@JohnRennie: Hi sir. Good morning :-)

@GuruVishnu hi :-)

May I ask a small doubt about $J$ from thermodynamics sir?

@GuruVishnu yes

Ok sir.

For the following question:

The solution is the following:

Of course the final answer attained by this method is correct. But I don't understand what exactly is $J$? Earlier I used it for joules as well as the mechanical equivalent of heat.

But here I don't understand why how they equated nothing to $J/\text{cal}$

Or in other words, where did that $J$ come from?

* Oops, there's a typo in the graph. It must be $0.02~\rm{m^3}$ instead of $0.2~\rm{m^3}$.

@JohnRennie: Hi sir.

I think it's just a poor choice of symbols.

@JohnRennie Ok sir. What is the $J$ the author is referring to? Is it something else?

I think the question means if the internal energy increases by 5000x calculate the value of x

i.e. J is a variable and you need to calculate its value. They shoukd have written 5000 J not 5000 J.

But finally the value of $J$ turns out to be its measure of $1$ calorie. Somewhat close to $4.186$ J.

Is it a coincidence?

This is a spectactularly badly written question ...

In fact I don't understand the question at all ...

Ok sir ...

@GuruVishnu Oh, wait

I think I understand what it means

Ok sir. Then could you explain?

The system does work of 6000 joules between a and b, and no work between b and c, so the total work done by the system is 6000 joules. Yes?

@JohnRennie Yes sir.

So if no heat is supplied the internal energy will decrease by 6000 joules, but we are told the internal energy increases by 5000 joules. So that means 11000 joules of heat must have been added. OK so far?

@JohnRennie Yes sir.

And the question says 2625 calories of heat was added, so 2625 calories = 11000 joules i.e. you are just being asked to calculate how many joules there are per calorie i.e. J is the conversion factor between joules and calories.

Ok sir.

You already spotted that the answer for J was indeed the number of joules per calorie.

@JohnRennie Yes sir. So here it's neither the mechanical equivalent nor the symbol for joule.

But this question was terribly confusing even though it looks very simple.

I agree. I've seen some bad questions but this is is especially awful.

Hm. It's quite surprising that this is from Concepts of Physics by H.C. Verma.

@JohnRennie: Hi sir :-)

@GuruVishnu hi :-)

I'm having a small doubt in finding the number of degrees of freedom from the adiabatic index $\gamma$. The values of $\gamma$ for CO2, H2O and CH4 are 1.30, 1.31 and 1.31 respectively. And after this table the book says:

> The table suggests that the degree of freedom of polyatomic molecules CO2, H2O and CH4 is 6.

Using $\frac{n+2}{n}=\gamma$

For $\gamma=1.30$, I get $n=6.667$ instead of $6.000$. Could you please explain how the author found the exact value of the index using 1.30 and generalised the same for all polyatomic molecules?

(The values of $\gamma$ was given for gas at $15^\circ~\rm C$)

The book uses the word suggests so this isn't an exact calculation.

The problem is that the vibrational modes are only slightly excited at room temperature.

If we take a non-linear triatomic then it has 3 translations DOFs and 3 rotational DOFs. Yes?

@JohnRennie Ok sir. Since it lies close to 7 than 6, can we say DOF=7 instead of 6? Or is there any kind of rule which must be followed to assign DOF when the calculation gives some intermediate number?

@JohnRennie Isn't it 4 rotational DOFs?

4 rotational DOFs? Where would the fourth axis of rotation be?

@JohnRennie Three axis in the plane of the triatomic molecule and the fourth axis perpendicular to the plane.

The axes of rotation lie along $\hat i$, $\hat j$ and $\hat k$.

There isn't room for a fourth axis unless you're in a four dimensional space.

The three axes of rotation are the principal axes.

Ok sir. I thought we're supposed to count the number of independent axis around which the molecule could rotate instead of finding the number of possible coordinate axes. I haven't used rotational part beyond two atoms, so I don't know what the convention is.

The rotation axis must pass through the centre of mass of the molecule.

So you can't have multiple parallel rotation axes.

@JohnRennie Yes of course. I don't see this fact to be violated in the 4 possible axis.

May I draw a rough diagram to make the point clear or did you already understand what I'm telling?

You're going to have to draw the four axes because I don't understand what you are proposing.

Sorry for my excellent drawing skills. I hope the following makes the point clear. Power point will take years to load on my system.

Blue-atoms

Black-bonds

Red-axes

Those are the three axes for a triatomic.

:54042358 oops :-)

Ok sir. I understood you point.

In your diagram the three axes 1, 2 and 3 are not independent

That is a rotation about axis 3 can be made up from a sum of rotations about axes 1 and 2

So you only have two independent axes in the plane of the molecule.

@JohnRennie Ok sir. Now understood where I went wrong. Thank you sir :-)

So for the following message:

Yes sir.

OK, so $C_v = 6 \times \tfrac12 R = 3R$

And $C_p = C_v + R = 4R$ so $\gamma = 4/3$.

Yes sir.

But this is an upper limit because any excitation of the vibrational modes will decrease this value. So actually for a triatomic $\gamma \le 4/3$.

Suppose we had seven degrees of freedom then we would get $\gamma \le 1.286$. Yes?

($\gamma \le 9/7$)

@JohnRennie Yes sir.

But the measured gammas are 1.30 and 1.31, and both are greater than 1.286.

Ah. Yes and this means DOF=6. Now I get how it's assigned.

Did I get it sir?

So the conclusion is that we have 6 fully active DOFs plus a contributuon from partial excitation of additional DOFs.

@GuruVishnu yes, though don't take this too seriously as it isn't intended to be a precise calculation.

@JohnRennie "... from partial excitation ..." - so DOF can be a non-integer too, sir?

@JohnRennie Ok sir :-)

Suppose the first vibrational excited state has an energy $E$, then the probability of finding the molecule in that state would be $e^{-E/kT}$. Yes?

@JohnRennie Sorry sir. I haven't seen that formula yet.

Boltzmann distribution?

@JohnRennie Ah! Yes I know it. But don't remember the exact formula for dN/dv. Ok sir.

Anyhow the point is that at very high temp $e^{-E/kT} \approx 1$ so molecules are freely moving between the two states. That means there is a degree of freedom associated with the vibrational state.

Yes sir. Does the "state" here refer to "thermodynamic state" or some kind of molecular state (electron excited state)?

At very low temperatures $e^{-E/kT} \approx 0$ so there is no degree of freedom associated with that state.

@GuruVishnu I'm talking about the vibrational states.

Ok sir.

The vibrational states are quantised, so from the ground state you need to add an energy $h\nu$ to reach the first excited vibrational state, where $\nu$ is the frequency of the vibration.

I don't know if you do the quantum harmonic oscillator in JEE. If not don't worry as you don't need to know the details.

So it seems the energy of a molecule is not quantised in steps of $\frac 1 2 k T$. Earlier I thought this value steps up only by integral multiples.

We're getting sidetracked.

@JohnRennie Yes sir. It's not in our syllabus.

The point I am making is that at very low temps the vibrational mode contributes zero DOFs and at very high temp it contributes one DOF.

@JohnRennie Oh. Sorry for that. I didn't realise energy quantisation is irrelevant to vibrational states.

In between it's a bit messy. It kind of acts as a fractional DOF.

@JohnRennie Isn't it 2? One the kinetic energy of vibration and the other potential energy due to the forces?

Oops, yes, two DOFs.

$$\frac 1 2 \mu v^2+\frac 1 2 kr^2$$

@JohnRennie Ok sir :-)

Anyhow, you can get partial DOFs in this way. That's why $\gamma$ can have values in between 5/3, 4/3, etc

Ok sir. I got some idea on this. If possible could you tell something about the energy quantisation? :

You need to distinguish between the energy of a single molecule and the average energy of an ensemble of molecules.

Suppose we consider a diatomic (because that's simple)

Ok sir. I see that the above one is for a collection of molecules and not for an individual one.

@JohnRennie Ok sir.

It has a single vibrational mode and the energies of the vibrational states are $\tfrac12 hf$, $\tfrac32 hf$, $\tfrac52 hf$, etc.

The vibrational energies are quantised just as the electron energies are quantised for a hydrogen atom.

$h$ Planck's constant and $f$ frequency?

Yes

If you look at a single atom then it will be in one of the vibrational states.

Ok sir.

But suppose we have a hundred atoms and 1 is in the first excited state while the other 99 are in the ground state.

Then the average vibrational energy (energy above the ground state) is $0.01 hf$.

Ok sir. So far everything seems simple.

So we'd calculate that on average there were 0.02 vibrational degrees of freedom, and each one gets an energy $\tfrac12 kT$.

But there isn't really a fractional excitation. We only get a fraction because we are averaging over many molecules.

@JohnRennie Yes sir.

I need to go now. I'll be back in half an hour or so.

@JohnRennie Did we double 0.01 to 0.02 just because we're considering a diatomic molecule? Can we just extrapolate it to higher order molecules?

@JohnRennie Ok sir. Good bye.

@GuruVishnu I'm back if you want to pick this up again.

@JohnRennie: Hi sir :-)

Are you free now? And how long will you be available?

@GuruVishnu hi :-)

@JohnRennie Are you free now, sir?

@GuruVishnu hi. I was on the phone, but I'm free now.

Ok. No problem sir.

@GuruVishnu what do you want to ask?

@JohnRennie Some time ago I said:

> The values of $\gamma$ for CO2, H2O and CH4 are 1.30, 1.31 and 1.31 respectively.

Yes ... ?

Only now I see that CO2 is a linear molecule unlike H2O and CH4.

How come their DOF or gamma values are so close to each other?

Hmm, yes, I hadn't spotted that.

So for CO2 gamma should be 3.5/2.5 = 1.4

I noticed that when I revisited our conversation. Is there any reason for that?

@JohnRennie That's same as a diatomic molecule without vibration - DOF=5

I don't know is the simple answer. Maybe the vibrational frequencies are lower in CO2 so they contribute more to the degrees of freedom.

There is data for Cp here

It would be interesting to see how $\gamma$ changes with temperature

@JohnRennie decrease as vibration needs to be considered.

^ Sorry for that. Accidently deleted a part of the message:

> Only thing I could guess is as temperature increases, $\gamma$ must decrease as vibration needs to be considered.

It seems pressing the up arrow engages the edit window. Cool.

It looks smoother than I expected. I guess the vibrational modes come in so gradually that there is no obvious step between the values of $\gamma$.

Hm. Yes sir. Thank you very much for the graph :-)

I suspect it might be a misprint in my textbook. According to the graph for temp = 200°C, the value of gamma is more than 1.35. However, for temp=15°C, it's given to be 1.30 for CO2.

Those temps are in K not °C

Or maybe, gamma might have an increasing profile initially attains maximum then reduces.

@JohnRennie Oops. Ok sir.

@JohnRennie: A different question sir:

> Is a slow process always isothermal? Is a quick process always adiabatic?

For the first one, no. We can very well have a slow process in an adiabatic container which will not be isothermal.

But I think for the second one, a quick process is always adiabatic.

Are there any cases when a quick process is not adiabatic?

@GuruVishnu it's not a well define question. Whether a process is adiabatic or not depends on the relative rates of the process and thermal equilibration. If you had some system where thermal equilibration was very fast then even a fast process could still be isothermal.

Although basically any process will be adiabatic if you do it fast enough. It's just a question of how fast that has to be.

Ok sir. Thank you for the clarification.

@JohnRennie: Hi sir :-)

@GuruVishnu hi :-)

After some time, when opportunity window reopens, I'm planning to buy a new laptop as the current one I'm using is more than 5 years old and started giving few troubles, mainly - in the bottom part about the size of taskbar, red pixels have completely stopped working, only 50% of blue pixels are active and green pixels work well. Due to this the colour is kind of inverted.

Could you please guide me on how to choose the appropriate laptop which I most probably be using for reading and some simulations?

As of now, I'm having some threshold specifications in my mind like a 15.6" monitor with numerical keypad.

@JohnRennie: Shall we discuss about this now or have it for a future discussion?

It's surprisingly easy to choose a laptop because there are no bad laptops on sale any more.

You already know you want a 15" screen and a numeric keypad, and the only other thing that you must, must get is a a solid state disk (SSD). SSDs are the single most important factor in making a laptop fast.

Ok sir. Previously you mentioned a 8 GB RAM is mandatory for Windows 10 as I'm currently running on a 2 GB one. Is there anything else I need to look for, sir?

Yes, 8GB RAM is the starting point. 16GB is better, but you'd only find that in the expensive laptops and 8GB is fine for most purposes.

I quite like Lenovo laptops. I have a Lenovo Chromebook and my niece has a top end Lenovo laptop.

But Dell and HP are good as well.

I too like Lenovo mainly because I've been used to it for more than 5 years.

So, most probably, my domain is limited to this brand as of now.

@JohnRennie When I looked at the Lenovo site for laptops, I noticed some have both HDD and SSD. Some have only HDD and others only SSD. Which one to choose?

SSD

You must get an SSD.

Seriously, if you get the old type HDD you'll be regretting it for years.

Something like an Ideapad S340 15

lenovo.com/gb/en/laptops/ideapad/s-series/…

Yes sir. I understand that. But it seems SSDs have less memory to cost ratio than HDDs. So they are combined in this fashion - 1 TB HDD + 126 GB SSD. Do I need to prefer a pure SSD over a dual memory like SSD + HDD?

Pure SSD. You don't normally need a huge disk in a laptop. You can use an external disk if you need to keep really large files like loads of films.

Ok sir. What is the minimum amount of space needed for a laptop to function properly? BTW I don't watch films much, so I doubt what I'll using the memory for.

256GB should be fine, and that's pretty standard in laptops these days.

Ok sir. What kind of processors I need to looks for? Intel or AMD?

The former is more familiar name to me as the current laptop I'm using runs on an Intel Celeron.

I'd choose Intel.

But the way to do this is to look on Flipkart (when they reopen) and see what they have in your price range.

Ok sir. I'm actually looking on the official Lenovo website.

It's open currently, but it seems they'd start shipping only after 16th April.

lenovo.com/in/en/laptops/ideapad/s-series/…

Look at that one.

Ok sir.

@JohnRennie Looks better than the one I'm currently having. It seems the middle one matches with the "pure" SSD specification.

But I think I don't need a graphics card as I'm not going to play extreme games.

Isn't Integrated graphics sufficient for me compared to NVIDIA® MX230 2GB GDDR5?

Or does it serve some purpose other than games?

Yes, integrated graphics would be fine, but I'm not sure if Lenovo do the sort of laptop you want with only inegrated graphics.

(Shall we continue our discussion either later today if you are available or tomorrow? I'm going to have my dinner.)

OK. I'll be around tomorrow as usual.