\begin{align} \ce{N2O4(g)&<=>2NO2(g)} & \Delta_\mathrm{r}G^\circ &= \pu{4.7 kJ mol^-1} \end{align}
Knowing that the standard values for pressure and temperature are $\pu{1 bar}$ and $\pu{298 K}$ respectively, find:
a) the dissociation grade (noted as $\alpha$) of $\ce{N2O4}$ at equilibrium, in standard conditions
b) the dissociation grade of $\ce{N2O4}$ at equilibrium and at $\pu{298 K}$ / $\pu{10 bar}$
First, we find $K$ from $\Delta G$
Then we write the partial pressures as the molar fraction of the compound times the total pressure of the system (at equilibrium)
After that, we write the relation for $K_p$ using partial pressures
and I'm stuck...
\begin{align} \Delta_\mathrm{r}G^\circ &= -RT \ln K & \Rightarrow K&=0.15 \mathrm{~(unitless)} \\ P_{\ce{N2O4}} &= \frac{1-\alpha}{1+\alpha}\cdot P & P_{\ce{NO2}} &= \frac{2\alpha}{1+\alpha}\cdot P & K_p = \frac{(P_{\ce{NO2}})^2}{P_{\ce{N2O4}}} \\ \Rightarrow K_p &= \frac{4\alpha^2}{1-\alpha^2}\cdot P \\ \end{align}
My question is: how can I convert between $K$ (which is unitless) and $K_p$ (which is not) so that I can find the dissociation grade at $\pu{1 bar}$ and $\pu{10 bar}$?