Raoult's Law explained with thermodynamics/ free energy
Basically a material in a mixture will have a lower vapor pressure than if it were pure.  We know that the vapor pressure of a liquid can be determined by: 
$$\Delta G = \Delta G_v^o + RT \ln(P_x)$$
Given that $\Delta G$ is $0$ at equilibrium, we get: 
$$\Delta G_v^o = \Delta H_v^o -T\Delta S_v^o = -RT\ln(P_x)$$
Being an ideal mixture $\Delta H$ is $0$ and the equilibrium vapor pressure is determined by: 
$$-T\Delta S_v^o = -RT\ln(P_x)$$
$$\Delta S_v^o = R\ln(P_x)$$ 
Meaning that a higher difference in molar entropy ($\Delta S_g^o$) between the phases means there will be a higher equilibrium vapor pressure ($P_x$). Since gases have higher molar entropy ($S_g^o$) the process of vaporization creates an increase in entropy (positive $\Delta S$).  However since a mixture has a higher molar entropy ($S_m^o$) in the initial state than the pure material ($S_p^o$) the phase change to vapor has a lower change in molar entropy: 
$$S_g^o&gt;S_m^o&gt;S_p^o$$ 
and thus: 
$$(S_g^o-S_p^o)&lt;(S_g^o-S_m^o)$$
meaning the $\Delta S_g$ will be higher for a pure material than a mixture.

Q.E.D.