thank u ! have an online mid term but to be honest its harder with the open book aspect. it's easier being in the class room knowing exactly what u should memorize and the critical thinking questions rely just on those facts. having someones face watch u on zoom while you're looking up formulas feeling sketchy for no reason is always fun lol.
Here we are using the fact that S is extensive (additive) by construction, as we took in a previous video the log of the multiplicity when we defined it to go from W=Wa*Wb to ln(W) = ln(Wa) + ln(Wb). Regarding the pressures, if the initial pressures pj are all different the final pressure should be the mole fraction weighted harmonic mean of the initial pressures: 1/pj = sum(xj/pj), so if the initial pressures are the same, the final pressure is the same, which is intuitive but not obvious.
Hello dear Prof.stuart Is there any experimental data which validate this formula about ideal solutions? Excuse me if im wrong but about the ideal gas same as what vander waals did B gas molecules have their own volumes. If we can ignore them in the ideal gas it doesn't make sense to do the same for ideal solutions. Thanks in advance for your guidence
In this context, don't think of solutions as a gas whose particles occupy the entire container (and thus would seem quite non-ideal). To have an ideal solution, all you need is for the enthalpy if mixing to be zero -- i.e. for the heterogeneous interactions to be the same strength as the (average of the) homogeneous interactions. The entropy of mixing is really a statistical factor, reflecting the greater number of places that the particles can be in the mixed system. You can think of the entropy in terms of concentrations, rather than volumes, if that helps.
Good observation / question! If the pressures are not equal, then the cancellation (in pink) won't happen as it does in this derivation. If you leave the term inside the log as a volume fraction, rather than converting to a mole fraction, then that would still be a correct result.
You explained that really well. I dont know how to thank you.
@@afafsgdchdhdg No thanks are needed. I'm glad you found it useful.
Love from india ❤
thank u ! have an online mid term but to be honest its harder with the open book aspect. it's easier being in the class room knowing exactly what u should memorize and the critical thinking questions rely just on those facts. having someones face watch u on zoom while you're looking up formulas feeling sketchy for no reason is always fun lol.
:-( Remotely proctored online exams are no fun.
Easiest lecture on TH-cam Thank you Respected Sir ❤️
You're welcome, I'm glad you found it easy to understand
It was a very helpful video ! Really hope this channel grows ❤️
Thanks, I appreciate it
Here we are using the fact that S is extensive (additive) by construction, as we took in a previous video the log of the multiplicity when we defined it to go from W=Wa*Wb to ln(W) = ln(Wa) + ln(Wb).
Regarding the pressures, if the initial pressures pj are all different the final pressure should be the mole fraction weighted harmonic mean of the initial pressures: 1/pj = sum(xj/pj), so if the initial pressures are the same, the final pressure is the same, which is intuitive but not obvious.
Good annotations, thanks
Excellent video, as always! May I know if you happen to have a video on enthalpy of mixing?
For an ideal solution, the enthalpy of mixing is zero. There's some information in this video: th-cam.com/video/Reb6gRL_gWs/w-d-xo.html
Thanks alot...please keep up the good work
There are more videos coming. Thanks for the support!
6:27..Sir will the pressure of system be the same before and after expansion?
Yes, if the initial pressures on both sides are the same. (This derivation assumes that is true.)
@@PhysicalChemistry And sir if pressures are not same then we will get volume fractions in place of mole fractions.
@@ishaan9017 That's right
Hello dear Prof.stuart
Is there any experimental data which validate this formula about ideal solutions? Excuse me if im wrong but about the ideal gas same as what vander waals did B gas molecules have their own volumes. If we can ignore them in the ideal gas it doesn't make sense to do the same for ideal solutions. Thanks in advance for your guidence
In this context, don't think of solutions as a gas whose particles occupy the entire container (and thus would seem quite non-ideal).
To have an ideal solution, all you need is for the enthalpy if mixing to be zero -- i.e. for the heterogeneous interactions to be the same strength as the (average of the) homogeneous interactions.
The entropy of mixing is really a statistical factor, reflecting the greater number of places that the particles can be in the mixed system. You can think of the entropy in terms of concentrations, rather than volumes, if that helps.
How does this equation change when factoring in situations that prevent or alter mixing, like immiscible liquids or fluids of different densities?
excellent, sir.
Thanks, I'm glad you thought so
so the pressure and temperature are constant before and after the process ?
Yes, in this example we're assuming that is true
@@PhysicalChemistry thank you
Thanks so much
You're welcome
Are the pressure of A and B equal at the beginning? If they are not, will the equation change?
Good observation / question!
If the pressures are not equal, then the cancellation (in pink) won't happen as it does in this derivation.
If you leave the term inside the log as a volume fraction, rather than converting to a mole fraction, then that would still be a correct result.
Did you actually write all of it in its mirror image while standing on the other side? If yes, then that's truly amazing!
No, I'm not nearly talented enough for that! See this video for an explanation: th-cam.com/video/YmvJVkyJbLc/w-d-xo.html
Why it it not clear
great
thanks
are we to always use this formula under ideal gas condition
This equation can be used for the entropy of mixing of ideal gases (or ideal solutions)
Don't we make it necessary that the pressure of the individual gases before mixing has to be equal to their combined pressure after mixing .
Yes, that's correct. Otherwise cancellation in the step at the top right won't work.
Pls make video on Miller indices
Good suggestion, thanks. This is on the to-do list
Thanks a ton
Is there any video about how to derive dq= nRT/V .dv ?
Try this one: th-cam.com/video/aVLU2s-OCRc/w-d-xo.html
Great video!
I have a question. May I know your E-mail ?
See the channel's "about" page
It was really helpful thank you ❤️🤍
You are welcome, I'm glad to help