Problem Solving Strategies

@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.

user8718165

5:52 AM

@yuvrajsingh ah! sorry for interrupting :) You guys continue :)

yuvraj singh

OK sir.

John Rennie

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.

yuvraj singh

@JohnRennie my reason is incorrect?

John Rennie

Friction doesn't act directly on the air molecules. Friction heats the rubber and the hot rubber heats the air.

yuvraj singh

Why does the rubber get heated?

@JohnRennie

John Rennie

5:56 AM

When you deform a solid some energy is always converted to heat. OK so far?

yuvraj singh

Sorry but I do not know that.

@JohnRennie

John Rennie

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.

yuvraj singh

OK.

@JohnRennie continue sir.

John Rennie

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.

yuvraj singh

Got it.

@JohnRennie I have one question related to ideal gas equation.

John Rennie

6:03 AM

Yes?

yuvraj singh

Actually how can we derive the real gas equation from non ideal gas. Equation by making suitable changes

@JohnRennie

John Rennie

The ideal gas equation is $PV = RT$. And we can rewrite this as:

$$ \frac{PV}{RT} = 1 $$

OK so far?

yuvraj singh

Yes.

John Rennie

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 $\rho$.

yuvraj singh

OK.

John Rennie

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.

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...

yuvraj singh

Yes sir.

John Rennie

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.

yuvraj singh

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.

John Rennie

@yuvrajsingh if a collision between two gas molecules is inelastic where does the energy go?

yuvraj singh

Heat?

@JohnRennie

John Rennie

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.

yuvraj singh

I do not know? @JohnRennie

John Rennie

The point is that there is nowhere else for the energy to go, so the collision must be elastic.

yuvraj singh

@JohnRennie How the real gas is different than ideal gas?

John Rennie

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 $b$.

yuvraj singh

Yes.

@JohnRennie

John Rennie

6:36 AM

@yuvrajsingh yes?

yuvraj singh

I am confuse why in the Vander wall equal we something in pressure, and subtract something from volume.

@JohnRennie

John Rennie

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. $V = V' - b$.

yuvraj singh

Yes

Is that the volume in which gas allow to move?

John Rennie

So our equation $PV = RT$ becomes $P(V' - b) = RT$ where $V'$ is the measured volume.

yuvraj singh

Due to attractive or repulsive force.

John Rennie

6:42 AM

@yuvrajsingh the volume reduction is due to repulsive forces.

yuvraj singh

OK sir.

John Rennie

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?

yuvraj singh

Yes.

John Rennie

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?

yuvraj singh

Yes

John Rennie

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.

M. Guru Vishnu

@JohnRennie, Good Morning Sir :-) Your new hat looks great!

John Rennie

@M.GuruVishnu morning :-)

Actually I need to change the hat. I normally change it every day.

M. Guru Vishnu

7:06 AM

@JohnRennie Is this known as a dynamic equilibrium? :-)

John Rennie

:-)

M. Guru Vishnu

@JohnRennie: Are you free now sir?

John Rennie

@M.GuruVishnu I need to work for about 15 minutes ...

M. Guru Vishnu

@JohnRennie Ok sir. No problem :-)

John Rennie

7:30 AM

@M.GuruVishnu OK, I'm free for about half an hour now.

M. Guru Vishnu

@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

In 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.

John Rennie

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.

M. Guru Vishnu

@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.

John Rennie

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.

M. Guru Vishnu

@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.

John Rennie

7:40 AM

It's easy to get mixed up with the signs. I recommend applying common sense! :-)

M. Guru Vishnu

@JohnRennie Yes sir. Thank you for the advise! :-)

@JohnRennie, I know paper is made of cellulose. Is cellulose electrically a dipole?

John Rennie

@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.

M. Guru Vishnu

@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?

John Rennie

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.

M. Guru Vishnu

@JohnRennie Ok sir. Now understood. Thank you :-)

John Rennie

8:00 AM

I need to work now for half an hour or so ...

user8718165

9:06 AM

Hello sir @JohnRennie

John Rennie

@user8718165 hi :-)

user8718165

@JohnRennie how are you sir?

I want to ask you a little qn.

John Rennie

@user8718165 yes ... ?

user8718165

@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?

John Rennie

Yes

user8718165

9:10 AM

@JohnRennie thank you very much sir...

@JohnRennie sir please come in general chat :)

yuvraj singh

9:36 AM

@user8718165 sorry brother my just hang.

It was by mistake.

Jasmine

10:32 AM

@JohnRennie Hello

:-)

@JohnRennie please help with amperian loop

John Rennie

@Jasmine hi

@Jasmine hello ??

Jasmine

10:48 AM

@JohnRennie hello

John Rennie

Hi

What's the question?

Jasmine

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

John Rennie

@Jasmine yes

Jasmine

@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

John Rennie

You can choose any loop you want ...

Jasmine

10:51 AM

But I just wanted to know if a square can do as well

John Rennie

A square loop would work but would be a nightmare to integrate along.

Jasmine

@JohnRennie yup but it should give value of B

@JohnRennie isnt B constant

Ohh

John Rennie

@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.

Jasmine

B is not constant just then angle between B and dl is constant and that is 0

@JohnRennie yup my bad

Silly doubt

John Rennie

No problem :-)

Jasmine

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

John Rennie

Yes. It's the integral of B round the path that is constant i.e. $2 \pi r B$ is a constant.

Jasmine

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

John Rennie

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.

Jasmine

@JohnRennie ok

John Rennie

Is this the correct layout?

Jasmine

@JohnRennie yup not sure what r,y is representing though

John Rennie

@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 $d\mathbf y \times \mathbf B$.

Jasmine

11:15 AM

@JohnRennie ok

John Rennie

And you're going to get the total force by integrating $d\mathbf F$.

Jasmine

@JohnRennie ok

John Rennie

Is that enough of a hint?

Jasmine

B(r) can be calculated , dy= dr

limit from -(x^2+(a/2)^2)^1/2 to +(x^2+(a/2)^2)^1/2

John Rennie

$d\mathbf y$ 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 $dr$.

Jasmine

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

John Rennie

@Jasmine I'm just answering a question in another room. I shouldn't be too long ...

Jasmine

@JohnRennie OK

And sorry it should be rdr=ydy

@JohnRennie I need to leave in 8 minutes