# Raoult and Henry's law

According to my textbook, for Raoult's law, we have:

$$μ_i(\mathrm{vap}) = μ_i^* (p_i^*,T) + RT \ln⁡ x_i$$

where $$μ_i^* (p_i^*,T)$$ is the chemical potential at $$T$$ and $$p_i^*$$, the pressure of pure $$i$$. So, when $$x_i=1$$, we have $$μ_i (vap)=μ_i^* (p_i^*,T)$$. That's okay and makes sense.

For Henry's law, we have:

$$μ_i(\mathrm{vap}) = μ_i^∞ (p_j^*,T) + RT \ln⁡ x_i$$

where $$μ_i^∞ (p_j^*,T)$$ is the chemical potential at $$T$$ and $$p_j^*$$, the pressure of pure $$j$$ (the other component of the mixture). So, when $$x_i = 1$$ we have $$μ_i^∞ (p_j^*,T) = μ_i^∞ (p_j^*,T)$$ and that doesn't make any sense, since we have $$i$$ pure and the equation says that the chemical potential is the chemical potential of pure $$j$$.

Am I interpreting something wrongly?

• Your last equality has two identical symbols on both sides of the = sign ! You simply repeat the same symbols on both sides. These formula cannot be different. It is like writing an equation : a = a. – Maurice Feb 15 '20 at 9:03
• In addition to Maurice's comment, your last statement "we have i pure and the equation says that de chemical potential is the chemical potencial of pure j" is inconsistent with the equations you have written, which describe the chemical potential of i (not j). – Buck Thorn Feb 15 '20 at 10:34