- #1
AbedeuS
- 133
- 0
Hey, just a small mathematical enquiry about why temperature tends to average out in, say a box of gases. Let's just use the standard "Partitioned vessel" view of it, with One side having 1 mole of Helium at..say 398k and the other side with 1 mole of helium at 298K. The heat capacity is (rounded up) 21 J/Mol K and I'm assuming it's the same over the temperature range.
Now to calculate the final temperature when both gases are mixed I'm just going to use [tex]q=nC_{v}\deltaT[/tex] to find the total thermal energy then calculate for T:
[tex]q_{298K} = 1*21*298 = 6258J[/tex]
[tex]q_{398K} = 1*21*398 = 8358J[/tex]
So:
[tex]T= \frac{q_{298K}+q_{398K}}{2*21} = 348K[/tex]
Now with the final temperature, and discounting any residual T = 0 crystal entropy the entropy of that mixture is calculated by:
[tex]\int^{398}_{0} \frac{C_{v}}{T} dT = C_{p}ln(\frac{398}{0})[/tex]
Stuck here, ln(0) is not possible I did have a grand plan to try and prove it to myself but how do you deal with this part in the integral?
Now to calculate the final temperature when both gases are mixed I'm just going to use [tex]q=nC_{v}\deltaT[/tex] to find the total thermal energy then calculate for T:
[tex]q_{298K} = 1*21*298 = 6258J[/tex]
[tex]q_{398K} = 1*21*398 = 8358J[/tex]
So:
[tex]T= \frac{q_{298K}+q_{398K}}{2*21} = 348K[/tex]
Now with the final temperature, and discounting any residual T = 0 crystal entropy the entropy of that mixture is calculated by:
[tex]\int^{398}_{0} \frac{C_{v}}{T} dT = C_{p}ln(\frac{398}{0})[/tex]
Stuck here, ln(0) is not possible I did have a grand plan to try and prove it to myself but how do you deal with this part in the integral?