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5:36 AM
Morning sir are you free now I have a question @JohnRennie
 
@yuvrajsingh morning :-)
 
Alesha here.
 
I'm free right now but I haven't checked if there is anything needing doing at work yet.
 
So can I ask a question?
@JohnRennie
 
Yes
 
5:38 AM
I have box, which contains a gas name X.
And we place the box in space.
 
OK ...
 
Or in vacuum assumption.
Now sir as the gas particles move, with certain k. E
Will the temperature of the box increase.
@JohnRennie
 
@JohnRennie Good Morning sir :-)
@JohnRennie sir, I figured out the solution. It's correct. :-)
 
@yuvrajsingh why would the temperature of the box increase?
@user8718165 cool :-)
 
When the gas molecules collide, with themselves and with others there will be lost of some energym
@JohnRennie
 
5:46 AM
@yuvrajsingh in ideal gas, no :)
 
The energy can't just disappear in a collision.
When gas molecules collide they can exchange energy with each other, but the total energy stays the same.
 
So I mean thus energy will not get converted into heat?
 
@JohnRennie sir total energy of the system will remain the same...isn't it :)
 
@user8718165 ywes
 
OK, so when the Tyre move on the ground why does tire get heated up then?
@JohnRennie
 
5:48 AM
@yuvrajsingh do you mean the gas moecules will heat up the material the box is made from?
 
@JohnRennie okay sir...thanks
 
Yes. @JohnRennie
 
@yuvrajsingh the box is friction-less....
 
@user8718165 friction does nothing just increases the kinetic energy of the air molecules. So as the k. E increases the pressure also increases.
Temperature of Tyre increase.
@JohnRennie yes?
 
@yuvrajsingh if the box starts out at a lower temperature than the gas then the gas will heat the box until the box And the gas are at the same temperature.
 
5:52 AM
@yuvrajsingh ah! sorry for interrupting :) You guys continue :)
 
OK sir.
 
In a car tyre the tyre is heated by friction as it rolls. The weight of the car deforms the tyre, and as the tyre rolls the rubber is continually stretching and shrinking. As this happens some energy is converted to heat.
Then the heated rubber heats the air inside the tyre.
 
@JohnRennie my reason is incorrect?
 
Friction doesn't act directly on the air molecules. Friction heats the rubber and the hot rubber heats the air.
 
Why does the rubber get heated?
@JohnRennie
 
5:56 AM
When you deform a solid some energy is always converted to heat. OK so far?
 
Sorry but I do not know that.
@JohnRennie
 
If you have a perfectly elastic material then you can do work on it to compress it, and when you let it expand again you get all the work back. No energy is lost.
In real materials you never get back all the work you put in. Some energy is always lost because it is converted to heat as the material deforms.
 
OK.
@JohnRennie continue sir.
 
And that is what happens in rubber. As the car moves the tyre rolls and the weight of the car deforms it then the rubber bounces back. But some energy is converted to heat as the rubber deforms then springs back.
That's why a car tyre gets hot as it rolls.
 
Got it.
@JohnRennie I have one question related to ideal gas equation.
 
6:03 AM
Yes?
 
Actually how can we derive the real gas equation from non ideal gas. Equation by making suitable changes
@JohnRennie
 
The ideal gas equation is $PV = RT$. And we can rewrite this as:
$$ \frac{PV}{RT} = 1 $$
OK so far?
 
Yes.
 
For technical reasons it's usually more convenient to use the density $\rho = n/V$, where $n$ is the number of moles. So for one mole we get $\rho = 1/V$. Then the equation is:
$$ \frac{P}{RT\rho} = 1 $$
For a real gas the fraction $P/RT\rho$ isn't constant but instead it changes with the density of the gas. We can approximate this by writing:
$$ \frac{P}{RT\rho} = 1 + B\rho + C\rho^2 + ... $$
i.e. the right hand side is a power series in $\rho$.
 
6:13 AM
The constants $B$, $C$, etc are called virial coefficients and this method of approximating the gas behaviour is called a virial expansion.
The classical virial expansion expresses the pressure P {\displaystyle P} of a many-particle system in equilibrium as a power series in the number density: Z ≡ P R T ρ = A + B ρ + C ρ 2 + ⋯ {\displaystyle Z\equiv {\frac {P}{RT\rho }}=A+B\rho +C\rho ^{2}+\cdots...
 
Yes sir.
 
The values of $B$, $C$, etc can be calculated from the interactions between gas molecules.
Well, in principle they can be calculated. In practice the calculation may be very difficult.
But I'd be surprised if you need to know this for the JEE. Though I think they might ask you about the Van der Waals equation of state.
 
Actually, way I was thinking about, to drive the we can consider the collision to be non elastic,.
@JohnRennie yes, sir derivation is not there.
And I am going to ask Vander wall equation after that.
 
@yuvrajsingh if a collision between two gas molecules is inelastic where does the energy go?
 
Heat?
@JohnRennie
 
6:23 AM
What we call heat is the collective motion of many molecules.
If you have just two molecules the only energy they have is their kinetic energy.
If the total kinetic energy is less after the collision than before, where could that energy have gone? There isn't any place for the energy to go.
 
I do not know? @JohnRennie
 
The point is that there is nowhere else for the energy to go, so the collision must be elastic.
 
@JohnRennie How the real gas is different than ideal gas?
 
In an ideal gas there are no forces between the gas molecules. In a real gas there are forces between the molecules. At close distaces molecules repel each other and at long distances they attract each other.
It's these forces that cause the virial coefficients to be non-zero.
The Van der Waals equation of state approximates these forces using its two constants $a$ and $b$.
 
Yes.
@JohnRennie
 
6:36 AM
@yuvrajsingh yes?
 
I am confuse why in the Vander wall equal we something in pressure, and subtract something from volume.
@JohnRennie
 
In the VdW equation we assume that the equation is $PV = RT$ where $P$ and $V$ are the real pressure and volume not the pressure and volume we measure.
Consider the volume first as that's simpler.
We measure some volume $V'$.
But the gas molecules aren't points. The molecules themselves have some volume and we'll call this volume $b$. Then the real volume is the volume we measure $V'$ minus the volume taken up by the molecules $b$ i.e. $V = V' - b$.
 
Yes
Is that the volume in which gas allow to move?
 
So our equation $PV = RT$ becomes $P(V' - b) = RT$ where $V'$ is the measured volume.
 
Due to attractive or repulsive force.
 
6:42 AM
@yuvrajsingh the volume reduction is due to repulsive forces.
 
OK sir.
 
At close distances the molecules behave like hard balls. That is you can't push them any closer to each other than twice their radii.
So that's why in the VdW equation we modify the volume term by subtracting $b$ from it. OK so far?
 
Yes.
 
Now, the $a$ term comes from the attractive forces between the molecules.
If you start with an ideal gas it has some pressure $P$. But if you now make the molecules attract each other then this reduces the pressure because it pulls the molecules inwards towards each other while the pressure acts outwards away from each other. OK so far?
 
6:49 AM
So the pressure we measure $P'$ is less that the real pressure $P$: $P = P' + \Delta P$
where $P'$ is the measured pressure and $\Delta P$ is the reduction in pressure due to the attractive forces.
But $\Delta P$ depends on the volume because the attractive forces decreases as the molecules get farther apart. So at large volumes $\Delta P$ is small and at small volumes $\Delta P$ is large. So we write $\Delta P = a/V^2$ for some constant $a$.
Then the real pressure is $P = P' + a/V^2$.
So now our equation looks like:
$$ (P' + \frac{a}{V^2}) (V' - b) = RT $$
where $P'$ and $V'$ are the measured pressure and volume.
And that's the VdW equation.
 
@JohnRennie, Good Morning Sir :-) Your new hat looks great!
 
@M.GuruVishnu morning :-)
Actually I need to change the hat. I normally change it every day.
 
7:06 AM
@JohnRennie Is this known as a dynamic equilibrium? :-)
 
:-)
 
@JohnRennie: Are you free now sir?
 
@M.GuruVishnu I need to work for about 15 minutes ...
 
@JohnRennie Ok sir. No problem :-)
 
7:30 AM
@M.GuruVishnu OK, I'm free for about half an hour now.
 
@JohnRennie I was just on the main site, so I was lucky to receive your ping. I hope you would have noticed this question:
1
Q: Confusion in the sign of work done by electric field on a charged particle

M. Guru VishnuIn my book it is given that, the work done to transport a charge $q$ through a potential difference $\Delta V$ is $q \Delta V$. Or mathematically, it can be written as follows:$$W_{\text{electric}}=q\Delta V.\tag{1}$$ We know that potential difference is defined as $\Delta V=\frac{\Delta U}{q}$ ...

One user had suggested to read a linked answer, but I was unable to understand that answer.
 
The sign of work is always confusing because it's often unclear whether we mean work done by the charge of work done on the charge.
 
@JohnRennie Here, I think in both the cases, it's work done on the charge. But I could be wrong. I faced some trouble in thermodynamics too on the same part where different books in physics and chemistry used different conventions.
 
To be honest I usually figure out the sign from the final result. If you're pushing a charge $+q$ towards a charge $+Q$ the PE is obviously increasing because you are increasing its energy. That is, if you let go of the charge it will accelerate away as its PE changes to KE.
 
@JohnRennie Yes sir. I too think it this way - are the force and displacement in the same direction or opposite direction? I tried to approach it a bit theoretically and I got into trouble.
 
7:40 AM
It's easy to get mixed up with the signs. I recommend applying common sense! :-)
 
@JohnRennie Yes sir. Thank you for the advise! :-)
@JohnRennie, I know paper is made of cellulose. Is cellulose electrically a dipole?
 
@M.GuruVishnu Cellulose is a carbohydrate so it contains only carbon, hydrogen and oxygen atoms.
The OH bond is polarised so it has a dipole moment. Indeed this causes the hydrogen bonding that holds cellulose molecules together.
But cellulose as a whole doesn't have a net dipole.
 
@JohnRennie, We know paper is an insulator (valence electrons are bound to the nucleus) and now we've discussed it has no net dipole moment. Then why does paper pieces get attracted to an electrically charged rod due to induction sir?
 
Nothing is a perfect insulator.
Paper has a very high resistance, so any currents that flow in it in response to a voltage are very small.
But the charge induced by the charged rod is very small, so the currents needed to produce that charge separation are also very small. So even though paper is a very poor conductor it still conducts enough to get some charge separation and hence an attraction to the rod.
 
@JohnRennie Ok sir. Now understood. Thank you :-)
 
8:00 AM
I need to work now for half an hour or so ...
 
 
1 hour later…
9:06 AM
Hello sir @JohnRennie
 
@user8718165 hi :-)
 
@JohnRennie how are you sir?
I want to ask you a little qn.
 
@user8718165 yes ... ?
 
@JohnRennie Sir suppose a jar of gas is moving with a constant v, can we say that all the gas particles are having a net velocity of v?
 
Yes
 
9:10 AM
@JohnRennie thank you very much sir...
@JohnRennie sir please come in general chat :)
 
9:36 AM
@user8718165 sorry brother my just hang.
It was by mistake.
 
10:32 AM
@JohnRennie Hello
:-)
@JohnRennie please help with amperian loop
 
@Jasmine hi
@Jasmine hello ??
 
10:48 AM
@JohnRennie hello
 
Hi
What's the question?
 
For a plain wire carrying current i we need to find the value of B at a distance x from wire
We can do it using Ampere circuital law
 
@Jasmine yes
 
@JohnRennie I want to know if the choice of loop can be a square along cross section
I know the best choice will be a circular loop along cross section
 
You can choose any loop you want ...
 
10:51 AM
But I just wanted to know if a square can do as well
 
A square loop would work but would be a nightmare to integrate along.
 
@JohnRennie yup but it should give value of B
@JohnRennie isnt B constant
Ohh
 
@Jasmine the trouble is that $B$ varies with distance from the wire, and your square loop varies the distance from the wire as you go round it.
 
B is not constant just then angle between B and dl is constant and that is 0
@JohnRennie yup my bad
Silly doubt
 
No problem :-)
 
10:55 AM
1 min ago, by Jasmine
B is not constant just then angle between B and dl is constant and that is 0
@JohnRennie is this correct ^
I feel at times my ability to visualise is decreasing with age
 
Yes. It's the integral of B round the path that is constant i.e. $2 \pi r B$ is a constant.
 
Okay :-)
@JohnRennie one more question
 
Yes?
 
A hint would do
 
@Jasmine OK ...
 
11:00 AM
There are two wires, one infinitely long , other with length a with current i in both
They are skew lines
Mutually perpendicular
Distance between them x
Need to find force of interaction
 
Let me draw a diagram ...
@Jasmine I've drawn the infinite wire normal to the page where the red dot is, so the field lines from the finite wire are normal to the page.
Then the finite wire of length $a$ is the solid line in the plane of the page a distance $x$ away from the solid wire.
 
@JohnRennie ok
 
Is this the correct layout?
 
@JohnRennie yup not sure what r,y is representing though
 
@Jasmine consider a distance $y$ away from the centre of the finite line.
I've drawn a field line that passes through this point on the finite wire. It's a circle of radius $r^2 = x^2 + y^2$
So the field of the infinite wire at this point is $B(r)$ and in the direction tangential to the circle.
The force on the wire at that point will be $d\mathbf F = (\mathbf B(r) \times d\mathbf y)I$
@Jasmine OK so far?
Oops, I think I got my cross product the wrong way round. It should be $d\mathbf y \times \mathbf B$.
 
11:15 AM
@JohnRennie ok
 
And you're going to get the total force by integrating $d\mathbf F$.
 
@JohnRennie ok
 
Is that enough of a hint?
 
B(r) can be calculated , dy= dr
limit from -(x^2+(a/2)^2)^1/2 to +(x^2+(a/2)^2)^1/2
 
$d\mathbf y$ is the element measured along the finite wire i.e. in the same direction as the distanec $y$ I've drawn. So $d\mathbf y$ is not equal to $dr$.
 
11:25 AM
@JohnRennie I differentiated this
x is constant
@JohnRennie please write the final equation
The calculation can be left but I am not sure if I am getting the correct thing
 
@Jasmine I'm just answering a question in another room. I shouldn't be too long ...
 
@JohnRennie OK
And sorry it should be rdr=ydy
@JohnRennie I need to leave in 8 minutes
The doubt I have is what will be dy×B
 
dy x b points normal to the page
 
@JohnRennie yup direction is constant
 
It's magnitude is $dy B \sin\theta$, where $\tan\theta = y/x$
 
11:39 AM
@JohnRennie what is B
Isnt B varrying with y
B is varrying with r
So its going to be too lengthy
 
$$ B = \frac{\mu_0 I}{2 \pi r} $$
Yes?
And $r = \sqrt{x^2 + y^2}$
 
Yup
Even angle between dy and B is changing
Too much variables
 
And $\sin\theta = y/r$ so:
 
Ok its integrable
Got it
 
$$ B\sin\theta = \frac{\mu_0 I y}{2\pi r^2} = \frac{\mu_0 I y}{2 \pi (x^2 + y^2)} $$
 
11:45 AM
@JohnRennie :-) yup I got it
 
@Jasmine cool :-)
 

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