# Ideal gas-thermodynamics, thermal equilibrium

1. Apr 2, 2013

### oxman

1. The problem statement, all variables and given/known data
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

2. Relevant equations

U=NVCvdT

3. 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

2. Apr 2, 2013

### voko

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.

3. Apr 2, 2013

### oxman

thanks voko, ur the man, i solved a and b since ive 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?

4. Apr 2, 2013

### voko

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.

5. Apr 2, 2013

### oxman

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...

6. Apr 2, 2013

### voko

In #1, you gave the full expression for U. Differentiate it, paying attention to the conditions of the problem.

7. Apr 2, 2013

### oxman

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 im not quite sure

8. Apr 2, 2013

### voko

Even if there were a volume dependence, it would not matter: all volumes are constant in this problem.

9. Apr 2, 2013

### oxman

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. Apr 2, 2013

### voko

That's exactly how you should solve this. All that alphabetical soup before dT/T is the constant c I mentioned earlier.

11. Apr 2, 2013

### oxman

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 theres gotta be some way in which i can clean it up

12. Apr 2, 2013

### oxman

i cant say c1=c2=c can i? because N1 doesnt equal N2

Last edited: Apr 2, 2013
13. Apr 2, 2013

### voko

That is right, they are different. All of the gas constants in this problem seem to be different.

14. Apr 2, 2013

### oxman

Ok,

so final answer i should get

A=c1ln((c1T1+c2T2)/(c1+c2)-T1)+c2ln((c1T1+c2T2)/(c1+c2)-T2)

i dont see a way to break this up further or clean it up other than writing it as ln(Tf/T)

15. Apr 2, 2013

### oxman

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

16. Apr 2, 2013

### voko

You are welcome.

17. Apr 2, 2013

### Staff: Mentor

Your TA couldn't have meant that. He must know that for an ideal gas, dU is always equal to CvdT.

18. Apr 2, 2013

### oxman

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