Recent content by always_tired

I finally figured it out. Only one of H or S can be defined in terms of ni such that the heat given by ΔH(T,P,n1,n2,...) ≠ TΔS(T,P).

I looked over that book and I'm still confused. It says that a constant T,P and no non-expansion work, dG < 0 for an irreversible process. I just don't understand how ΔHp can differ for a reversible and irreversible process since its a state function. If we look at your example, the entropy of...

I'm not sure I understand how that can show dG < 0. Can you write out the equations? Thanks.

By Clausius' inequality δq - TdS ≤ 0. For a constant T,P process in a closed system and no non-expansion work my text states that dG = dH - TdS = δq - TdS ≤ 0 but this seems incorrect to me. If pressure is constant such that dH = δq, doesn't this mean that δq = δqrev since dH is a state function...

No, it gains electrostatic potential energy. deltaU = -W; work is negative in this case; potential energy increases. From "University Physics" by Young and Freedman: "Negative charge moves in the direction of the electric field, field does negative work, potential energy increases" (pg 756)...

? It gains potential energy.

There is a negative point charge with a positively charged plate above it and a negatively charged plate below it. Define up as positive and down as negative. When the point charge moves towards the positively charged plate it does negative work. Why? The force the charge experiences is upward...