Asked by Emily
The ∆G° of vaporization for butane at 298 K and 1.00 atm is -2.125 kJ/mol. Calculate the pressure, in atm, of butane vapor in equilibrium with butane liquid at 298 K.
Answers
Answered by
skylar
I get 22.9 kJ . ... (1) Liquid water @ 100∘C boil −−−−→ Water vapor @ 100∘C ... The 1 atm pressure is just for context, and says you are at everyday conditions. ... of a "latent heat of vaporization" ( ΔHvap,w ) in the following equation: ... J/g⋅K=1.841 J/g⋅∘C is the specific heat capacity of water vapor.
Answered by
izzyloaaa
Hi skylar,
The question asks for the pressure, in atm, not the kJ.
Thanks.
Isabelle Loaisiga
The question asks for the pressure, in atm, not the kJ.
Thanks.
Isabelle Loaisiga
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