime is two n small n prime is
nothing but small n and s prime electronic
00:24:17.639 --> 00:24:24.779
is nothing but two into s electronic and therefore
this is the expression that we would get here
00:24:24.779 --> 00:24:31.990
and if i now compare the expression for s
prime to the expression for s that is the
00:24:31.990 --> 00:24:39.710
entropy we immediately conclude that s prime
is equal to two s so as expected the sackur
00:24:39.710 --> 00:24:49.510
tetrode equation correctly preserves the extensive
property of entropy as derived through the
00:24:49.510 --> 00:24:53.799
prescriptions of molecular thermodynamics
00:24:53.799 --> 00:25:02.099
now if i look a little closely enough i understand
that this entire analysis depends very much
00:25:02.099 --> 00:25:09.840
on the presence of this factor capital n here
ok now where did this factor capital n come
00:25:09.840 --> 00:25:18.399
from if we go back and look at the derivation
we will find that incorporation of this factor
00:25:18.399 --> 00:25:26.739
n factorial in the expression for capital
q is the reason why in the final expression
00:25:26.739 --> 00:25:35.289
of s the tom capital n appeared here now what
would have happened if we just did not
00:25:35.289 --> 00:25:41.659
write down this n factorial we just wrote
capital q is equal to small q to the power
00:25:41.659 --> 00:25:49.789
of m which means that the corresponding expression
for entropy would have been given by something
00:25:49.789 --> 00:25:51.940
like this
00:25:51.940 --> 00:26:00.279
now what would have happened if you wanted
to check for the extensivity property of entropy
00:26:00.279 --> 00:26:07.429
you would replace n by two n here also you
will replace n by two n and you will replace
00:26:07.429 --> 00:26:24.999
v by two v and then s prime will turn out
to be two nkb ln two pi m kbt by h square
00:26:24.999 --> 00:26:40.559
to the power of three by two into two v into
e to the power of five by two n plus two nkb
00:26:40.559 --> 00:26:50.460
s electronic so immediately you can see s
prime is not equal to s and that is because
00:26:50.460 --> 00:26:57.320
there will be an extra ln two term here that
will destroy the equality and therefore we
00:26:57.320 --> 00:27:05.169
have made everything correct by incorporating
this term n here which in turn was introduced
00:27:05.169 --> 00:27:13.729
in this picture by the famous scientist
gibbs who pointed out that well unless and
00:27:13.729 --> 00:27:21.249
until you are having this factorial n then
you are not going to get the correct thermodynamic
00:27:21.249 --> 00:27:22.330
properties
00:27:22.330 --> 00:27:32.729
the other reason why why gibbs wanted
to introduce this factor factor n factorial
00:27:32.729 --> 00:27:39.149
is that obviously unless you do so you are
not taking into account of the fact that you
00:27:39.149 --> 00:27:45.659
have identical particles constituting the
gas ok we have already discussed it many times
00:27:45.659 --> 00:27:52.649
that in order to avoid over counting of entropy
over counting of the microscopic states of
00:27:52.649 --> 00:28:01.429
a system we should be dividing by n factorial
in the case where i have identical particles
00:28:01.429 --> 00:28:08.139
now let us next have a look at the entropy
of mixing of ideal gases and here also we
00:28:08.139 --> 00:28:13.999
are going to use the sackur tetrode equation
and check if the sackur tetrode equation can
00:28:13.999 --> 00:28:19.389
explain the entropy of mixing of ideal gases
correctly
00:28:19.389 --> 00:28:24.759
now the first case that we think about is
the irreversible free expansion of an ideal
00:28:24.759 --> 00:28:31.450
gas that are comprised of capital n particles
so we are basically looking at the situation
00:28:31.450 --> 00:28:40.609
where my initial system is something like
this so i have a partition separating the
00:28:40.609 --> 00:28:46.639
entire volume into two equal volumes v on
this side and v on the other side and this
00:28:46.639 --> 00:28:53.669
part is evacuated so if i now withdraw this
partition the gas will undergo a free expansion
00:28:53.669 --> 00:28:59.779
which is an irreversible expansion and the
final volume of the gas is going to be two
00:28:59.779 --> 00:29:08.769
v now in this particular case what we have
we have to consider the fact that in the initial
00:29:08.769 --> 00:29:15.679
state the number density is capital n particles
present in a volume v of the gas of the blue
00:29:15.679 --> 00:29:20.799
gas and therefore the initial number density
is n by v
00:29:20.799 --> 00:29:27.429
so using the sackur tetrode equation i can
write down the entropy of the system in the
00:29:27.429 --> 00:29:34.239
initial equilibrium state now what is the
entropy of the system in the final equilibrium
00:29:34.239 --> 00:29:41.200
state so for that i will have to know the
final number density of the gas here i have
00:29:41.200 --> 00:29:47.809
now capital n particles occupying a volume
of two v therefore i see that the number density
00:29:47.809 --> 00:29:54.200
has half and if i put it back here and since
this is essentially joules experiment where
00:29:54.200 --> 00:29:59.089
the temperature remains constant i can argue
that the lambda values are going to remain
00:29:59.089 --> 00:30:05.210
the same and within the temperature range
that we are talking about s electronic changes
00:30:05.210 --> 00:30:08.889
let us assume that they are going to be very
very small
00:30:08.889 --> 00:30:15.629
now what do you expect we have already seen
that under such circumstances we in from thermodynamics
00:30:15.629 --> 00:30:23.309
we expect an increase in entropy and by sackur
tetrode equation if i use it i find that delta
00:30:23.309 --> 00:30:30.950
s is nkbln two which is a positive number
and therefore correctly predicts the increase
00:30:30.950 --> 00:30:38.200
in entropy when i allow the ideal gas to
undergo an irreversible free expansion which
00:30:38.200 --> 00:30:44.950
anyway takes place under isothermal condition
now if i have two different gases on two
00:30:44.950 --> 00:30:51.820
sides of the partition if i withdraw the partition
it is as if as far as the yellow gas is concerned
00:30:51.820 --> 00:30:56.419
there was no yellow gas on the other side
of the partition so it is an irreversible
00:30:56.419 --> 00:31:02.299
free expansion of the yellow gas just like
the irreversible free expansion for the blue
00:31:02.299 --> 00:31:09.769
gas and therefore for each ideal gas i can
once again repeat the same argument and therefore
00:31:09.769 --> 00:31:16.299
i should be able to say that well here the
change in entropy is going to be positive
00:31:16.299 --> 00:31:23.940
and it is going to be twice the amount of
we have seen from the case one and that
00:31:23.940 --> 00:31:28.049
is correctly predicted by the sackur tetrode
equation
00:31:28.049 --> 00:31:35.599
so this brings us to the final and most interesting
case and this is a case of reversible mixing
00:31:35.599 --> 00:31:42.740
of two identical ideal gases so what is it
that we are trying to do here i have in the
00:31:42.740 --> 00:31:50.269
initial equilibrium state the same gas on
both sides of the partition and i am reversibly
00:31:50.269 --> 00:31:57.460
withdraw the partition present in the two
halves of the box now what will happen what
00:31:57.460 --> 00:32:03.210
do you expect since both the gases are
they are the same and you havent done anything
00:32:03.210 --> 00:32:08.090
from outside rather than just removing the
partition you do expect that there should
00:32:08.090 --> 00:32:13.690
not be any change in entropy or in other words
delta s is going to be zero
00:32:13.690 --> 00:32:21.619
now let us see what is predicted by the sackur
tetrode equation we find that the initial
00:32:21.619 --> 00:32:30.309
state i have capital n particles of the ideal
gas on this side and capital n particles of
00:32:30.309 --> 00:32:37.089
the gas on the other side and therefore i
have a total of two n particles of the blue
00:32:37.089 --> 00:32:43.419
type divider and present in a volume of two
v therefore the initial number density is
00:32:43.419 --> 00:32:52.029
n by v and accordingly that initial entropy
of the system is given by this now when i
00:32:52.029 --> 00:32:58.799
allow the reversible mixing under isothermal
condition what is the final density the final
00:32:58.799 --> 00:33:05.619
number density once again is n by v and therefore
sackur tetrode equation correctly predicts
00:33:05.619 --> 00:33:12.559
that the delta s in such case is going to
be equal to zero and therefore you once again
00:33:12.559 --> 00:33:20.279
see the importance of the appearance of the
capital n term inside the log and unless you
00:33:20.279 --> 00:33:28.450
have that correction its not possible to reproduce
all these informations and finally if one
00:33:28.450 --> 00:33:34.919
compares the estimated values of entropy as
obtained from the sackur tetrode equation
00:33:34.919 --> 00:33:40.200
to the ones obtained experimentally as you
can very well see from the data presented
00:33:40.200 --> 00:33:48.409
here that there is an astonishing agreement
between the theoretical prediction and the
00:33:48.409 --> 00:33:50.409
experimental results
00:33:50.409 --> 00:33:58.440
so this brings us to the conclusion that the
formalism that we have proposed through the
00:33:58.440 --> 00:34:05.899
molecular thermodynamics so statistical thermodynamics
is actually an accurate one it not only gives
00:34:05.899 --> 00:34:14.109
us a molecular derivation of the ideal gas
law the form of chemical potential of the
00:34:14.109 --> 00:34:21.490
an ideal gas or why the temperature dependent
term why the internal energy will be dominated
00:34:21.490 --> 00:34:28.440
by a term proportional to temperature at normal
temperatures it is also capable of reproducing
00:34:28.440 --> 00:34:34.270
the essential properties of entropy of an
ideal gas and it agrees very well with the
00:34:34.270 --> 00:34:41.609
experimental data so that for any theoretical
framework is an enormous achievement and in
00:34:41.609 --> 00:34:49.359
the next class we will see how these considerations
can be extended to understand the properties
00:34:49.359 --> 00:34:58.260
of diatomic gases ideal gases where the structure
is also playing underlying structure is also
00:34:58.260 --> 00:35:05.160
playing an important role by having this additional
contribution of rotational and vibrational
00:35:05.160 --> 00:35:06.160
h structure
00:35:06.160 --> 00:35:06.339
thank you