FEC: Raoult's Law

Raoult's Law

Raoult's Law is a special case of the general vapor-liquid equilibrium expression(s):

$\hat f_i^L = \hat f_i^v$

$\mu_i^L = \mu_i^v$

$x_i \gamma_i f_i^o = y_i \hat \phi_i P_{tot}$


Raoult's Law is based on the assumptions that the vapor phase behaves as an ideal gas, while the liquid phase behaves as a (Lewis-Randall) ideal solution.

From the combination of these equations and assumptions, we will start with the following:

$\mu_i^{ig} = \mu_i^{is}$

From our previous discussion we can replace these terms to yield the following:

$\mu_i^{ig} = g_i^{ig} + RT \ln[y_i] = g_i^{is} + RT \ln[x_i] = \mu_i^{is}$

Rearranging gives:

$g_i^{ig} - g_i^{is} = RT \ln\left [\frac{x_i}{y_i} \right ]$

The left hand side of this equation requires us to take a three step path in order to evaluate the change in $g$:

3 steps

The $\Delta g$ from each of these steps is respectively:

$\Delta g_i = g_i^{ig} - g_i^{is} = v_L(P^{sat}-P) + 0 + RT\ln\left [\frac{P}{P^{sat}} \right ]$


Two of three of these expressions comes from the relation that $\displaystyle{\left (\frac{\partial g}{\partial P} \right ) = v}$, while the third simply recognizes that there is no $\Delta g$ associated with a simple phase change.

If we further assume that our system is at low enough pressure, $P$, that

$v_L(P^{sat}-P) \approx 0$

our expression reduces to:

$g_i^{ig} - g_i^{is} = RT\ln\left [\frac{P}{P^{sat}} \right ] = RT \ln\left [\frac{x_i}{y_i} \right ]$

Which can be simplified and rearranged to yield Raoult's Law:

$y_iP = x_iP_i^{sat}$


The value of $P = P_{tot}$, so this expression is a simple case of the general one at the top except with both activity coefficient ($\gamma_i$) and fugacity coefficient ($\hat \phi_i$) equal to 1 (their idealized values). Finally, we also assume that the pure species fugacity (under L-R conditions) is given as $P^{sat}$, which holds true for low pressure problems.


Perform bubble-point and dew point calculations using Raoult's Law