Recent content by argali

Sorry to revive what I thought was a settled question, but in light of my new confusion regarding pressureexternal vs pressureapplied, I wanted to point out another dilemma I am having. If work does indeed equal the product of applied pressure (which I take to be roughly analogous to the force...

Actually that was it. The answer they have was that the reversible process does more work. I'm guessing they didn't actually want you to calculate anything, but instead just wanted me to remember that reversible processes always do the most work. My confusion stems from the more general...

Thank you so much! I am still, however, somewhat confused by why work equals the external pressure times volume change. If the pressure being exerted by the system is only equal to the external pressure, wouldn't there not be a volume change in the first place because the two forces would cancel...

Ah! So is the reason you can use w = pΔV for expansions that the external pressure will not change due to the immensity of the surroundings?

The pressure will change, right?

I calculated the work done by both processes and found that the irreversible process does more work than the reversible process. Doesn't the reversible process always do more work?

Sorry. Typo on my part; I meant reversible. I will edit when I'm back on my computer.

I am trying to teach myself the basics of thermodynamics using a book I bought from a university book store. My question relates to the amount of work done by reversible vs irreversible processes. An example in the book (Chemical Principles by Atkins & Jones) reads the following: "A cylinder...