4:20 AM
good morning sir :-) @JohnRennie

@user8718165 morning :-)

@JohnRennie sir I want to ask something based on our yesterday's discussion :-)

@user8718165 yes ... ?

@JohnRennie sir suppose there is a piston with a gas and it expands from V1 to V2...the gas pressure on the piston will gradually decrease...is it ?
there is vacuum outside....

@user8718165 what process are you considering? For example are you applying a force to the piston to hold it in place, then gradually decreasing that force to allow the piston to slide out slowly?

4:28 AM
@JohnRennie yes sir...at V1 I'm stopping the piston from moving...then I remove hand

When you say you remove your hand, do you mean you just let go and allow the piston to move freely?

@JohnRennie yes sir (it isn't making sense...right)
@JohnRennie

@user8718165 suppose the piston area is $A$, and the pressure inside is $V$, then the force on the piston is just $F=PA$, and the acceleration of the piston is $a = PA/m$, where $m$ is the mass of the piston.

@JohnRennie okay
pressure inside is P....(that was a typo....) :-)

Like that.
@user8718165 oops, yes.

4:39 AM
@JohnRennie okay sir

Assuming this is an adiabatic expansion then the pressure and volume are related by $PV^\gamma = C$ for some constant $C$, so $P=C/V^\gamma$. Then our equation becomes:
$$\frac{d^2x}{dt^2} = \frac{C}{V^\gamma}\frac{A}{m}$$
And the volume is $V = Ax$ so we end up with:
$$\frac{d^2x}{dt^2} = \frac{C}{(Ax)^\gamma}\frac{A}{m}$$

@JohnRennie okay sir....got it
@JohnRennie It will be a function of position as everything else is constant

$$\frac{d^2x}{dt^2} = \frac{C}{A^{\gamma-1}m}\frac{1}{x^\gamma}$$
@user8718165 yes

@JohnRennie thank you sir...for this equation....:-)
Sir one last qn

You could solve that equation to get $x$ as a function of time, then use this to calculate the pressure as a function of time. But I doubt the equation has a simple analytic solution.

4:46 AM
@JohnRennie gamma will make things complicated...
cp/cv

Yes

@JohnRennie okay sir...got it
@JohnRennie another qn

@user8718165 yes ... ?

5:02 AM
@JohnRennie okay

@JohnRennie okay sir

@user8718165 I only drew that so I could include it in a comment to the question. See the comment I've just posted.
@user8718165 I think he means the diagram on the right, and it's a surprisingly complicated question.
I have to say I don't think your answer does it justice.
I might post an answer to the question, though that will have to be ina while as it will take a while to right.

@JohnRennie okay sir....Then I'm sure....yours will be accepted and have the highest votes XD
@JohnRennie okay sir...I want to ask just one last qn....

@user8718165 yes ... ?

5:18 AM
@JohnRennie sir in that piston...the gas will get colder...right

@user8718165 if it is an adiabatic expansion then yes the gas will get colder

@JohnRennie now suppose I try to lift the piston upwards so that at each moment accln of the piston will be higher than what it was in our case when only the gas was pushing on the piston...I won't pull it fast enough so as to create joule expansion...
@JohnRennie sir

I need a few moments. I'm dealing with a problem at work.

@JohnRennie absolutely sir....totally fine :-)

5:55 AM
@JohnRennie still working sir

@user8718165 I'm free now

@JohnRennie okay sir....

@user8718165 strictly speaking the pressure is only defined if the expansion is reversible, and an expansion is only reversible if it is infinitely slow.

@JohnRennie why is it so sir

Though in practice we just just need the speed the piston moves to be small compared to the speed of the gas molecules.
If we look closely at a gas then it's a collection of molecules all whizzing around at several hundred metres per second.

6:04 AM
yes....

The reason there is a pressure on the piston is that gas molecules hit the piston and bounce off it. That means the momentum of the gas molecule changes (on average) by $2mv_{rms}$ and that momentum is transferred to the piston.

@JohnRennie yes sir

If the piston is stationary then the collision is elastic i.e. the gas molecule bounces off the piston with the same speed it hit the piston.
But if the piston moves then some of the kinetic energy of the gas molecule is transferred to the piston and the gas molecule bounce off with a lower velocity than it hit. And if the gas molecule velocity falls that means the temperature falls.

@JohnRennie sir...momentum conserved...right

This is the basic reason the gas cools when it expands against an external force.

6:08 AM
@JohnRennie yes sir...got it

So in the instant after a collision the molecules that have just bounced off the piston have a lower velocity than the molecules far away from the piston i.e. our gas no longer has a uniform temperature.
And if the temperature isn't uniform the pressure isn't uniform either.
OK so far?

@JohnRennie okay sir...got it...that's why we say pressure isn't defined for fast piston movements

@user8718165 yes, because the pressure isn't uniform so there is no longer a single pressure for the whole gas.

@JohnRennie okay sir

In practice gas molecules move very fast, so the slower molecules near the piston quickly collide with the hotter molecules far from the piston and the velocities even out again.
So if the piston is moving slowly compared to the speed of the gas molecules then we can treat the pressure as uniform to a very good approximation.

6:14 AM
@JohnRennie sir does it mean that if a gas molecule bounces against a very fast moving piston...the molecule will have a very low KE in opposite direction...right
after the collision

@user8718165 yes

@JohnRennie that means I can change temperature of a volume of gas by moving the piston very fast...but that concept isn't much of a use in thermodynamics...right sir?

In real life gas molecules move so fast that the non-uniformity is too small to matter. However you are right that strictly speaking equilibrium thermodynamics is an idealised case that is only an approximation to the real world.
There is a discipline called non-equilibrium thermodynamics that deals with situations where the non-uniformity is big enough to matter.
Non-equilibrium thermodynamics is a branch of thermodynamics that deals with physical systems that are not in thermodynamic equilibrium but can be described in terms of variables (non-equilibrium state variables) that represent an extrapolation of the variables used to specify the system in thermodynamic equilibrium. Non-equilibrium thermodynamics is concerned with transport processes and with the rates of chemical reactions. It relies on what may be thought of as more or less nearness to thermodynamic equilibrium. Non-equilibrium thermodynamics is a work in progress, not an established edifice...

@JohnRennie In most of the cases I've encountered during expansion...only the gas applies the pressure to the piston and gets colder...no extra force is used to pull the piston....
I was just imagining varois cases....so I asked you
now its okay :-)

@user8718165 yes, JEE questions are almost exclusive concerned with equilibrium thermodynamics.

6:22 AM
@JohnRennie yes sir....got it...
@JohnRennie Thanks for telling :-)

@user8718165 :-)

6:37 AM
@JohnRennie hello

is it? XD

Aha. Why so

The electron is attracted to the positive charge and repelled from the negative charge.
So we need to do work to push it away from the positive charge and towards the negative charge. Yes?

6:47 AM
Yes

And when we do work on the electron we are increasing its PE. So the closer the electron is to the negative charge the higher its PE will be.

Okay. I can understand the logic behind this

You can write down the equation for the PE to show this mathematically ...

Let me try
Yeah it's true if we analyse by potential energy equation too
At B it is more positive than at C, D
At A, it is zero potential energy

@JohnRennie sir can you please tell me

6:59 AM
@JohnRennie I need some help clearing some concepts up about a lab I have on Monday. Whenever you've got time (and are done helping others).

7:12 AM
@user8718165 OK, you next then @Kyle
@kylecampbell do you want to ask now?

okay
one sec
If you go to the bottom of page 3, it gives a diagram of the circuit of interest. The input voltage is the function generator, but I'm a bit confused what the output voltage is. Is this just the voltage of the capacitor?

Yes, it's just the voltage across the capacitor
The aim is going to be to measure the amplitude and phase shift relative to the input voltage.

We'd expect it to be leading the input voltage by $\pi/2$ then, right?
or would it be lagging...

I always have difficulty remembering if the capacitor voltage leads or lags. Likewise with inductors, though I remember that inductors and capacitors are opposites i.e. one leads and the other lags.

I remember voltage leads current in an inductor. So the voltage must lag in a capacitor...

7:26 AM
Makes sense. The voltage on the capacitor builds up as it charges so you'd expect the voltage to lag.

OK, so in that case, one of the questions states: "Does the voltage across the resistor and capacitor add to the voltage from the function generator? Is this what your expected? If they do not add, what is the correct relationship between these terms?" - I'd expect not, because now we have them just adding as vectors.
The other question would be: how would I measure the phase difference from the oscilloscope?
I know there's no button that just lets you measure the phase difference, so I'd have to make some measurements of the time or something but I'm not sure exactly what of.
@JohnRennie the most important part I'm trying to figure out is Q11 - am trying to sketch a rough idea of what I'd expect the phasor diagram to look like

As you say the voltages will add as vectors, well phasors.
The instantaneous voltages do add, but the amplitudes $A$ in $A\sin(\omega t + \phi)$ don't simply add to give the total amplitude.
I guess the question is a bit unclear whether when they say voltage they mean the instaneous voltages or the amplitudes.

good point

To measure the phase lag you have to make measurements off the oscilloscope display. I would look at the points where the voltage is zero and measure the distance between those points to get the phase lag. Scopes normally have a grid on the display for you to make the measurement.

as in, you'd measure the distance (in units of time) between the two points and then relate that to the angle?

7:41 AM
Yes

something like the time (measured/full period) x (2pi) = phase difference?

yes

OK cool
the last thing I need help with is the phasor diagram
we already determined that the phasor for the output voltage is just the voltage across the capacitor, so that's done. is the phasor for the input voltage "in sync" with the phasor for the voltage across the resistor?
no it can't be
the current should be in sync with the voltage across the resistor though right?

@kylecampbell yes. For the resistor the voltage and current are in sync i.e. the phase lag is zero.
For the capacitor the voltage lags the current by 90°

7:52 AM
When you add the resistor and capacitor together using your phasor diagram you get the impedance for the pair of them, and the phase angle for the pair is the phase difference between the voltage and current flowing through the battery.

@JohnRennie does that look accurate?

@kylecampbell that doesn't look like the phasor diagrams I'd expect ...

how do I fix it?

This is what a phasor diagram normally looks like.
It's for calculating the total impedance Z.
This is a general diagram. In your case $L=0$ so there's no upper bit to the diagram.

Is $V_{in} = V_{resistor}$?

8:00 AM
It's not calculating voltages or currents, it's a way of calculating the total impedance of the components.
Do you know about complex impedance?

I don't
I've used Euler's relation in other contexts though

OK, look at that phasor diagram and imagine it was the complex plane i.e. the horizontal axis is the real axis and the vertical axis is the imaginary axis. Then the resistance would be a real number i.e. just $R$.
The inductance would be an imaginary number $iL$ and the capacitance would be an imaginary number $-iC$. OK so far?

So the resistance, inductance and capacitance are adding just like complex numbers. Obvious they aren't really complex numbers but they add as if they were. The complex number is just a way of encoding the resistance and the phase into a single quantity.
Anyhow we can write complex impedances as:
$Z_R = R$
$Z_L = i \omega L$
$Z_C = 1/(i \omega C)$
And these complex impedances add just like regular resistances do. e.g. for an inductor, capacitor and resistor in series you just have $Z = Z_R + Z_L + Z_C$

8:09 AM
That's exactly what the phasor diagram is doing. It's doing the addition graphically.

why are they asking us to draw the phasor diagram like that then?
I see what you're saying now

Hmm, I'm not sure what q11 is asking ...

okay, that's alright. I found it a bit strange too

Well if you draw the input voltage on the $x$ axis then the output voltage lags by 90° so it will be vertically downwards.

where is the voltage across the resistor with respect to the function generator (input) voltage?

8:16 AM
The current phasor will lag by some angle $0 > \phi > -90°$ so it would be a vector pointing to the lower right.
The voltage across the resistor is just $V_R = IR$ so it points in the same direction as the current phasor.

current leads voltage in a capacitor though
wouldn't it be in the 4th quadrant in that scenario?

We'll have to pick this up later. I need to work for about an hour now.

for sure
I'm gonna go to bed - can we finish it off tomorrow or the next day?

Tomorrow then

2 hours later…
10:15 AM
@JohnRennie sorry sir...I'd gone for classes :-(

@user8718165 hi

@JohnRennie hello
@JohnRennie Sir that potential energy thing... How to get the mathematical expression sir?

@user8718165 for a single charge $Q$ the potential is just $V(r) = kQ/r$. Yes?

@JohnRennie yes sir

If you have more than one charge the potentials for each charge just add, so we get $V = kQ_1/r_1 + kQ_2/r_2 + ... + kQ_n/r_n$

10:24 AM
@JohnRennie yes sir.....okay

If you have a continuous charge distribution then the sum just turns into an integral.

@JohnRennie $V=k\int\sigma(r)dr/r$
I will be wrong..I suppose

Suppose you have a charge density $\rho(\mathbf r)$. Note that we need a vector to specify the position because the charge may vary in a random way so it isn't enough to just make it a function of distance.

@JohnRennie yes sir

And consider a small volume element $dV$. We could use a cube of sides $dx$, $dy$ and $dx$ in which case $dV = dxdydz$, or some other shape of volume element. It doesn't matter.

10:33 AM
@JohnRennie okay...got it

The charge inside that volume element is $dQ = \rho(\mathbf r)dV$

yes.....

So the potential is $dU = k\rho(\mathbf r)dV/|r|$
And the total potential is then:

@JohnRennie aha....that's where I got it wrong......
I should've written $ds$ instead of $dr$

$$U = \int_V \frac{k \rho(\mathbf r)}{|r|} dV$$

10:35 AM
also that absolute symbol

where the integral is over the volume $V$ containing the charge.

@JohnRennie got it sir

Aladdin's question was about an electron, so the charge is $-e$. If we consider the point $A$, with distance to the charges $r_1$ and $r_2$, then we just add the potentials for the two charges. So we get:
$$V = \frac{k(-e)(-q)}{r_1} + \frac{k (-e) (+q)}{r_2}$$

yes sir....got it...thank you so much
@JohnRennie sir one question

@user8718165 yes ... ?

10:50 AM
$f(x)=\log(\sin(x)-2)$
@JohnRennie sir this function doesn't make sense...right?

Well the argument is always negative so there are no real values of the function. The function can be defined if you allow complex numbers, though taking logs of complex numbers gets a little complicated.

@JohnRennie okay sir...got it...I was blindly differentiating it in the hope of getting the correct answer...XD

11:07 AM
@JohnRennie Sir but after finding derivative...I got a function which makes sense...how is that possible sir?
does it make sense in complex plane as we discussed a while ago

I have no idea

@JohnRennie I'm sure...that's nonsense...I agree sir :-)

But consider the simple function $y(x) = ln(x)$. The derivative of this is $dy/dx = 1/x$

@JohnRennie okay...

So the derivative has a real value everywhere except $x=0$ while the integral has no real values for $x \le 0$.

11:10 AM
@JohnRennie yes.....got it