Which of the following must be true for adiabatic processes?

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Which of the following must be true for adiabatic processes?

Thermodynamics
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A) \(C_v = C_p\) B) \(\Delta H = 0\) C) \(\Delta U = 0\) D) \(\Delta S = 0\) E) \(q=0\)
I put B which is wrong. I don't think any of these answer choices given are right though.
An adiabatic process is one in which no heat is gained or lost by the system.

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There is one that is right there
I agree with you but I think the correct answer should be \(\Delta q = 0\)
In an adiabatic change, q = 0, so the First Law becomes ΔU = 0 + w.
Yes it is The first law of thermodynamics with Q=0 shows that all the change in internal energy is in the form of work done. This puts a constraint on the heat engine process leading to the adiabatic condition shown below. This condition can be used to derive the expression for the work done during an adiabatic process.
I thought Q was a constant, so the change in Q is 0. @JoannaBlackwelder why isn't it \(\Delta U = \Delta q + \Delta w \) and \(U=q+w\) ?
@taramgrant0543664 I guess I'm not familiar with this notation, since when I think of adiabatic I think the heat content is not zero but the change in the heat content is 0. Does what I'm asking make sense?
I'm pretty sure ΔU=Δq+Δw is used for isobaric conditions
Take a look here: http://www.chem1.com/acad/webtext/energetics/CE-2.html#S3A
The a diabetic system doesn't allow heat in or out so its not a change it would be just 0
When q is positive, heat is added, when q is negative heat is taken out, and when q is 0 there is no heat transfer.
Well I know what adiabatic means, that's not the problem, I don't think the notation is correct. For example when we substitute in things in the differential form: \[dU=dq+dw\] we can substitute in \[TdS=dq\] or \[pdV=dw\] to get \[dU=TdS+pdV\] But this is different, if you're writing \[dU=q+w\] that's a different thing altogether.
So what I'm saying is \(\Delta q=0\) should be the correct answer not \(q=0\) because: \(q=0\) means there is no heat content. \(\Delta q=0\) means there is no change in heat content. Which is what adiabatic means to me, since none was transferred. Does this help clear up what I'm asking specifically @JoannaBlackwelder and @taramgrant0543664 ?
I get where you're coming from it is a little confusing I never really looked at it like that
The notation q=0 doesn't mean no heat content. It means no change in heat content.
I guess I should just accept that this is what the notation means. In thinking about if the equation was actually \(U=q+w\) (not \(\Delta U\) ) then that would mean the internal energy U is the sum of its internal heat energy q and then what's w? While it might make sense to think of q as the internal heat energy, I guess w is more like potential energy? Or pressure energy? I guess it falls apart to think of it like this. Thanks for your help both of you @taramgrant0543664 and @JoannaBlackwelder I guess I'll try to sort this out a little more on my own unless you have more you'd like to add! :D

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