[KZ99]

E_j =h[²/8π²I] x J(J+1)

N_j/N_o = g_je^(-E_j/kT)

My molecule is CO and these are the two equations I need to workout I think.

If I know the energy which is the rotational constant (B value)

How could I possibly workout the temperature, the image shown below is something I found from a book.
Problem is I do not know N_j, N_o and g_j

[Corribus]

What's your question, exactly?

[KZ99]

Thanks for the reply
Question is that a space probe is analysing the rotational spectrum of CO in the atmosphere of saturn. Im only given the bond length as data. It is asking me how could the experiment estimate the temperature of the atmosphere

[Corribus]

Well why don't you describe your thinking.  If you've got nowhere to begin, why don't you start by thinking about how temperature might affect the rotational spectrum of a gas?  And if you don't know where to begin with that, think about where spectral lines come from and what a real spectrum might look like at absolute zero.

You might also want to look up what a partition function is.  This will set you on the right path.

[KZ99]

Thanks for the reply I found partition function

qr = Σ^∞_J=0 (2J + 1)e– ^{βhcBJ(J + 1)}

As to my first equation
N_j = population of higher energy levels
N_o= Population in the lowest energy level

I found that Intensity depends on the relative population.
The thing is we have not been taught on partition function so I can't use it.
we have been taught on Boltzmann only and to use the Boltzmann equation I only know the energy, k , don't know anything else. The question is quite distressing. 

[Corribus]

First of all, don't be distressed.  It's just chemistry. 

The partition function is related to your Nj/No formula and describes all the energy levels that can be populated at a certain temperature.  They're both derived from the same concept.  But forget about equations for a moment.  Always think about concepts first, then you can calculate.  Calculation is meaningless without understanding what you're calculating.

So, let's think about it.  (Well, you think; I'll listen.)  Do you understand why, in principle, the rotational spectrum should be able to give you information about temperature? 

[KZ99]

Do you understand why, in principle, the rotational spectrum should be able to give you information about temperature?

This is where I am stuck, I have been googling this but I can't find a source that actually answers what you written!

This is what i have found

http://books.google.co.uk/books?id=zzxLTIljQB4C&pg=PA307&lpg=PA307&dq=boltzmann+distribution+rigid+rotor&source=bl&ots=waeNIDm_gH&sig=hF0XjRlf2mw1yMA4NzPY2yK3d84&hl=en&sa=X&ei=1ktZUevSHMqJ0AWZ8oDwBA&ved=0CD0Q6AEwAg#v=onepage&q=boltzmann%20distribution%20rigid%20rotor&f=false

http://www.chem.unt.edu/~mschwart/chem5210/files/hdout-chap-4-5210.pdf slide 49 onwards

http://books.google.co.uk/books?id=yE3PoVoBDy8C&pg=PA152&lpg=PA152&dq=boltzmann+distribution+rigid+rotor&source=bl&ots=Hlw62bicEz&sig=-7po9k6zeBrcUjeiENa1rj-Hy6k&hl=en&sa=X&ei=1ktZUevSHMqJ0AWZ8oDwBA&ved=0CFUQ6AEwBw#v=onepage&q=boltzmann%20distribution%20rigid%20rotor&f=false

another problem is that the question is asking me how it could estimate! With the word estimate just wondering if I really need to do any calculations 

[Corribus]

http://www.chem.unt.edu/~mschwart/chem5210/files/hdout-chap-4-5210.pdf slide 49 onwards

OK, well this presentation does tell you almost exactly what you need to know.  Especially slides 52, 53 and 56.  But it doesn't explain anything, just shows a sample calculation.

Since you weren't provided with an spectral information, my guess is that they only want you to describe how you would go about calculating temperature from spectral data.  But that doesn't necessarily make the problem easier.

Let's start with a few concepts and build up to this answer. 

First, rotational spectroscopy - as the figure in your opening post shows, there are a number of allowed rotational states for a molecule, and selection rules indicate that only transitions between states that are ΔJ = ±1 are allowed.  (That is, you can go from J = 1 to J = 2, but not J = 1 to J = 3.  This means that in a spectrum, there is exactly one transition that originates from J = 1, one transition state that originates from J = 2, and so forth.  So what you get in your spectrum is a series of lines, each separated by 2B in energy (B is the rotational constant, specific to the molecule of interest) and each corresponding to a transition from a single state with a single energy.  The intensity of each transition is related to the proportion of absorbing molecules at an instant of time that are originating from the state corresponding to that transition.  So for instance, the intensity of the lowest energy (first) transition line depends on the number of molecules that are in the J = 0 state at the time the spectrum is acquired.  The intensity of the second lowest (second) transition line depends on the number of molecules that are in the J = 1 state at the time the spectrum is acquired.  And so forth.  In spectroscopic language, we would say that the transition intensity depends on the population of the originating state; this is a general phenomenon of spectroscopic transitions, although there are often other factors that impact transition intensity as well.  For pure rotational spectra and for transitions with ΔJ = ±1, though, we can assume that the intensity is directly proportional to the population of the originating state.

So: knowing now what you know about rotational spectra, how would TEMPERATURE impact the intensities of spectral lines?  How would TEMPERATURE impact the population of rotational states?  If you aren't sure, what would the populations of the rotational states look like at absolute zero temperature, do you think?  (Think classical physics: it takes energy to make molecules rotate.)

[KZ99]

Thank you very much for your lengthy reply. I replying late to you as i took a 2 day break from spectroscopy. I have calculated B values for this question, so I think it is more than just an explanation. I know boltzmann is definetly involved because that the only thing we have been taught regarding tempeerrure. Apart from that I honestly cannot find how temperature affect spectral lines. I have tried google and I found about doppler effect it says

Another use of the Doppler effect, which is found mostly in plasma physics and astronomy, is the estimation of the temperature of a gas (or ion temperature in a plasma) which is emitting a spectral line. Due to the thermal motion of the emitters, the light emitted by each particle can be slightly red- or blue-shifted, and the net effect is a broadening of the line. This line shape is called a Doppler profile and the width of the line is proportional to the square root of the temperature of the emitting species, allowing a spectral line (with the width dominated by the Doppler broadening) to be used to infer the temperature.

The above para is from Wikipedia

Also found from one of the slides in my previous comment that

N_j is at a maximum for d_Nj/d_N0=0

After all the question is how could the experiment estimate the temperature of the atmosphere and my mind keeps saying a calculation is definitely involved with Boltzmann.

Would it be possible for you to guide me a bit more please?

[Corribus]

Ok, I see you're going far afield.  Temperature can impact spectral lines through Doppler broadening, but the answer is much simpler, so let me help you out a bit more.

Let's suppose you have a closed vessel with your gas of interest with rotational constant B.  If there is heat available (i.e., nonzero temperature), the molecules will be rotating (and moving and vibrating, but forget that for a moment: rotation only right now).  The frequencies at which they rotate are quantized - only specific frequencies are allowed, which are specified by the quantum number J and the rotation constant B, which is related to the inertial mass and bond length.  J can be anywhere from 0, 1, 2, 3, ...∞.  There are an infinite number of rotational states, and the higher J is, the more energy the molecule has (faster rotation). 

Now, suppose you were able to zoom in with a special magnifiying glass and look at all the individual molecules in the vessel, and I asked you to make a table where you counted how many molecules in the sample were rotating with each J value, what would you expect to find?

Well, it should be obvious what you would NOT expect to find.  You would expect to find that  every molecule wouldn't be rotating with the same speed: there would be a distribution of values.  Some would be faster, some would be slower.  And there would be a mean value. 

Ok, let's do a little experiment.  Let's cool your vessel of gas down to absolute zero Kelvin and then use your magic magnifying glass again.  What do you find?  With no energy in the environment, you'd find that the molecules were no longer rotating.  NONE OF THEM.  Because there's no energy to be fed into rotational states.   Every molecule will have J = 0.

Now, let's add a little heat into the system.  Just a little.  The temperature is no long zero Kelvin it's 0.00000001 Kelvin.  Use your magnifying glass again.  What do you see?  Now there is just a tiny bit of heat, enough so that a handful of molecules start to rotate.  Most of them are still at J = 0, but now you see a few in J = 1 state, maybe even one or two at J = 2. 

Now add a bit more, and a bit more, and a bit more heat.  Each time the temperature goes up a little bit, the average rotational energy of the molecules will increase, which means the average J value increases.  There will always be a distribution of energies (except in the one case of T = 0 K, where all the molecules are at J = 0), but I think you can see now that there will be some kind of correlation between the mean J value in the sample and the temperature.

Onward, to Boltzmann!  The Boltzmann constant is essentially a way to relate the distribution of molecular energies (levels populated) with the temperature.  You should know by now that at a given temperature, there will be a distribution of kinetic energies: not all molecules move at the same speed because there are all kinds of collisions going on.  This will slow some down and speed some up.  This translates into the natural distribution of molecular states observed at any one time - and it works for all kinds of states: vibrational, rotational, whatever.  The Boltzmann constant is a statistical constant that allows you - through equations like you posted above and the partition coefficient - to say, 'If the temperature is T, what proportion of the molecules will be in state S with respect to all the possible states there can be?'  That probability will be a function of the temperature and the energy of the state.  The higher the temperature, the more energy is, the more higher level energies are likely to be populated at any time.  The Boltzmann constant allows you to quantitatively determine these probabilities.

So, with all that in mind, what can we say of spectroscopy?  I will say one more thing here and then let you come up with the rest.  The lines in a rotational spectrum originate from transitions between J and J + 1.  The difference between any J and J + 1 level is given by the energy formula for a rigid rotor (under that approximation), and the difference between any two lines should be about 2B.  If EVERY rotational level were equally populated when an experiment was performed, the rotational spectrum would be an infinite number of lines, each separated by 2B and starting at the energy difference between J = 0 and J = 1 (also 2B, so the spectrum would be an infinite number of lines: 2B, 4B, 6B, 8B, . ∞).  For example, if B is 100 whateverunits, then the spectrum would have a line at 200, 400, 600, 800, ... ∞.  The relative intensity of any line is (in principle) proportional to the number of molecules in the rotational state associated with the line at the time of the experiment.  So if two times the number of molecules are at J = 2 than are at J = 3, then the J = 2 to J = 3 line will be double the intensity of the J = 3 to J = 4 line.  (In practice line intensities are affected by other things as well, but let's keep it simple.)

At absolute zero, only the lowest rotationl level can be populated, so the spectrum at 0 K would have exactly one line: at 2B.  No other lines could be observed at this temperature because no other states can be populated - you won't see the characteristic line for the J = 3 to J = 4 transition because there are no molecules that start at J = 3 when the temperature is zero Kelvin.  You have to have molecules at a given rotational level to see the spectral line that originates from this rotational levels.

Alright.  We've already looked through your magnifying glass at what happens to molecules when you increase the temperature a little bit above absolute zero.  Of course, we don't have magic magnifying glasses.  We have spectrometers!  So, what does your spectrometer see when you start at absolute zero and increase the temperature just a little bit?  You will no longer have a single line at 2B originating from J = 0 to J = 1.  So what do you have?  And how can you generalize this to what you might observe at any temperature?

The best answer will invoke the kind of equation you posted in your opening post, but let's still forget equations.  Just think about how the spectrum will change as temperature increases. Then we can discuss how to use the equation in more detail if you wish to calculate temperature from spectral data - but you sould get yourself to the point how you understand HOW it could work before you start to even think about equations.   

[KZ99]

Fantastic Explanation, Thank You Very Much!!!!