## Delta U as 0

isochoric/isometric: $\Delta V = 0$
isothermal: $\Delta T = 0$
isobaric: $\Delta P = 0$

Ekayana Sethi 2K
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### Delta U as 0

For which different conditions is Delta U 0 making work=-q. I think deltaU is 0 in an isolated system, isothermal system, at equilibrium and when initial and final conditions are same. Is there any other time when delta U is 0? correct me if im wrong.

Thuy-Anh Bui 1I
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### Re: Delta U as 0

Those are all correct. What these conditions all have in common is that there is no change in temperature. This fact can be given if the problem says the process is explicitly isothermal, or no q is added/removed (adiabatic), or you could possibly deduct no change in temperature from PV=nRT.
Last edited by Thuy-Anh Bui 1I on Wed Feb 14, 2018 2:35 pm, edited 1 time in total.

Vincent Tse 1K
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### Re: Delta U as 0

ΔU = 3/2 nRΔT for monoatomic ideal gases, so anytime ΔT = 0, that's true. Even if it weren't monoatomic, only the 3 changes to some other number, so the ΔT = 0 condition still holds true.

ΔU also stays the same if a multi-step process (the PV diagrams!) returns to its original starting point, if that's what you mean by initial + final conditions being the same.

Yea Eun Lee 1H
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### Re: Delta U as 0

how come delta u is zero for an isothermal expansion?