@JohnRennie hi

@AshishAhuja Hi :-)

In the following question, can $TV^{\gamma - 1} = c$ be applied? I don't think it can since its an irreversible process but the solution uses it to arrive at the answer.

Yes it can be applied.

..so it can be applied for irreversible processes or is this not an irreversible process?

It can be used for all adiabatic processes whether irreversible or not. That's because for all adiabatic processes we know dQ = 0 regardless of how the change happned.

@JohnRennie Hi !, Is this similar to the one you explained yesterday ? youtube.com/watch?v=-jJ5PPcLUw8

@KavinIshwaran The train and lightning bolts experiment is one of the original thought experiments used in the development of special relativity, and it's often taught to students.

yes, so its the same that you explained yesterday ?

However it isn't usually analysed using the metric approach that we talked about yesterday. It's usually analysed using the principle that the speed of light has to be the same for both observers.

If we see it mathematically through the metric, We can see that my time frame and your time frame is not the same when I move relative to you, I wonder if it can be seen that way in the video

@JohnRennie yes...

You could analyse it using the metric approach, though I don't want to do that right now as it's a bit involved and I have other stuff to do.

@JohnRennie sir I dont think that is true

from what I remember $PV^{\gamma}=C$ holds for a reversible process only

@JohnRennie No problem :)

@satan29 Hmm, now I think about it a Joule expansion is adiabatic and in a Joule expansion the temperature is constant.

So I guess you're correct.

So in that case the question is wrong.

@AshishAhuja Ignore me, I got this wrong!

what would be the correct way of solving the question? Or is there too little information?

There is too little information. You'd need to know the mass of the piston and the area of the piston so you could calculate how the piston accelerates.

can we calculate dW?

ah yes ok. I have one more question to ask:

OK, I'll try to get this one right :-)

@satan29 I don't think we can..

@satan29 The work includes the change in the KE of the piston.

hmm

@JohnRennie if the piston is assumed to be massless?

ok but yeah then we would still need to know the external pressure, which we don't know

Then the expansion is instantaneous (approximately) and it is a Joule expansion.

dW = P_ext dV and we don't know what P_ext is

Yes

yes precisely..

do note "gas", not "ideal gas".

I took a guess that T_p > T_q which turned out to be correct but I'm not sure why..

I don't think the question can be answered unless you add the extra condition that the two states are in equlibrium.

@AshishAhuja hang on, a massless piston means effectively adding the constraint pext=pint no?

@satan29 no, that's only true for reversible/quasi-static processes..

@JohnRennie equilibrium with what?

With each other

@JohnRennie why is such a condition necessary for the question to be answered? And if you assume this condition, then how would you answer it?

If you have two isolated systems with different values of Cp and/or Cv and we know nothing else about them I don't see how the question can possibly be answered.

But if you place them in contact you could consider what happens if one sample of gas transfers heat to the other or does work on the other.

The question probably means that you have a gas at a certain temperature T_p in a closed vessel, for this T_p the Cp - Cv value is R. When you change the temperature of the system to a temperature T_q, then Cp - Cv = 1.1R

I'll have to leave this for now I'm afraid.

hmm ok

@JohnRennie sorry, one more thing: if you have sodium stearate in oil at a certain concentration/temp such that micelles are formed, would the alkyl part point out of the sphere or towards the centre of the sphere? The exact question is:

The micelles are in an oil solution so the medium around the micelle is hydrophobic. Yes?

yes

So the hydrophobic alkyl groups will point outwards into the hydrophobic oil.

The hydrophilic COONa groups will point inwards away from the oil.

that's what I think too, but the given answer is (1) :-/ anyway, I need to go for lunch now, thanks.

There's an error in the question. It probably meant a solution in water. I'm not even sure sodium stearate forms micelles in an oil solution.

@JohnRennie Sir can you explain that when you are free ?

@JohnRennie hi

Two bodies of masses m1=40 kg and m2=60 kg are attached to the end of the string of negligible mass and suspended from massless pulley. The acceleration of the bodies(g=10) is-

$(T-m1g)-(T-m2g)=(m1+m2)a$

Now I want to ask can I take a as-a in eqn because I know acceleration is downwards? Or should I leave it as a?

and solve as it is. Which one will give correct answer?

I suppose correct answer is -2 which can only come by taking -a.

A chain of mass M and length L hangs half over the table and half hanging. The friction coefficient between surface of table and chain is $\mu$. We need to find the final speed with which it falls off the table

@JohnRennie i used Work energy theorem but my ans seems a little off from the actual answer

@cOnnectOrTR12 you can take any way

if you take it -a you already take it to moving down so $|a|=2$

if you take it a

going up that is it will be -2, indicating that it's not moving along a but opposite to it

i'm one of the bad explainers here, JR might explain in a better way

@JohnRennie we solved the chain one, don't worry about it :-)

But the solution has simply used angle of repose: $\tan 45^{\circ}=\mu=0.3x$ and I don't understand why they are using this...

@Wolgwang It is just $mg\sin\theta=\mu mg\cos\theta$... -_-

Ignore the above question.