Molar Gibbs Energy for an ideal gas
Your final result is almost correct, although your derivation is somewhat questionable, and, in particular, this equation is wrong: $$\Delta G_m = V_m\Delta P = RT\ln\left(\frac{P_f}{P_i}\right).$$ For one, as you've mentioned, $V_m$ isn't constant, so we can't use the finite difference implied by $\Delta$, and I don't understand how you would have gotten the second equality without having integrated. 

Instead, we should have $$\Delta G_m = \int_{G_{m,i}}^{G_{m,f}}\mathrm{d}G_m = \int_{P_i}^{P_f}V_m\,\mathrm{d}P = RT\int_{P_i}^{P_f}\frac{\mathrm{d}P}{P} = RT\ln\left(\frac{P_f}{P_i}\right) = RT\ln\left(\frac{V_i}{V_f}\right).$$ The third equality follows from $V_m = RT/P$, as you note, and the final equality follows from substituting $P = nRT/V$, the ideal gas law, and cancelling out terms.