Problem Solving Strategies

Ajay

00:16

@JohnRennie When you're back on, shall we go through the theory?

John Rennie

03:58

@Ajay Hi :-)

I'm around now for a while so ping me when you want to discuss this.

John Rennie

04:33

@cOnnectOrTR12 Hi :-)

cOnnectOrTR 12

05:40

Hi John ! How’s the weather?

John Rennie

It's really nice in the UK at the moment. Bright and sunny and about 20°C.

Warm (by UK standards) but not too hot.

cOnnectOrTR 12

@JohnRennie wow! That’s a very good balanced temperature.

John Rennie

Chester is a really nice place to live. It never gets too cold or too hot, and it doesn't rain that much (for the UK).

cOnnectOrTR 12

@JohnRennie so should I pack my bags?

John Rennie

:-)

cOnnectOrTR 12

05:49

Where will I live? Does your place have some space?

John Rennie

I suspect most people from India would find it too cold in winter. For about three months the temperature stays below 10°C.

Although it doesn't go below 0°C that often. Or at least not much below 0°C.

cOnnectOrTR 12

No problem 😉

naturallyInconsistent

@JohnRennie weather is that nice that far up north? london mind blown

John Rennie

@naturallyInconsistent Chester is in the North West of England and that part of England is warmed by the Gulf Stream.

naturallyInconsistent

@JohnRennie Yes, I was looking on the map. Very near Liverpool. I didn't know about Gulf Stream being warm. That's nice.

John Rennie

05:56

Also Chester is in between two sets of mountains, the North Wales mountains and the Peak district. That means it's in a rain shadow whether the wind is from the west or east so it doesn't get much rain.

@naturallyInconsistent One of the concerns about global warming is that melting of Arctic ice could stop the Gulf Stream, and that would catastrophically lower temperatures in the UK.

So global warming would actually make Chester a lot colder!

naturallyInconsistent

nuts!

John Rennie

@RR. Hi :-)

The net linear force is zero but the two forces exert a torque on the sphere and this torque would not be the same if the forces acted at the centre of the sphere.

I would take moments about the point of contact with the ground. If you calculate the torque about this point then divide by the MOI that will give you the angular acceleration. Then multiply by the radius of the sphere to get the linear acceleration.

Ajay

06:11

@JohnRennie Hi :)

John Rennie

@Ajay Hi :-)

How do you want to do this? Shall I start talking and you interrupt me if you have questions?

Ajay

Sure.

John Rennie

OK, we'll discuss this circuit as it's the one your teacher suggested.

We have a signal generator that generates an AC voltage:
V(t) = V₀ sin(ωt)
where ω = 2𝜋f.

And this voltage causes a current:
I(t) = V(t)/Xₜ
where Xₜ is the total reactance of the circuit

Does this make sense so far?

Ajay

Yes

John Rennie

The problem is that with capacitors and inductors the voltage and current are out of phase by 90°. If we measured the voltages Vr, Vl and Vc across the resistor, inductor and capacitor we'd get something like:

Vr(t) = Vr₀ sin(ωt)
Vl(t) = Vl₀ sin(ωt + 90°)
Vr(t) = Vc₀ sin(ωt - 90°)

Then the total voltage is a really complicated expression:
V(t) = Vr₀ sin(ωt) + Vl₀ sin(ωt + 90°) + Vc₀ sin(ωt - 90°)

But luckily there is a really easy way to add up the voltages to get the total, and that's what I need to explain.

@Ajay Does this make sense so far? Am I OK to start explaining the simple way to add the voltages?

Ajay

06:24

Yes

John Rennie

OK, I have to start by explaining that there is a relationship between circular motion and AC. This may seem a bit odd if you haven't heard this before but it's really important in understanding circuits. Here's a diagram I found through Google that shows how it works:

On the right we have a vector rotating about one end at an angular frequency ω

If you take the position of the red dot to represent the voltage then we get that position as:
y = y₀ sin(ωt)
Yeas?

Ajay

Yes.

John Rennie

Which is the same equation as the voltage from our signal generator. So you can represent the AC voltage as a rotating vector.

And this is the key part, if you have several voltages, like in our case Vr, Vl and Vc, you can write all three as rotating vectors. Like this:

This is at time t = 0 so ωt = 0. The red arrow is the voltage across the resistance

We know the voltage across the capacitor lags by 90° so we draw it rotated by -90° relative to the resistor.

And we know the voltage across the inductor leads by 90° so we draw it rotated by +90° relative to the resistor.

Does this make sense so far?

Ajay

Yes

John Rennie

This is the hardest part of the explanation so if anything isn't clear please feel free to ask.

Ajay

06:32

Ok

John Rennie

OK, so we have the three vectors, all at different angles, and to add up all the voltages we just have to vector add the three vectors.

It's just straightforward vector addition.

i.e. calculating the resultant of the three vectors.

OK so far?

@Ajay You've gone quiet. Have I gone a bit too fast?

Wolgwang

@JohnRennie Hi!

John Rennie

@Wolgwang Hi :-)

Ajay

@JohnRennie No, I understand.

You can continue.

John Rennie

What I've actually drawn on the diagram are the reactances. That's because we have the same current through all three components, and for all three components the voltage is given by V = IX. So we get:
Vᵢ = IXₜ = IR + IXl + IXc
and if we divide through by I we get:
Xₜ = R + Xl + Xc

So we can vector add R, l and Xc to get the total reactance Xₜ

Now, Xl and Xc are along the same axis but in opposite directions, so when we vector add them we get:
Xlc = ωL - 1/(ωC)

OK so far?

Ajay

06:46

Yes

John Rennie

So we have ωL - 1/ωC in the y direction and R in the x direction, and the magnitude of the vector sum is just given by Pythagoras' theorem:
Xₜ² = (ωL - 1/ωC)² + R²
Yes?

Ajay

Yes

John Rennie

Then the current is given by:
I = Vin/Xₜ
and the voltage through any of the three components is given by:
V = IX
so the voltage across the resistor is:
Vr = Vin R/Xₜ = Vin R/√((ωL - 1/ωC)² + R²)

And that is what I am plotting in the spreadsheet.

Does this all make sense so far?
If so I'll explain why we get a peak in the voltage at the resonant frequency.

naturallyInconsistent

detects smoke emission from ears

Ajay

Yes, this makes sense.

John Rennie

06:55

OK. Look at the denominator in the fraction.
√((ωL - 1/ωC)² + R²)

If we choose the frequency ω such that ωL = 1/ωC then we get:
ωL - 1/ωC = 0
and the denominator simplifies to just R. Yes?

Ajay

Yes

John Rennie

And that gives Vr = Vin
i.e. *all* the voltage is across the resistor and the voltage drop across the inductor and capacitor is zero.

This is the peak in the plot of Vr against frequency.

At any other frequency (ωL - 1/ωC)² is greater than zero

That means the denominator increases so Vr decreases.

@Ajay Is this all OK so far?

Wolgwang

@JohnRennie In the meantime, can I ask a question?

John Rennie

Yes, go ahead

Wolgwang

$$\theta_A=\theta\sin(\sqrt{\frac gl}t -\theta)\\\theta_B=\theta\sin(\sqrt{-2\frac gl}t +\theta)\\ \theta_A-\theta_B=\pi$$

Ajay

07:09

@JohnRennie yes.

Wolgwang

I am getting some complicated answer in terms of theta but the given answer is independent of it.

John Rennie

@Ajay And that's it! If you look at my spreadsheet that's how I calculate Vr, and if we plot Vr against ω we get the graph:

And that nicely shows the peak.

We get the peak when:
ωL - 1/ωC = 0
or:
ω = 1/√(LC)

Ajay

So I’m investing his frequency affects the voltage of a resistor?

John Rennie

Yes. In the experiment you measure Vr as a function of frequency and you should get the peak at the frequency ω = 2𝜋f = 1/√(LC)

Ajay

Ok 👍🏽

John Rennie

07:18

The diagram with the three reactances on it is called a phasor diagram.

If you go on to learn electronics at college you'll use phasor diagrams a lot!

Wolgwang

@JohnRennie Free?

John Rennie

@Wolgwang I have a long running question that I'm answering on WhatsApp but I think it's nearly finished.

naturallyInconsistent

@Wolgwang You should be getting $$\phi_A=-\theta\cos t\sqrt{\frac g\ell}$$ $$\phi_B=+\theta\cos t\sqrt{2\frac g\ell}$$ $$\phi_A=\phi_B$$

and then you can see that the answer must be independent of $\theta$

The 2 is with t and outside of the sqrt

Wolgwang

07:38

@naturallyInconsistent Is phase difference 0 in the starting?

naturallyInconsistent

@Wolgwang With choosing the minus sign in front of A and not in front of B, it is clear that there is a phase difference of $\pi$ at the start

John Rennie

to see the phase difference write it as:
$$\phi_A = \theta \sin(\omega t - \pi/2)$$
$$\phi_B = \theta \sin(\omega t + \pi/2)$$

Wolgwang

@naturallyInconsistent Thanks :) got the answer. I was mixing angle between string and phase difference.

@JohnRennie Yeah. Got it :)

John Rennie

Sorry I was tied up but as exams get near things are getting very busy. @naturallyInconsistent thanks for jumping in :-)

ronak jain

08:38

@JohnRennie Hi sir.. Good Afternoon

John Rennie

@ronakjain Hi :-)

ronak jain

Sir.. I was solving a problem on interview bit. But since I can only write the function of class.. I'm not getting how to proceed

interviewbit.com/problems/unique-paths-in-a-grid

int Solution:: result(vector<vector<int>>&matrix, vector<vector<int>>dp, int row=0, int col = 0){
    if(dp[row][col] != -1){
        return dp[row][col];
    }
    if(row >= matrix.size()){
        return 0;
    }
    if(col >= matrix[0].size()){
        return 0;
    }
    if(matrix[row][col] == 1){
        return 0;
    }
    if(row == matrix.size()-1 && col == matrix[0].size()-1){
        return 1;
    }
    int a = result(matrix, dp, row+1, col);
    int b = result(matrix, dp, row, col+1);

@JohnRennie I'm not getting what should be the syntax for declaring another function here

John Rennie

That looks like it ought to work.

I'll need a few moments to create an account on the site ...

ronak jain

Should the static function not declared inside the class first ?

@JohnRennie ok sir :)

John Rennie

Hmm, they want my mobile number and I'm not happy with that.

ronak jain

08:51

I hope email ID will work

Google sign will work

John Rennie

@ronakjain Ah, I see, yes the result() function would need to be in the class definition and they don't give you access to the that.

Can you just declare it as a non class function?

i.e. just:

int result(vector<vector<int>>&matrix, vector<vector<int>>dp, int row=0, int col = 0)

ronak jain

But then how will I call this function inside the uniquePath.....

John Rennie

Just call it like any function.

ronak jain

Can the class access the function outside of the class ? I was not aware of it..

John Rennie

Yes it can.

ronak jain

08:54

It worked sir... Thanks :)

John Rennie

:-)

RR.

@JohnRennie Hi sir!

Are you free now?

@JohnRennie Since the linear force becomes zero why can't we directly find the acceleration to be 5 m/s^2?

John Rennie

09:13

@RR. Hi :-)

The forces produce a couple on the sphere. Yes?

RR.

Yes sir

John Rennie

Since we are told the sphere does not slip it can only accelerate downhill by increasing its angular velocity. But the way the forces are drawn they produce an anticlockwise couple i.e. they would tend to make the sphere rotate anticlockwise and roll uphill not downhill. Yes?

RR.

Yes

John Rennie

If the forces are zero the sphere will just roll downhill n the usual way.

As we increase the forces they will slow the acceleration downhill until we reach a point where the torque from the forces balances out the torque due to gravity and the sphere will remain stationary.

Then if we keep increasing the forces the sphere will start to accelerate uphill instead of downhill.

Does this make sense?

RR.

Yes...

John Rennie

09:18

Do you want to go through the calculation?

RR.

Yes...

John Rennie

Suppose we take moments about the point of contact with the slope. i.e. we take this as the instantaneous centre of revolution.

The torque due to the weight is mg times the normal distance r sinθ so it's:
τ = mgr sinθ
Yes?

RR.

Yes

John Rennie

The MOI of a sphere about its centre is I = ²⁄₅mR²

And to get the MOI about the point of contact with the ground we use the parallel axis theorem i.e. we get:
I = ²⁄₅mR² + mR² = ⁷⁄₅mR²
Yes?

RR.

Yes

John Rennie

09:25

Then α = τ/I = mgR sinθ/⁷⁄₅mR²

And the linear acceleration of the centre of the sphere is a = Rα = mgR² sinθ/⁷⁄₅mR²

So we end up with:
a = ⁵⁄₇ g sinθ

This is all without the two extra forces. Does this look OK so far?

RR.

Yes sir...

John Rennie

The bottom force produces an anticlockwise torque -FR/2 and the top force produces an anticlockwise torque -F3R/2 so the total torque from the two forces is -2FR.

Then the total torque on the sphere is:
τ = mgR sinθ - 2FR
Yes?

RR.

Yes

John Rennie

As before α = τ/I and a = Rα so we end up with:
a = (mgR² sinθ - 2FR²)/(⁷⁄₅mR²)

a = ⁵⁄₇ (g sinθ - 2F/m)

Does this all make sense?

RR.

Yes sir

John Rennie

09:33

sinθ = ¹⁄₂, F = 1N, m = 1kg, g = 10 m/s² ...

a = ¹⁵⁄₇

Does that match the answer?

RR.

No sir...

Answer given is 20/7

John Rennie

:-(
What's the answer?

RR.

20/7

I think I spotted why we went wrong...

John Rennie

Hmm, it's close because it has the 7 in the denominator that comes from the MOI of the sphere.

RR.

The acceleration is asked about the COM

We've calculated at the point where it touches the surface... Would that cause a different answer?

@JohnRennie Yes sir

John Rennie

09:37

No, we calculated the torque about the point of contact, then we calculated the angular acceleration about this point from the torque and the MOI.

Then the linear acceleration of the centre of the sphere is the angular acceleration α times the distance of the centre from the point of contact R.

i.e. a = αR

Hmm, I cannot see the error ...

RR.

@JohnRennie Ahhh... I understood where I had understood it wrong

@JohnRennie This is the solution that they have given

John Rennie

Aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaarghhhhhhhhhhhhh :-(

I wrote the torques as -FR/2 and -3/2FR/2

But they are +FR/2 and -3/2FR/2 i.e. the bottom torque is clockwise.

So when you add them you get FR not 2FR as I wrote.

Yes, that gives a = ²0⁄₇

RR.

@JohnRennie How come? Isn't the bottom torque producing an anti-clockwise turn?

John Rennie

@RR. That's the mistake I made. Remember that the pivot is the point of contact with the ground, not the centre of the sphere.

So this is the torque from the bottom force.

naturallyInconsistent

Actually, by considering the couple, you could see that it has to be FR regardless of choice of pivot.

John Rennie

09:48

Yes, I should have spotted that as well :-(

Oh well ...

But the important thing is you see how the method works.

The reason we use the point of contact as the ICOR is that there is an unknown frictional force between the bottom of the sphere and the slope.

But if we use the point of contact as the ICOR then this frictional force passes through the ICOR so the torque it produces is zero.

(ICOR = Instantaneous Centre Of Rotation)

Does this make sense?

RR.

I kinda understood the whole question... but the torque part is confusing me a bit... Rotation ain't my strength😅

naturallyInconsistent

When the universe rotates, nobody has any strength

RR.

@JohnRennie Ohhh...

John Rennie

Do you know what we mean by the Instantaneous Centre Of Rotation?

RR.

@JohnRennie Not too sure... I'm guessing it's the point we temporarily assume as the COM aka the point of contact?

John Rennie

09:53

Hmm, this obviously got missed during your JEE preparation.

@RR. No, it has nothing to do with the centre of mass.

RR.

@JohnRennie Yes sir... I'd skipped rotation for its complexities for mains... haunts me now =(

John Rennie

Do you want to discuss it now?

RR.

@JohnRennie Ohhh...

Yes sir... Please... If u don't mind

John Rennie

Give me a moment to draw a diagram ...

RR.

Sure sir

John Rennie

09:58

Imagine we have a sphere rotating about a point like this:

I've drawn an arrow to show the direction the centre of mass of the sphere is moving.

Does the diagram make sense so far.

RR.

Yes sir

John Rennie

Note that in the first diagram the COM is moving horizontally. Yes?

RR.

Yes

John Rennie

Now compare this with a sphere rolling on a flat surface:

The sphere rolling on the flat surface also has its COM moving horizontally. Yes?

RR.

Yes

John Rennie

10:03

So if you look at the two red circles, the velocities of the spheres are exactly the same.

That is, at the instant shown by the red circle on the left the velocity of the sphere is exactly the same as the velocity of the sphere rolling on the flat surface.

Is this OK?

RR.

Yes

John Rennie

That means we can treat the sphere rolling on the flat ground as if it was rotating about the point of contact with the ground i.e. about the blue dot.

RR.

Yes

John Rennie

Are you still happy with this?

So if there are any forces acting on the sphere we can calculate their torque about the point of contact i.e. about the blue dot.

RR.

Yes... Now I'm able to connect it with the previous question

@JohnRennie Ohhh... Makes sense...

So, the blue dot becomes the ICOR?

John Rennie

10:09

Yes.

Now imagne the sphere on a slope:

The weight mg acts downwards from the centre of the sphere, so we can see that it will produce a clockwise torque about the blue dot. Yes?

RR.

Yes

John Rennie

That torque will be τ = mgR sinθ
Yes?

RR.

Yes

John Rennie

And if there is a clockwise torque about the blue dot then there must be an angular acceleration about the blue dot.

That is, the sphere is instantaneously accelerating about the blue dot with an angular acceleration α given by: α = τ/I

Does this make sense so far?

RR.

Yes

John Rennie

10:13

OK :-)

And you know that linear and angular acceleration are related by a = rα
Yes?

RR.

Yes sir

John Rennie

So that tells us that the instantaneous linear acceleration of the centre of the sphere has to be:
a = Rα = Rτ/I

So we can use the idea of an instantaneous centre of rotation to calculate the linear acceleration of the sphere down the slope.

RR.

Yes

John Rennie

Does this all sound reasonable?
This is what I (and the solution) did in the answer to the question.

We use the idea of an ICOR where the sphere touches the ground, then find the torque about the ICOR.

In the question we have the weight, as above, but we also have the two extra forces producing an additional torque.

RR.

Yes sir... I just realised how a small and kind of simple concept played such an important role in that question

John Rennie

10:18

Yes. The ICOR is very useful when you have objects rolling on planes, whether they are flat planes or inclined planes.

RR.

@JohnRennie So, weight produced clockwise torque, the bottom force also clockwise and the top force anticlockwise...

John Rennie

Yes! :-)

RR.

@JohnRennie This is the first time I've heard of ICOR and I'm glad I learnt it today =)

@JohnRennie =)

John Rennie

OK :-)

RR.

Thank you sir =)

John Rennie

10:20

You're welcome :-)

Yuv Singh

13:22

@JohnRennie hey sir! Could you please help me with a chemistry topic?

John Rennie

14:26

@YuvSingh I'll help if I can, but I don't know a lot about chemistry.

Ajay

@JohnRennie I have a question.

Two, actually(probably more)

What frequency range should I use?

How do I measure the voltage using the oscilloscope?

How do I linearise the curve?

Currently it's like a triangle, but i'm not sure how to resolve this.

John Rennie

14:58

@Ajay Hi :-)

The peak frequency is at f = √(LC)/(2𝜋)

You want to choose your values of L and C to make f some convenient frequency e.g. about 1 kHz. I've chosen the values in the spreadsheet to do this. You use the spreadsheet to see what the graph would look like for any values of R, L and C.

The curve cannot be linearised, and indeed you aren't trying to linearise it. You're trying to show the curve peaks at the frequency f = √(LC)/(2𝜋)

@Ajay This is the picture you posted of the oscilloscope screen:

2 days ago, by Ajay

The display is basically a graph with the y axis being voltage. At the bottom left of the screen it tells you what the scale is:

So or example for the yellow curve (channel 1) each major division is 5.00 volts.

So the yellow curve goes from about -6V to +6V i.e. it is V(t) = 6 sin(ωt)

Yuv Singh

15:18

@JohnRennie why does the pressure remain constant during liquification once some liquid is formed, till the time all the vapour is coverted into liquid?

John Rennie

@YuvSingh Didn't we talk about this before? I remember explaining it so someone using the concept of chemical potential.

Yuv Singh

chemistry.stackexchange.com/q/173808/134170?sem=2

I am new here

Just discovered this room today

John Rennie

OK :-) Welcome to the Stack Exchange. Your question didn't look like homework to me so I'm not sure why they closed it. Oh well, I'm happy to explain it here.

Do you know what Gibbs Free energy is?

Yuv Singh

Barely any idea haven't studied thermodynamics yet

John Rennie

OK, you don't need to know the details. You just need to know that it's the chemical equivalent of potential energy.

And just like any system will try to find the lowest potential energy, in chemistry your system will always try to find the lowest Gibbs free energy.

Yuv Singh

15:25

I think I can't get it now

Will have to study thermo first

I thought it could be explained using simple equilibrium concept

John Rennie

@YuvSingh Well it can!

That equilibrium concept is that the gas and the liquid have the same GFE when they are in equilibrium.

Yuv Singh

Thermodynamics again lol

John Rennie

Just like in physics things are in equilibrium when they have the same potential energy.

Yuv Singh

Wait

I'll write what I understand

John Rennie

OK ... ?

Yuv Singh

15:29

Btw when we say we increase the pressure what do we mean? Do we mean we compress the gas, hence molecules get closer resulting in greater number of collisions and thus greater pressure exerted by the vapour on the walls of container or we mean we apply pressure by compressing it?

Btw what is the difference between the pressure that we apply on the gas by using piston like stuff and the pressure exerted by the vapour on the walls of container?

John Rennie

The PV diagram in your question shows that the pressure remains constant while the volume is being reduced, so we are reducing the volume of the gas e.g. by pushing in a piston into the container where the gas is.

Yuv Singh

Yeah

John Rennie

Yuv Singh

Right but once a liquid drop is formed the pressure remains constant

Here exactly my confusion lies

John Rennie

e.g. where the blue arrow is the volume is being reduced and the pressure remains constant.

@YuvSingh That's exactly what the Gibbs Free energy explains.

And you don't need to know the details about how the GFE works. You just need a few key facts that I can tell you.

Yuv Singh

15:35

I have been kinda bothered about this for a few days

John Rennie

Do you want me to try and explain?

Yuv Singh

I am scared your explanation might have a use of thermodynamics

Which I haven't studied yet

John Rennie

Shall I try and explain and you can stop me if it gets too complicated?

Yuv Singh

Ok sir

Sure

John Rennie

As a rough approximation the GFE is a measure of the energy of the system, and you can increase the energy by doing one of two things:
1. add heat by raising the temperature
2. do work by compressing the system

Is this OK so far?

Yuv Singh

15:38

Yup

John Rennie

Now our curves are isotherms i.e. constant temperature, so option (1) doesn't apply because the temp doesn't change. So we can only change the GFE by doing work i.e. by compressing the system. Yes?

Yuv Singh

Yup

John Rennie

But liquids are (approximately) incompressible so we can't change the GFE of the liquid even by compressing it. Basically the GFE of the liquid is fixed and we can't change it all.

OK so far?

Yuv Singh

Wait we are compressing gas ryt?

John Rennie

The liquid and gas have the same pressure because they are in equilibrium. So if you apply a pressure to the gas then the liquid has the same pressure.

So we are applying the same pressure to both, but the liquid is incompressible so the pressure doesn't affect it.

Yuv Singh

15:42

Yup

John Rennie

But the gas is compressible so as we reduce the volume we do work on the gas and we increase its GFE. Yes?

Yuv Singh

Yup

John Rennie

Liquefaction starts when we compress the gas so much that its GFE becomes higher than the GFE of the liquid.

Since the system always adopts the lowest energy configuration, if the GFE of the gas is higher than the GFE of the liquid the gas turns into liquid.

i.e. we get condensation of the gas into liquid.

Does this make sense so far?

Yuv Singh

Yup

Understood

John Rennie

But when gas turns into liquid we have less gas in the same volume, so in effect we are decompressing the gas and the GFE reduces again.

So we end up with a mixture of liquid and gas, and both have the same GFE.

Yes?

Yuv Singh

15:47

Wait

John Rennie

Yes ... ?

Yuv Singh

Lemme understand this

How are we effectively decompressing the gas?

John Rennie

Suppose we have 1 mole of gas in our container. We compress it and the GFE increases, so 0.1 moles of gas condense to liquid. So now we only have 0.9 moles of the gas. Yes?

Yuv Singh

Yup

John Rennie

So by condensing some of the gas we have less gas in the same volume so the pressure decreases. And the GFE is related to the pressure. That means condensing some of the gas has lowered the GFE of the gas that remains.

Yuv Singh

15:53

Less gas in the same volume?

Then how does pressure decrease?

Ok

Got it

Go ahead

John Rennie

OK :-)

So the point is that if you reduce the volume again even more gas condenses and reduces the number of moles of the gas again.

Yuv Singh

Yup

John Rennie

The GFE of the gas cannot get any higher than the GFE of the liquid because if it des the gas condenses to liquid.

And the GFE of the liquid is constant because it is unaffected by pressure.

Yuv Singh

Ok

John Rennie

That means the GFE of the gas stays constant as well, and the GFE of the gas is related to the pressure, so the pressure stays constant.

All that happens as you decrease the volume is that gas condenses to liquid.

Yuv Singh

16:01

Sir you said GFE of gas can't get higher than that of liquid. Because if it does, the gas condensed to liquid. But here gas is condensing to liquid. Isn't it?

Did I miss something?

John Rennie

We are reducing the volume of the gas and normally that would increase the GFE, but the gas condenses to liquid and that reduces the GFE of the gas to back down to equal the GFE of the liquid.

Yuv Singh

Yup right. But I think I should study these things first

I can't feel these things

John Rennie

OK

Yuv Singh

Sir could I ask you questions from now onwards ?

I am just a beginner jee aspirant

So, please don't mind if I end up asking silly things lol

John Rennie

Yes, you're welcome to ask, though I only know much about physics so I can't help with chemistry

Yuv Singh

16:08

I expect help in physics only

John Rennie

OK :-)

Yuv Singh

I am preparing online. So, I don't have anyone to clear my doubts

I end up wasting a lot of time due to this

Btw where can I send you pictures?

Of questions?

John Rennie

You can upload pictures here if you're using a laptop, though I'm not sure if that works from a phone.

Alternatively upload them to Imgur and post the link.

But note that I am mainly here from about 10 a.m. to 5 p.m. Indian time and I don't normally answer questions this late.

Yuv Singh

I don't have my own laptop. Will be purchasing one in 2-3 months

John Rennie

Use Imgur then.

You can upload the picture there and post the link here.

Yuv Singh

16:14

They'll take down each question of mine lol

John Rennie

Yes, JEE questions aren't allowed on the main site but they are fine here.

Yuv Singh

Ok sir

John Rennie

I need to go now but I will be back tomorrow.

Bye :-)

Yuv Singh

Bye

See ya

Transcript for

Problem Solving Strategies