I agree with Päällikkö.
F=M(d^2R_cm/dt^2)=dP/dt=o
To have even a uniform motion of CM,what we need is a motion of CM from the rest.That must involve an acceleration.
Since acceleration of CM is zero,we must conclude that CM did not move at all.
OK,it is clear that the complete picture is not clear to me.Otherwise,I need not have asked the questions here.
Regarding the Maxwell's speed distribution.I am ASSUMING that just at the start of free expansion,(when the KE of the "dm" gas molecules starts to increase) all the gas molecules did...
denverdoc,let me finish the uncomfortable part very first.
Actually,I was in a tremendous bad mood while firing against AM.I was appearing an exam when AM's post,particularly some interesting parts I happened to like.I thought over it for considerable time but found (at that time) it to be...
You are saying self contradicting things.If temp is the same,From Kinetic Theory of Gases, the dm molecules must have the same KE as previous.Also you did not justified why dm molecules' KE will be gtreater than the rest.What you are saying is not enough to understand that this KE will "exceed"...
AM,I am believing that you have left the discussion.If you are less interested,
I have nothing to say.But,you should confirm whether you approved or were not satisfied by my logical development in my last thread.
Agree.But I believe the "dm" molecules have the same range of energy as their inmates behind.I will justify it later in this post.
OK.Lets proceed assuming the total system is isolated.
Do not agree.
Let us first consider the "chember" molecules.In equilbm,they were colliding with each...
AM,I may say sorry as I insulted you in my last thread.
However,I am NOT going to agree with what you have said.
What you are talking about the kinetic energy is not clear.Ofcourse,the gas molecules moves in a bigger space,but that does not help at all.Kinetic energy=(1/2)mv^2.How can you...
Maximum,I may thank you for the 1st part.
The 2nd part cannot be possible as as self work is always zero,and a gas cannot work on itself.The d(W_ad) is same for all paths.Be it reversible or be it irreversible.
The 3rd part was even miserable.You said that free expansion is not a random process.
1.As W_ad refers only to reversible (or, irreversible,for the sake of arguement,say) adiabatic process,dissipative configuration work will not be included in d'(W_ad).If included,d'(W_ad) will be non-unique by virtue of non-uniqueness of dissipation.So,it is not.Then,it is not also included in...
Let me approach systematically:
1.dissipative configuration work means a dissipative work that changes the confoguration,say,volume.Suppose,in the path 'a' to 'c' as referred to in my fourth thread we assumed free expansion without a dissipative work that changes configuration.However,if we...
I hope I got your point.Because when we apply more external pressure,workdone is greater,as mentioned by you,but the process is irreversible,irreversible W_ad> reversible W_ad.So,there may be arbitrary no. of ways of irreversible W_ad where irreversible work done will be different.
So,W_ad...
I am afraid we are gradually shifting out from my question.
Let me clarify it once more.Sears Salinger consider a system undergoing an adiabatic process between two eqlbm. points 'a' & 'b' having the same kinetic & potential energy.First a free expansion(configuration work zero) from 'a' to...
I agree.Yet, for the sake of understanding,I am insisting to clarify that the
'anything that is not reversible and changes the system' includes non-quasistatic irreversible configuretion work as well as dissipative configuration work.Is it?