Ideal gas-thermodynamics, thermal equilibrium

AI Thread Summary
The discussion centers on solving a thermodynamics problem involving two ideal gases in thermal contact, requiring the calculation of final temperature and changes in energy. The final temperature is derived as Tf=(N1Cv1T1+N2Cv2T2)/(N1Cv1+N2Cv2), and the total change in internal energy is expressed as dU=N1Cv1(Tf-T1)+N2Cv2(Tf-T2). For the quantity A, the approach involves integrating dA=dU/T, leading to A expressed in terms of natural logarithms of temperatures. Participants clarify that for ideal gases, the relationship dU=CvdT holds true, emphasizing the importance of recognizing the path independence of heat and work in this context. The discussion concludes with the confirmation that the calculations align with the properties of ideal gases, despite initial confusion regarding the TA's comments.
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Homework Statement


Two ideal gases are separated by a partition which does not allow molecules to pass from one volume to the other. Gas 1 has: N1, V1, T1, Cv1 for the number of molecules, volume it occupies, temperature in kelvin, and specific heat per molecule at constant volume respectively. Gas 2 has: N2, V2, T2, Cv2. The two gases are in thermal contact and reach a final temperature

a) find the final temperature and the total change in energy of the combined system. Check your answer for the final temperature when N1=N2, V1=V2. Cv1=Cv2

b)Evaluate the total change ina quantity H whose differential change is dH=dU+Vdp for each component and for the entire system

c)evaluate the total change in a quantity A whose differential change is dA=(dU+pdV)/T for each component and for the entire system

Homework Equations



U=NVCvdT

The Attempt at a Solution



I already solved for the final temperature for part a, and when evaluated at equal N and V i got Tf=(T2+T1)/2

i am having trouble finding b and c
 
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Note that the volume of each component as well as the entire system is constant. That means ## \frac p T = k = const ##. Use this to express both differential quantities via ## dT ##, then integrate, using the results of a) for the limits.
 
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 said something about how this is path dependent and i need to prove that, any help?
 
Since ##U = cT##, where ##c## is some constant, ##dU = c dT##. ## p dV ## is indeed zero because the volume does not change. So all you have is ## dA = \frac {dU} T = \frac {c dT} T ##. I am sure you can integrate that.
 
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...
 
In #1, you gave the full expression for U. Differentiate it, paying attention to the conditions of the problem.
 
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
 
Even if there were a volume dependence, it would not matter: all volumes are constant in this problem.
 
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?
 
  • #10
That's exactly how you should solve this. All that alphabetical soup before dT/T is the constant c I mentioned earlier.
 
  • #11
oh ok thanks alot,

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
 
  • #12
i can't say c1=c2=c can i? because N1 doesn't equal N2
 
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  • #13
That is right, they are different. All of the gas constants in this problem seem to be different.
 
  • #14
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)
 
  • #15
thanks again for being so helpful, you truly are the man
 
  • #16
You are welcome.
 
  • #17
oxman said:
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...

Your TA couldn't have meant that. He must know that for an ideal gas, dU is always equal to CvdT.
 
  • #18
thats what i thought, I think he might have not realized that it said ideal gases in the problem
 
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