ΔU

[jagged mason]

Question says "same condition" meaning ΔT=0 so ΔU= f(T) so why in this case ΔU is not 0?

[hushed zinc]

<@&1227988359910260737>

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[languid peak]

i dont think formula mei delta T hota hai

its deltaH= deltaU + delta(ng)RT

T will be 373 K

[jagged mason]

but change in internal energy puchha h toh usme delta T hota h

[languid peak]

it is delta T when the substance is not undergoing a chemical reaction or phase change

[jagged mason]

i got ur logic @languid peak but delta U aur temp ke perspective se agar soche toh zero nhi hoga?

[languid peak]

yeah but dusron ko bhi toh consideration mei lena hai na

[languid peak]

this is only when jab reaction nhi ho rahi hai so no ng will be there

[orchid canopy]

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

[orchid canopy]

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

[void flume]

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!)

@jagged mason

[jagged mason]

Thank you guys

+solved @languid peak @orchid canopy @void flume