Here is the question, I am stuck on part B:


Basically, this is my thought process so far, as $x_a \rightarrow 0$, the Henry's law line becomes tangential to the actual partial pressure curve. Graphs Henry's law states that:
$$ p_{\alpha} = Hx_{\alpha} $$

Similarly, using the activity coefficient model: $$ p_{\alpha} = \gamma_{\alpha}x_{\alpha}p^*_{\alpha} $$

Where H is Henry's constant and $p^*_{\alpha}$ is the vapour pressure of the pure component. So at small $x_{\alpha}$ values one can assume: $$ H = \gamma_{\alpha}p^*_{\alpha} $$

From this I can get values for the $\gamma$ (I've been given values for pure component pressure). Then I can get values for $\ln \gamma_1, \ln \gamma_2$. My next step would be to substitute in values of $x_2$ and $x_1$ into the given correlations. So, for $\gamma_1$ to be correct, $x_1$ must be very small, assume it is zero therefore $x_2 \approx 1$, therefore:

$$ \ln \gamma_1 = \frac{A \times 1^2}{\bigg( (1-1)^2\frac{A}{B} + 1 \bigg)^2} \implies \ln \gamma_1 = A $$

Clearly, this is wrong because if I run through the same process for $\ln \gamma_2$, I get a different A value. Can someone please advise, thanks.


There was an error in the question, the model in the question is the van Laar activity model, the coefficient for $\ln \gamma_2$ on the numerator should be B, in this case, it is easily solvable.

  • 1
    $\begingroup$ Thanks for updating. I was trying to work this out, but I couldn't find anything wrong with what you were doing. I was looking up common models for this system, but I forgot about van Laar! $\endgroup$
    – J. Ari
    Mar 31 '17 at 14:03

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