- #1

AbedeuS

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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?**