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00:01
I feel so sleepy though I slept for like 3 hours in the evening, probably due to lack of socialization . It's 2 am now.
 
5 hours later…
05:26
@JohnRennie Hi
@HannahMontana hi :-)
Can you please look into my above questions ?
If you heat the gas by a temperature ΔT then the heat that goes into heating the gas is given by its specific heat i.e. Q = C ΔT. Yes?
@JohnRennie Yes
And that heat is just the change in internal energy. So if the specific hear is constant (it is constant for an ideal gas, not for a real gas) then at a temperature of T in Kelvin the internal energy would just be U = CT.
As I discussed above, for a monatomic ideal gas this energy is stored as the KE of gas molecules.
When you define the internal energy as U = CT it makes no reference to the velocity of the container. It's just the heat needed to heat it from absolute zero up to the temperature T. Yes?
05:40
Yes.
@JohnRennie Please ping.
Now, we can also define the specific heat as the KE of the gas molecules. Each molecule has an energy ³⁄₂kT so if we have one mole, i.e. Na molecules, then the total energy is U = Na k T.
And Na k = R so the total energy is ³⁄₂RT.
And comparing this with the previous definition we see that ³⁄₂R = C i.e. the specific heat is ³⁄₂R. And you should recognise this as the value of Cv for an ideal monatomic gas.
@JohnRennie Yes exactly
So both approaches give the same result for the molar internal energy i.e. U = ³⁄₂RT.
But, this only works when we take the KE of the gas molecules in the COM frame i.e. the frame in which the gas as a whole is at rest.
OK so far?
05:45
@JohnRennie Yes , ok.
That's why we need to use the COM frame. If the gas as a whole was moving at some speed v there would be an extra KE ¹⁄₂mv² and the energy calculated from the KE would no longer be the same as the energy calculated from the specific heat.
is the question does change in frame affect internal energy of ideal gas?
@JohnRennie Hmm.. yes ok.
@ChristinaMelita it must ..
Thanks @JohnRennie :))
05:50
:-)
:58384777 no, the molecules have random directions so vx, vy and vz are not all equal
But the average KE is ³⁄₂kT so the average modulus of the velocity is given by ³⁄₂kT = ¹⁄₂mv²
oh isnt it part of kinetic theory of gases? to assume the gas have equal velocity in all three directions.
where v² = vx² + vy² + vz²
@ChristinaMelita no
because my faculty said that and used it to prove KE
@JohnRennie oh
The average velocity is zero in all three directions if that's what you mean.
its probably a primitive proof for us to understand.
@JohnRennie umm no like we assumed velocity of every molecule to be same while solving.
05:57
Do you mean the velocities of all the atoms are the same but they are in random directions?
ya
but the rms iis same in x y and z direction
Ah, yes, that's true.
oh ok
thank you
And i didnt understand the conclusion of the doubt the previous person asked
what was the final conclusion?
@ChristinaMelita Hannah originally asked:
> Why we define internal energy in COM frame ?
06:04
And the answer is that the internal energy calculated from the KE only matches the energy calculated from the specific heat if we measure the KE in the COM frame.
@JohnRennie Hi! :-)
oh is it that if we measure from COM frame internal energy=Kinetic enrgy?
Yes
@Wolgwang hi :-)
In Boolean logic, What does $X\overline{Y}\cdot (X+Z)$ mean?
$(X\overline{Y}) \wedge (X+Z)$
or $(X)\wedge (\overline{Y})\wedge (X+Z)$?
i think both are same is it not?
06:08
I think so.
(A^B^C)=(A^B)^C=A^(B^C) right?
thank you @JohnRennie
We can skip $(X\overline{Y})$ while making truth table, right?
wdym skip i dont understand.
NVM I got it :-D Thanks
06:19
oh ok xD
 
3 hours later…
09:02
Why this could have got a downvote?
I don't know. People aren't necessarily logical in their downvoting. I have also had downvotes for no obvious reason.
I upvoted it to compensate for the unfair downvote :-)
Thanks XD
09:15
@PrateekMourya A lot of people would nuke remotely homework qns . . .
2
09:39
@JohnRennie Can anything be proved in Physics?
@Wolgwang this is a somewhat philosophical question. While questions like this can be fun to argue about mist physicists are not very interested in them.
The current notion of the scientific method is down to a philosopher called Karl Popper.
His idea is that you create a hypothesis and then you do an experiment to test it. If the experiment shows your hypothesis is wrong then you go away and create a new and improved hypothesis, then you test the new hypothesis and so on.
But suppose the experiment doesn't disprove your hypothesis, does that mean your hypothesis is true? Well not necessarily. After all there are many different experiments you could do to test your hypothesis and any one of those experiments could disprove it.
So to prove your hypothesis you'd have to do every possible experiment. Then, and only then, could you state confidently your hypothesis is true.
@Wolgwang OK so far?
Yes.
So no theory is permanent, there is always a possibility that it can be disproved?
Well look what happened with Newton's laws. Until 1905 we thought they were always true. Then we discovered that at high speeds we get relativistic effects and Newton's laws don't apply.
So it took over 300 years to disprove Newton's laws.
Ah!
Is there any difference between model and hypothesis?
Now we think quantum mechanics is true because so far it has never been disproved, but quantum mechanics is less than a hundred years old. Give it another two hundred years and who knows?
8
@Wolgwang I'm not sure the terms model and hypothesis are precisely defined. I would consider them to mean much the same, but maybe philosophers of science have different definitions for them.
09:54
Thanks :-)
 
1 hour later…
10:58
@Wolgwang in my unsolicited amateur opinion, i think hypothesis (which i think is synonymous with postulates) is when scientist assume/guess (in Feynman's own words) where they just guess what is most likely to happen and model is when they give a description of how the things involved
You can think of comparing and contrasting Bohr model and Bohr postulates as an example
Thanks :-)
@Wolgwang My pleasure. Also if you are more interested, i suggest looking up one of Feynman's lecture at MIT on the scientific method where he talks about how theories are made
@JohnRennie Good evening sir. Can you tell what is meant by Va- Vd in this case, with respect to the work done by the Electric field and the external force. Like from where to where?
I got the correct answer which 20V, but I am not sure if my concept is write. I simple did Work= Force*perpendicular distance with F=electrical force (cause work per unit charge is V and force per unit unit charge is E)
11:42
@AdilMohammed you've got to use force*parallel distance. you were lucky this was a square
$$\int{\textbf{F}\cdot d\textbf{l}}=W_{done}$$ the dot product signifies that the magnitude of force along the length (or the magnitude of length along the force travelled) is the work done
12:31
Oh yes you are right, i should have written parallel distance but what about the rest?
 
1 hour later…
13:36
@PrateekMourya Yes
14:03
@Wolgwang I haven't seen the linked comment to this one but the post you have shown here is french( I think)
@napstablook Yes.

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