Which of the following must be true for adiabatic processes?

Question

anonymous · Answer

A) $C_v = C_p$
B) $\Delta H = 0$
C) $\Delta U = 0$
D) $\Delta S = 0$
E) $q=0$

anonymous · Answer

I put B which is wrong. I don't think any of these answer choices given are right though.

taramgrant0543664 · Answer

An adiabatic process is one in which no heat is gained or lost by the system.

taramgrant0543664 · Answer

There is one that is right there

anonymous · Answer

I agree with you but I think the correct answer should be

$\Delta q = 0$

joannablackwelder · Answer

In an adiabatic change, q = 0, so the First Law becomes ΔU = 0 + w.

taramgrant0543664 · Answer

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.

anonymous · Answer

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 $U=q+w$ ?

anonymous · Answer

@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?

taramgrant0543664 · Answer

I'm pretty sure ΔU=Δq+Δw is used for isobaric conditions

joannablackwelder · Answer

Take a look here: http://www.chem1.com/acad/webtext/energetics/CE-2.html#S3A

taramgrant0543664 · Answer

The a diabetic system doesn't allow heat in or out so its not a change it would be just 0

joannablackwelder · Answer

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.

anonymous · Answer

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.

anonymous · Answer

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 ?

taramgrant0543664 · Answer

I get where you're coming from it is a little confusing I never really looked at it like that

joannablackwelder · Answer

The notation q=0 doesn't mean no heat content. It means no change in heat content.

anonymous · Answer

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