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Raoult's Law explained with thermodynamics/ free energy
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Kai Rabenstein
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:
$$S_g^o>S_m^o>S_p^o$$
and thus:
$$(S_g^o-S_p^o)<(S_g^o-S_m^o)$$
meaning the $\Delta S_g$ will be higher for a pure material than a mixture.
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: $$S_g^o>S_m^o>S_p^o$$ and thus: $$(S_g^o-S_p^o)<(S_g^o-S_m^o)$$meaning the $\Delta S_g$ will be higher for a pure material than a mixture.
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: $$S_g^o>S_m^o>S_p^o$$ and thus: $$(S_g^o-S_p^o)<(S_g^o-S_m^o)$$ meaning the $\Delta S_g$ will be higher for a pure material than a mixture.
Q.E.D.
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: $$S_g^o>S_m^o>S_p^o$$ and thus: $$(S_g^o-S_p^o)<(S_g^o-S_m^o)$$meaning the $\Delta S_g$ will be higher for a pure material than a mixture.
Q.E.D.
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