How do I solve for the equilibrium constant if given temp & partial pressure...?

Question

I have read the book cover to cover and can't find a method for answering this type of problem:

The vapor pressure of water at 80. ºC is 0.467 atm. Find the value of K(subscript c) for the process

H2O (l) <--> H2O (g)

at this temperature.

Answer 1

Kp=p(H2O)=0.467

Kp=Kc(RT)^Δn

0.467=Kc(0.0821 x 353)

Kc=0.0161

Answer 2

Since H2O(g) is the only gas in the equation,

Kp = P H2O = 0.467 atm.

Kp takes the same form as Kc except partial pressures are used for gaseous substances instead of molarities. That is, if I have the reaction:  A(g) + B(g) =&gt; C(g), then

Kp = (P C) / (P A)(P B).

Kc and Kp are related:   Kp = Kc(RT)^delta n

where delta n = gas moles product - gas moles reactant.

In our problem,

0.467 = Kc x ((0.0821)(355 K))^1

0.467 = Kc (29.1)

0.0160 = Kc