[mortalapeman]

Ok the question I am working on is:

Calculate the final temperature of a sample of CO2 (16g) that is expanded reversibly and adiabatically from 0.5 L at 298.15 K to 2.0 L.

I know that adiabatic means the delta q = 0, however I'm really sure how that helps me with this problem. It seems more like the key word here is "reversibly". From my current understanding of this problem, that keyword means to solve this I need to hold pressure constant and just algebraically solve for T2. Is my understanding of this problem correct or am I missing something?

Thanks in advance,
ApeMan

[mortalapeman]

I meant to say "I'm not really sure how that helps me", not "I'm really sure how that helps me". And btw where is the edit post button on this forum? I went straight to this page from my book marks and tried to edit it but i saw no link to do so.

[Jorriss]

Next to quote there is a button that says modify, that's edit. Except, for me, it's only there about half the time...

[Borek]

You can modify your posts only for a limited amount of time.

[mortalapeman]

Ah and i just realized that i forgot to say what i was asking to find. I'm looking for w,q, delta U and delta H. So since nobody seems to have an idea what i'm talking about, maybe someone could solve for one of those variable and it would answer my question xD

[MrTeo]

If a transformation is reversible you can evaluate the work done using this expression:

$$ W=\int_{V_i}^{V_f}pdV /$$

And no, the pressure is not constant... here is the formula for a reversible adiabatic:



where



Now you have all the infos you need to solve your problem... use for CO₂ the c_V and c_p values of a biatomic gas (if I remember well...).