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
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