Recent content by oxman

phi= dN/dAdt ***

Molecular flux--thermodynamic kinetics Homework Statement A volume is divided into two equal parts by a partition. The left volume has N atoms. The atoms have a mass m. A hole of area A is made in the partition at time t=0 A)Develop an expression for the number of white atoms on each...

so since PV=NkbT H=N1Cv1(Tf-T1)+N1kb(Tf-T1)+N2Cv2(Tf-T2)+N2kb(Tf-T2) is there a relationship between kb and Cv?

Homework Statement i solved for H=N1Cv1(Tf-T1)+V1(Pf-P1)+N2Cv2(Tf-T2)+V2(Pf-P2) my TA says that i have to simplify this, i don't really see a good way to go about it Homework Equations Pf=(Tf/T1)P1 Pf=(Tf/T2)P2 The Attempt at a Solution

solving for Tf i get, Tf= ((N1Cv1T1+N2Cv2T2)/(N1Cv1+N2Cv2)) from there i solved for dU1 and dU2 where dU1=N1Cv1(Tf-T1) dU2=N2Cv2(Tf-T2) i then added them together to get total change in energy for dA i solved for dA1 and dA2 integrated them and then added them together...

thats what i thought, I think he might have not realized that it said ideal gases in the problem

thanks again for being so helpful, you truly are the man

Ok, so final answer i should get A=c1ln((c1T1+c2T2)/(c1+c2)-T1)+c2ln((c1T1+c2T2)/(c1+c2)-T2) i don't see a way to break this up further or clean it up other than writing it as ln(Tf/T)

i can't say c1=c2=c can i? because N1 doesn't equal N2

oh ok thanks a lot, essentially i will get A=A1+A2 = c1(ln(Tf/T1)+c2ln(Tf/T2) where Tf=(c1T1+c2T2)/(c1+c2) seems ugly, but there's got to be some way in which i can clean it up

gotcha, thanks If i wanted to solve dA=dU/T is there a way to do it without using CvdT=dQ could i perhaps do: dA1=dU1/T=N1Cv1dT/T then take the integral from T1 to Tf, and then do the same for dA2 and then add the two together?

well in step (a) i got, Tf=(N1Cv1T1+T2Cv2N2)/(N1Cv1+N2Cv2), and for dU1=N1Cv1(Tf-T1) dU2=N2Cv2(Tf-T2) dU=N1Cv1(Tf-T1)+N2Cv2(Tf-T2) From here i can see there is no volume dependence, but I am not quite sure

thanks again voko for being the man, i did do that and i got, A=Cvln(Tf/T1) however my TA said that regardless since heat and work are path dependent i need to prove that dU=cdT since that is only applicable at certain times he said...

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thanks voko, ur the man, i solved a and b since I've posted this however, i have yet to solve c, one of the TAs told me that the value "pdV" goes to 0 in the expression dA=(dU+pdV)/T however from here i am lost. i know that dQ=dU+dW and if pdV= 0 than dQ=dU, and i know that dQ=CvdT...but he...