I am confused about the last equation on the slide that occurs at 5:35. The relationship of (Cv/T)dt = -(nR/V)dV followed by integration is only appropriate for adiabatic expansion according to McQuarie (see page 777). Yet, here we use this to deal with isochoric heating in the last equation. It appears on pg 822 of McQuarie that the combination of the heats from paths B and C, resulting in q_(B+C) allows us to play the substitution game that is written at the end of the slide. But then at the bottom of page 822 in McQuarie the same equation is written as you've written, where the substitution is just made for just deltaS_C, not deltaS_B+C. This seems hand wavy but gets us to the point of a cyclic entropy process. Is there any light that can be shed on this that would help explain why these actions can be taken?
I think I understand the confusion, and appreciate why it might arise. Let me note that if we take the temperature integrals in that final equation as "numbers" (which is what an integral is, after all), it must be correct. That is because entropy is a state function, and we've already established the entropy changes for paths A and B, so the "value" that is listed for path C is the one that is required in order for the sum of paths B and C to equal A. What is odd as written, though, is that a quantity V1 appears that is unrelated to path C. In the absence of thinking about all of the paths available, one would not know how to easily evaluate the temperature integral exclusive to path C -- this lecture in some sense shows how to do that (for an ideal gas).
@@ChemProfCramer Ah...I was thinking along these lines. So in essence deltaS_C is deltaS_B+C in that last equation because, deltaS_B = 0. That is, deltaS_B+C = deltaS_B + deltaS_C = deltaS_C. Therefore, considering that both paths B and C were followed for the expansion process, I can consider what V_1 is in the final deltaS_C expresion because it came from the beginning of path B (which is included in the overall entropy change ). Math/practical-logic mind bender. HA! Fun stuff. Thanks so much for your input (and if I got it wrong, let me know if you time).
@@ChemProfCramer As an educator I thought that you'd find this fascinating: Did you know Botzmann didn't come up with S = klnW?! Have a look if you have time. You don't have to comment back, just thought you'd find it interesting because you, like I, attribute the equation to him in our lectures. th-cam.com/video/Tr_gv5CKB1Y/w-d-xo.html
I am confused about the last equation on the slide that occurs at 5:35. The relationship of (Cv/T)dt = -(nR/V)dV followed by integration is only appropriate for adiabatic expansion according to McQuarie (see page 777). Yet, here we use this to deal with isochoric heating in the last equation. It appears on pg 822 of McQuarie that the combination of the heats from paths B and C, resulting in q_(B+C) allows us to play the substitution game that is written at the end of the slide. But then at the bottom of page 822 in McQuarie the same equation is written as you've written, where the substitution is just made for just deltaS_C, not deltaS_B+C. This seems hand wavy but gets us to the point of a cyclic entropy process. Is there any light that can be shed on this that would help explain why these actions can be taken?
I think I understand the confusion, and appreciate why it might arise. Let me note that if we take the temperature integrals in that final equation as "numbers" (which is what an integral is, after all), it must be correct. That is because entropy is a state function, and we've already established the entropy changes for paths A and B, so the "value" that is listed for path C is the one that is required in order for the sum of paths B and C to equal A. What is odd as written, though, is that a quantity V1 appears that is unrelated to path C. In the absence of thinking about all of the paths available, one would not know how to easily evaluate the temperature integral exclusive to path C -- this lecture in some sense shows how to do that (for an ideal gas).
@@ChemProfCramer Ah...I was thinking along these lines. So in essence deltaS_C is deltaS_B+C in that last equation because, deltaS_B = 0. That is, deltaS_B+C = deltaS_B + deltaS_C = deltaS_C. Therefore, considering that both paths B and C were followed for the expansion process, I can consider what V_1 is in the final deltaS_C expresion because it came from the beginning of path B (which is included in the overall entropy change ). Math/practical-logic mind bender. HA! Fun stuff. Thanks so much for your input (and if I got it wrong, let me know if you time).
@@bohdanschatschneider9962 spot on.
@@ChemProfCramer As an educator I thought that you'd find this fascinating: Did you know Botzmann didn't come up with S = klnW?! Have a look if you have time. You don't have to comment back, just thought you'd find it interesting because you, like I, attribute the equation to him in our lectures.
th-cam.com/video/Tr_gv5CKB1Y/w-d-xo.html