ΔU
Question says "same condition" meaning ΔT=0 so ΔU= f(T) so why in this case ΔU is not 0?
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@Dexter
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.i dont think formula mei delta T hota hai
its deltaH= deltaU + delta(ng)RT
T will be 373 K
but change in internal energy puchha h toh usme delta T hota h
it is delta T when the substance is not undergoing a chemical reaction or phase change
i got ur logic @trin but delta U aur temp ke perspective se agar soche toh zero nhi hoga?
yeah but dusron ko bhi toh consideration mei lena hai na
this is only when jab reaction nhi ho rahi hai so no ng will be there
if u see, delta U= n Cv delta T for an ideal gas only
yahan pe in ur Q, thr is phase change involved
so here deltau is not ncv delta T
rather its this for a rxn
same condn means, while water changes to steam, wht is the internal energy, and phase change of water to steam takes at 100 degree celsius so.... ig u get it
THIS is the main thing. When no change in phase/rxn.
From first principles, dU = q + w
here, q is dH, w = expansion work (since water gets vapourised, volume increases!)
@Say_miracle_shadow
Thank you guys
+solved @trin @Slembash @iTeachChem
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