Heat capacity under constant pressure or volume question

[Chestermiller]

From your last help, I tried something more familiar to me. I again wrote dE=dQ+dW but this time I wrote dW=μ*dN - W where W is the work done to push the gas out but in a form that implies a change in volume. To clarify, I wrote W=Pv*dN(as you did) which is like pushing small volumes v out of the fixed control volume. Also, μ is the chemical potential(like you said, the energy that gets transported outside the control volume) and it is equal to μ=dE/dN which gives μ=f/2*kT. So, dW=f/2*kT*dN + Pv*dN with v=V/N (note that I wrote +PvdN rather than -PvdN because when dN>0 then work is being done ON the system so dW>0 as opposed to the typical dW=-PdV).
These, along with dE=dQ+dW and dE=0(due to dV=0) and thus dE=f/2*k*(NdT+TdN) gives:
dQ+f/2*kT*dN +Pv*dN=f/2*k*(NdT+TdN) and so
dQ=-PvdN+f/2*k*NdT and because TdN=-NdT and PV=kNT
dQ=(f/2+1)*k*PV*(dT/T)
which is the correct answer.

I think this is correct. What really annoys me is that our professor thought that it was an easy task of writing the extra term of PvdN. This is conceptually very tricky and I think I would have never think of this if it wasn't for you..

Nice job.

As I said in a previous post, I think that your professor did you a disservice by giving you this problem, which is waaayyyy too advanced for you right now. The open system (fixed control volume) version of the first law is a little difficult to relate to at first, because of the very term you identified. Of course, once you have learned about it and got some experience applying it, it seems much easier.

I was very impressed with how you attacked this problem and with the questions you asked. I think you have a bright future in science.

Chet

 

[Joker93]

Nice job.

As I said in a previous post, I think that your professor did you a disservice by giving you this problem, which is waaayyyy too advanced for you right now. The open system (fixed control volume) version of the first law is a little difficult to relate to at first, because of the very term you identified. Of course, once you have learned about it and got some experience applying it, it seems much easier.

I was very impressed with how you attacked this problem and with the questions you asked. I think you have a bright future in science.

Chet

Wow, thanks! And I thought that you would think "ah, this kid did not understand the basics that I was saying to him"!
Thanks for your patience(I know I have taken a lot of your time; it's just that if I don't understand something exactly, then I can't connect the dots and find a solution that is also conceptually valid in my head).
Thanks again!