- #1
Abhishek Jain
- 9
- 0
I am learning physics on khan academy and they do a proof to show that delta G for a reversible reaction is negative and how for a irreversible reaction it is positive. However in the proof, they assume that the heat put in by the isotherm is less for an irreversible reaction compared with a reversible reaction. They state the the heat generated by friction that is produced reduces heat required to be added to the system. However, my intuition is that it would require more heat because it would require more work to reach the same height as the reversible piston due to friction. Wouldn't it need to do more work to counteract the work done by friction? Yes the work done by friction is generating heat which will require less heat to reach T1. However, since it requires more work to move the piston due to friction, wouldn't Qir+Qf>Qr, since Wir>Wr to reach same position? The only way for Qir to be less than Qr would be for Qf to account for the work done by friction + some more to make Qir smaller. However, since it is an isotherm delta U = 0, so work = q. so Wf = Qf? What am I missing here? Please help. This is a foundational concept and I feel it is necessary for me to understand this to understand this topic fully. The link to the video is below.
https://www.khanacademy.org/science...us-gibbs-free-energy-spontaneity-relationship
https://www.khanacademy.org/science...us-gibbs-free-energy-spontaneity-relationship