How Does Thermal Equilibrium Affect Ideal Gases in a Partitioned System?

oxman
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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
 
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Welcome to PF!

And what is your question?
 
i have no idea what is meant by parts b and c

i understand that U=NCvdT

so N1Cv1(Tf-T1)=-N2Cv2(Tf-T2)

and i think i understand how to solve for the total change in energy
 
oxman said:
i understand that U=NCvdT
That would be dU.

oxman said:
and i think i understand how to solve for the total change in energy
Please tell.

To get dH and dA, you'll have to integrate the equations from the initial conditions to the final conditions.
 
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

essentially for A i got


A=c1ln(Tf/T1) + c2ln(Tf/T2) where c1=N1Cv1 c2=N2Cv2

pdV goes to 0 because there is no change in any of the volumes

correct?
 
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