In the last point, Sal says that for any spontaneous process, deltaS is > 0. It should be emphasised that he means deltaS of the universe, because the change in entropy of a system can be negative (become more ordered) and still be spontaneous as long as deltaS of the surroundings is a higher positive value. In other words, if a spontaneous process creates a more ordered system, the entropy of the entire universe is still increasing and thus becoming more disordered.
Uhmmm, no, that qirr is less than qrev is because the work done is less, it is done in one step. When you are talking about rev and irrev processes you don't need to, nor should you talk about this "friction" for example now in your cylcle, how do you account for the heat in friction. Sorry, but I think you are incorrect on this explanation. I enjoy most of your lectures though.
Just 30 mins ago i started woundering about this and after i came to a conclusion i though of checking this comment and i was surprised to find that someone had commented just 24 mins ago
Uhmmm, no, that qirr is less than qrev is because the work done is less, it is done in one step. When you are talking about rev and irrev processes you don't need to, nor should you talk about this "friction"
Im studying for my DATs. You have helped me out more than my kaplan course !!!!
You are a genious SAL !
Thanx
In the last point, Sal says that for any spontaneous process, deltaS is > 0. It should be emphasised that he means deltaS of the universe, because the change in entropy of a system can be negative (become more ordered) and still be spontaneous as long as deltaS of the surroundings is a higher positive value. In other words, if a spontaneous process creates a more ordered system, the entropy of the entire universe is still increasing and thus becoming more disordered.
H is a state function , so are T and S ....then why ΔG is different for the two processes .
I have score top in ch like some basic concept of chemistry and classification of elements as I studies them from here
Iv been using Gibbs equation for so long and never really understood how the equation was formed. Good video!!!
Uhmmm, no, that qirr is less than qrev is because the work done is less, it is done in one step. When you are talking about rev and irrev processes you don't need to, nor should you talk about this "friction" for example now in your cylcle, how do you account for the heat in friction. Sorry, but I think you are incorrect on this explanation. I enjoy most of your lectures though.
do you have any intuitive reasoning for why is work done less in an irreversable process?
Yes you are right because of friction in irreversible system the heat absorbed will be more that that of reversible as there is no friction
🇮🇳🇮🇳
Just 30 mins ago i started woundering about this and after i came to a conclusion i though of checking this comment and i was surprised to find that someone had commented just 24 mins ago
great job! thanks a lot!
thanks
When Delta S is a state function, why does it depend on whether the process between the states is reversibel or irreversibel?
delta S can be defined as q/T when only when its reversible as q is a path function
delta S is best defined as k*ln'omega'
reservoir and reversible...first three letters are not the same.
Uhmmm, no, that qirr is less than qrev is because the work done is less, it is done in one step. When you are talking about rev and irrev processes you don't need to, nor should you talk about this "friction"
I HATE P/V DIAGRAMS!!!