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Cunsidercontaintrdiided into compartments Wst comoartment Land filled wich nitrogen 8as; the second compartment fillled with onEen gas; the last compartment filled ...

Question

Cunsidercontaintrdiided into compartments Wst comoartment Land filled wich nitrogen 8as; the second compartment fillled with onEen gas; the last compartment filled Lhtee companments are 100 bar pressure and 298K.When the partitions between compurtments are remov and the gases allowved mix, what is the change enCrey AG_u} and the = change entropy mxina (BS_? Treat these gases perfect:Gibbs

cunsider containtr diided into compartments Wst comoartment Land filled wich nitrogen 8as; the second compartment fillled with onEen gas; the last compartment filled Lhtee companments are 100 bar pressure and 298K. When the partitions between compurtments are remov and the gases allowved mix, what is the change enCrey AG_u} and the = change entropy mxina (BS_? Treat these gases perfect: Gibbs



Answers

Carbon dioxide gas at $320 \mathrm{K}$ is mixed with $\mathrm{ni}$ trogen at $280 \mathrm{K}$ in an insulated mixing chamber. Both flows are coming in at $100 \mathrm{kPa}$ and the mole ratio of carbon dioxide to nitrogen is 2: 1 Find the exit temperature and the total entropy generation per kmole of the exit mixture.

Here. Each gas expends into the other houses. The container, as total other gas would know. There therefore considers its gasto undergo off The extension process in which its volume doubled Now from the original stem trophy is twice for the single day. Artist. Don't ask difficult twice both. And, uh, natural logo B f up on me, I No. We can put the values here to into and little 10400 more deployment. It won't be one fall doing more. You will, Permal. Children multiply with natural law was proof. Now we can write it at the drop 0.507 caldron Wolper killed.

Hi friends. This is the problem based on changing and profit forgets expanding from initial value toe final volume me if which is given by and are long off final volume divided by initial volume. Now here and this problem it is given our guest expending so that its final body becomes two times off its initial value number off, more off hydrogen gets find euro for food and that off oxygen you it is 0.44 We have toe find changing and trophy off the system changing and trophy it's and art long off. Final volume upon initial volume and will be and one plus and two our loan off final volume upon initial volume substitute the value and one is given 0.44 and two is also point euro for for 8.31 Learn off two times off initial volume upon initial volume. So it is to be 0.50 seven dual per Calvin. That's all Thanks for watching it

Eso the entropy change can be calculated by assuming idol gas behavior. The entropy change for, um all off idol gas undergoing general process off changing the state from an initial state one the final state to So for the first part, the pressure changes from Pito be at United Calvin for offseason is s. So we need to find Delta s, which is minus in Are and be to by P one. Oh, which is, uh, one more mm to our state. Point 314 Too far more for Calvin and Ellen. The one by P two says to so which is 5.763 to a park kelvin. So the temperatures changed from 2 98. You will trade 50 tied to constant pressure off. So for the B bar Delta s his name CV Ellen, they to by 81. So it'll be a wine. C is 28.82 Um, do you one? It was 3. 55 divided by 2 98. Yes, it because 5.44 Jules birth Kelvin

Transferred to a gas mixture contained in a piston cylinder discussed in the previous problem with the total entropic change here. Also in Tripoli change. The ex energy destruction are to be determined for two places. So when we understand this case and totally generally during the process is determined by applying the intro ph balance on the extended system. Okay that includes the question cylinder device and its immediate surroundings so that the boundary temperature of the extended system of them, environment temperature at all times it remains the same. Therefore we can find out that. Yeah, As in -1 out. This question was to delta A. System that is from this weekend. Find it out like that. Q. In upon T boundary plus as jim Equals two Delta S. Yeah that is as jen equals to em off S. Two minus S. One minus You. M. seven. Then exerted destroyed during the process from its deficient. It's a girl strong it destroyed. That is caused your T. zero into estrogen. Now noting the total mixture pressure and thus the partial pressure of each gas remains constant. The entropy change of accompaniment in the mixture during the process is Delta as I equals two AM I. Into C. P. And then T. Two upon T. One minus R. L. N. P. Two upon P. One. I. It is equals two AM I. C. P. I. Ln T two upon T. One. So assuming the ideal gas behavior and the values of C. P. The delta S of H. Two and N. Two can be determined by Delta. As of H. two ideal was to 18.21 kg two kelvin & Delta s. n. two. I did. Is it calls to you? 4.87 you know, don't work. Okay. No. As gen Equals to 8.496. Hello to kelvin. So putting that in the excel energy destroyed. We get 2489. He looked to No. You think that Megan's law finding out there is that one is the best 1? Is it has to for the cases we get that. It's gin using a mega slow. We have done using. I'm like at slow we get as gentle B nine 9.390 kg joules per kelvin and X destroyed was 22751 kg two. Yes, So I hope you understood the question. Thank you.


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