Problem Solving Strategies

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sanya

06:16

Hi sir@JohnRennie

John Rennie

Hi :-)

sanya

For an isothermal reversible expansion of an ideal gas is change in enthalpy zero?

John Rennie

The definition of enthalpy is H = U + PV

sanya

Yes

John Rennie

And since PV = nRT, if T is constant that means PV is constant

And also internal energy is proportional to temperature, so if T is constant U is constant

sanya

06:19

Yes

John Rennie

So since both U and PV are constant that means H must be constant as well.

So yes, ΔH = 0 in an isothermal expansion.

sanya

So can we say PV work done is 0?

John Rennie

No. The product PV is constant, but work done is ∫PdV not just PV.

In an isothermal expansion the gas absorbs heat from the environment then uses that heat to do work.

The internal energy stays constant because the work done exactly equals the heat absorbed.

sanya

∆H =∆U + ∆(PV)

So work done is non zero then ∆PV = -P∆V?

John Rennie

Yes, but Δ(PV) is not equal to the work.

sanya

06:26

can we write ∆(PV) as ∆PV + P∆V?

John Rennie

You can write d(PV) = PdV + VdP

But to get Δ(PV) you need to integrate the right hand side.

Δ(PV) = ∫PdV + ∫VdP

And the first term gives the work.

That must mean ∫VdP gives the heat absorbed, though I hadn't thought about this before.

sanya

So in any case del H should remain 0. theres this question, 10g of argon is compressed isothermally and reversibly at a temperature of 27°C from 10l to 5l. Calculate q w ∆E ∆H

They have not mentioned if its an ideal gas or not but the solution booklet uses pv=nrt and also says that ∆E=0 . But they say that ∆ H = work done.

John Rennie

Argon is an ideal gas to a very good approximation.

Can you photograph and upload the solution?

sanya

Yes just a second

They have ignored ∫VdP

John Rennie

Yes, this is only true if the pressure is constant.

sanya

06:38

But its a reversible process...

John Rennie

Q: What would be the enthalpy change for a isothermal expansion?

I was doing a question on thermodynamics and came across this one stating During isothermal expansion of gas of an ideal gas : a) Internal energy decreases b) Enthalpy increases c) Enthalpy reduces to zero d) Enthalpy remains unchanged I cannot agree to any of these as $\Delta H = \Delta U + W$...

Can you post the question? Maybe I have misunderstood what the question is asking.

sanya

Okay

No.7

John Rennie

I think the answer is wrong.

sanya

Yes OK

If it were an irreversible process, it could have been correct ?

John Rennie

I'm always reluctant to assume the answer is wrong, because it usually isn't, but in this case I don't understand how ΔH can be non-zero.

sanya

06:50

Yes

John Rennie

@sanya I'm not sure to be honest. In any case it would depend on the exact details of the irreversible process.

sanya

Okay

Thankyou for the help:)

John Rennie

You're welcome :-)

2 hours later…

John Rennie

08:24

@KavinIshwaran Hi :-)

Kavin Ishwaran

@JohnRennie Hi !

Just one minute

The question is straight forward but.

My method does not give the correct answer

I just added the angular momentum of the disc in self rotation plus the angular momentum while it revolves

in the both case angular frequency is same since the radius are same.

John Rennie

Reading now ...

@KavinIshwaran I get (C)

Kavin Ishwaran

they have given 5

John Rennie

Ah, wait ...

The MOI of B as it rotates about A is ¹⁄₂MR² + M(2R)²

Kavin Ishwaran

Yes

John Rennie

08:39

So it's 4¹⁄₂MR² and therefore L = 4¹⁄₂MωR²

And the MOI of B rotating on its own axis is ¹⁄₂MR² so L = ¹⁄₂MωR²

Add them and you get 5MωR²

I missed a factor of 2 when calculating the MOI of B about the centre of A.

Does this make sense?

Kavin Ishwaran

Yes

John Rennie

So the answer is 5.

Kavin Ishwaran

@JohnRennie If we consider the axis through point of contact. Can we find the angular momentum ?

John Rennie

Possibly. Do you want to try it?

Kavin Ishwaran

Yes

Actually I did but I got 4MwR^2

John Rennie

08:45

I'm not sure it would work. It would mean we were no longer finding L about the COM of B, so then we can't use the COM of B to find L about the centre of A.

Kavin Ishwaran

Oh !

@JohnRennie Can you explain why it does not work a little in detail ?

John Rennie

I'm not sure I can explain ...

To be honest I have never been entirely comfortable about adding MOIs when we have multiple axes.

Kavin Ishwaran

Yeah..

John Rennie

But I would always use the COMs of the objects i.e. calculate L about the COM then use the COM to calculate L about the main axis.

Kavin Ishwaran

I was not sure if we could take COM as I thought we have to consider radius of gyration

2 hours later…

brainfreeze

11:14

@JohnRennie In this question, it is written that the COM of B is rotating about COM of A with ω. That should mean that the velocity of COM of B should be ω(2R). Since A isn't slipping on B, doesn't this mean that A is rotating about its own axis with 2ω and not ω? That would make L of B about its own axis as MωR².

John Rennie

11:29

@brainfreeze No, the angular velocity of B is just ω.

To see this sketch the diagram and put a spot or some other mark on B where it touches A, then imagine B rolling all the way round A so the spot rotates relative to the centre of B.

You'll find that when B returns to its original position the spot will have rotated round B just once.

Or try it in real life with two coins.

brainfreeze

12:06

https://youtu.be/FUHkTs-Ipfg?t=209 Any point on the boundary of the coin (or B) rotates through an angle of 4π by the time B reaches its original position. Velocity of COM of B is 2ωR, and its COM is rotating about COM of A in a circle with radius 2R. So it takes 2π/ω to complete one revolution. This implies that angular velocity of B about its axis should be 4π/(2π/ω) = 2ω.
Where am I going wrong here?

Bumblebee

12:53

@JohnRennie Hello:) Been a while.

I have a question regarding the derivation of the jeans mass

my notes set gives the Jeans mass as $$M_{J} = \frac{3 k_{B} T R}{2 G m}$$

However I do not know how this is derived

I worked from the equation $$\frac{G M_J m}{r} = \frac{3}{2} N k_{B} T$$

And I got $$M_J = \frac{3 N k_B T r}{2 G m}$$

Is it possible to transform this equation to the one given in my notes? If so, how?

4 hours later…

Kavin Ishwaran

16:52

@JohnRennie Hi !

John Rennie

Hi :-)

mechanist

@JohnRennie, hey! You got any further into that problem I shared?

John Rennie

The two cylinders problem?

mechanist

yes

John Rennie

I've given up on it. Sorry :-(

mechanist

17:00

No problem @JohnRennie

Kavin Ishwaran

@mechanist there is a solution online toppr.com/ask/question/…

@JohnRennie Why curl of a conservative force should be zero ?

John Rennie

Roughly, the curl of a vector field is kind of a rotation, and if there is any rotation that must mean points in the field move in circles.

So basically whenever the curl is non-zero it means you can find field lines that form loops.

Kavin Ishwaran

Oh !

John Rennie

And if any field line is a loop that means you can integrate the field round the loop back to your starting position and get a non-zero result.

And that means the field is non-conservative.

mechanist

@KavinIshwaran thanks for that, and you should read about this from feynman's book. His explanation is nice, especially since we aren't taught vector calculus yet.

if you have the xtra time ofc

Kavin Ishwaran

17:08

@JohnRennie electric field is conservative only in static condition or always ?

@mechanist Ok :-)

John Rennie

@KavinIshwaran The electric field is conservative if and only if dB/dt = 0

Any magnetic field that changes with time will induce a non-conservative electric field. In fact this is how electricity generators work.

Kavin Ishwaran

Ah Yes.

Suppose if we have a electric field changing with time, it wont be conservative ?

John Rennie

When we say a field is conservative we normally mean if we integrate round a loop in space at constant time we always get zero.

The at constant time bit mean we are effectively taking a snapshot of the field to do the calculation.

Kavin Ishwaran

Ok so we are taking a static condition.

John Rennie

So I don't think the field can change during the integration by definition.

Kavin Ishwaran

17:15

I see

@JohnRennie Thank you for the clarifications :-)

John Rennie

You're welcome. Bye :-)

Kavin Ishwaran

Bye :-)

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Problem Solving Strategies

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