Good morning sir :-) @JohnRennie

@user8718165 morning :-)

@JohnRennie Sir I want to ask some questions...are you busy now :-)

@user8718165 I'm working for about half an hour I'm afraid

@JohnRennie no problem sir..that's okay :-)

@user8718165 I'm free now if you're still around.

@JohnRennie hii

@user8718165 hi

@JohnRennie $w=\int Pdv$ .... That's the expression for work done by a gas....right

Yes, though it applies to anything that expands.

@JohnRennie But when the gas expands against constant external pressure...it's simply written $w=P\DeltaV$.....I don't understand this

Don't we have to take the net force on The piston (during expansion)$F=F_{gas}-F_{ext}$ @JohnRennie

Where f ext is a constant

No, it's not the net force that matters.

Look at is this way. Suppose the piston is expanding upwards, and the external force is due to a mass $m$ resting on top of the piston i.e. the external force is $mgh$. Is this OK or should I draw a diagram?

Yes sir...that's okay....I get it

Suppose the piston travels up a distance $x$ then the mass moves up a distance $x$ so the change in the PE of the mass is $mgh$.

It doesn't matter what the pressure inside the piston is, the work done on the mass to raise it a distance $x$ is just $mgh$.

So the work done by the expanding gas is $mgh$.

Okay Sir....got it....

Hii @JohnRennie

@user8718165 hi

@JohnRennie Hello sir :-) ....are you working Sir?

@user8718165 no, I've finished work now :-)

Sir 1 question.....

Yes ... ?

@JohnRennie A water dish is placed in an evacuated vessel and a manometer is connected the the vessel. Now the same experiment is repeated with a water filled test tube....at constant temp...

@user8718165 OK?

1.Setup a will reach equilibrium ahead of setup b 2.at equilibrium vapor pressure of both setups will be the same 3. Vapor pressure of a will be higher than b at equilibrium 4. The setups will reach equilibrium but the equilibrium vapor pressures may not be the same

@JohnRennie sir I think only 1 and 2....are correct..

If you put any water into an evacuated vessel the water will evaporate until the pressure of the water vapour equals the vapour pressure of water.

Yes sir....

So it doesn't matter whether there is a huge dish of water or a small test tube of water, as long as there is some liquid water left the pressure will be the same.

So, as you say (2) is correct and (3) and (4) are incorrect.

@JohnRennie Okay Sir what bout 1

As for (1), if you have a larger surface area of water the evaporation is likely to be faster, simply because the rate of evaporation per unit area will be the same. So the larger area will evaporate faster.

@JohnRennie okay Sir...got it... Though time reqd will be different, at equilibrium the properties of the setups will be identical....

@user8718165 yes

Do you know about Gibbs free energy in chemistry?

@JohnRennie I just learnt the formula last yr....but now I forgot it... I'll be comfortable with it in a few days :-)

Ok. I was going to explain using Gibbs free energy, but that needs to wait until after you've studied it.

As far as I remember it tells us bout the feasibility of a reaction

You can think of Gibbs free energy as the chemical equivalent of potential energy.

In physics spontaneous processes only happen if the potential energy decreases, and likewise in chemistry spontaneous processes only happen if the Gibbs free energy decreases.

@JohnRennie okay Sir.... Didn't know that.. Thanks:-)

@JohnRennie Yeah Sir....I remember having studied it in chemical kinetics a month or so ago:-)

@JohnRennie Sir... Will you be back in the evening?

@user8718165 possibly, but my enthusiasm for answering physics questions is at a low ebb in the evening.

@JohnRennie okay Sir.... :-) Thanks for answering now... :-)

@user8718165 :-)

@JohnRennie hello

@Aladdin hi :-)

I wanted to ask why 3°C rise in temperature is dangerous?

For the human body?

From what I gathered, it will make the natural disasters more dangerous... Is that what we are concerned

No the temperature of planet

Oh, you mean a 3°C rise in the average global temperature?

Yes

Averages hide a lot of detail. There is a degree to which trying to give an average temperature for the Earth is meaningless since it averages out the Sahara Desert and Antarctica.

3°C doesn't sound like much, but a small change in the global average is associated with much larger changes in the highest and lowest temperatures.

It could, or example, stop the monsoon. That would have some interesting effects on the Asian subcontinent.

Hmmm okay. So it can influence the climate in various regions

Yes. The climate change process can cause a great deal of variability in the weather.

Also, will the seas and others lifeforms affected

Stating an averaged out temperature rise is one way of tracking the changes, but it's a poor way as the changes will be far greater than the small average temperature change suggests.

So changes will be greater...... So It will be threat to farmers around the world

@Aladdin for example, calcium carbonate solubility in water is very temperature sensitive. It doesn't take a very large increase in temperature for shelled organisms to find their shells are dissolving.

Given that organisms generally grow shells to stop themselves from being eaten, that doesn't bode well for the long term survival of anything with a shell :-)

Aha yes

@Aladdin no-one knows in detail what would happen because the weather is a chaotic system and very hard to model. But any change is bad for farmers because farming is a very highly optimised process and any change is going to cause a lot of disruption.

The world as a whole doesn't have a huge surplus of food so if food production is affected even slightly there are going to be people starving.

True

Also bacterias will get harmed too... That will also be a problem I guess.

I don't know enough about the subject to say anything really authoritative, but basically any change is bad news.

Thanks for your time

I guess the effects are materializing now. Hopefully there are solutions...

Hello sir @JohnRennie

@user8718165 hi

@JohnRennie Sir....how can we define evaporation rate?

@user8718165 moles of water turning to vapour per second?

@JohnRennie okay Sir....

@JohnRennie imgur.com/3jbeaDV

sir are the plots correct

What are the plots intended to show?

Ah, I think I see. You're not graphing the net evaporation rate, you're graphing the transport from water to vapour and vapour back to water separately?

@JohnRennie rough sketch of the time dependence of rate of evaporation and condensation in a closed vessel and attainment of eqb. point..

So equilibrium is attained when the rates are the same?

and temperatures are different(t1>t2)

@JohnRennie yes sir

In that case the diagrams look reasonable.

You would expect the evaporation rate to be higher at higher temperature.

@JohnRennie okay sir...that means as you lower the rate of evaporation the eqb. point shifts to the right

I'm not sure it's as simple as that.

@JohnRennie why is it so?

As the temperature rises the evaporation rate increases, but at the same time the vapour pressure increases.

So you have to evaporate more water to reach equilibrium.

The rate of evaporation and the vapour pressure are not necesarily connected.

@JohnRennie aha...got it...

Suppose the rate increases slowly with temperature but the vapour pressure increases rapidly with temperature. The equilibrium would take longer to attain not shorter.

I don't think there's an easy way to know which effect dominates, the change in the evaporation rate or the change in the vapour pressure.

@JohnRennie yes sir...I see

@JohnRennie got it sir...thanks a lot :-)