14. Given the ellipse $C: \frac{x^{2}}{2}+y^{2}=1$ with its left and right foci being $F_{1}, F_{2}$ respectively, let $P$ be a point on the ellipse $C$ in the first quadrant. The extensions of $P F_{1}, P F_{2}$ intersect the ellipse $C$ at points $Q_{1}, Q_{2}$ respectively. Then the maximum value of the difference in areas between $\triangle P F_{1} Q_{2}$ and $\triangle P F_{2} Q_{1}$ is $\qquad$.
Given $F_{1}(-1,0), F_{2}(1,0), P(\sqrt{2} \cos \theta, \sin \theta)$,
then $P F_{1}: x=\frac{\sqrt{2} \cos \theta+1}{\sin \theta} y-1$,
solving $\left\{\begin{array}{l}x=\frac{\sqrt{2} \cos \theta+1}{\sin \theta} y-1, \\ x^{2}+2 y^{2}=2\end{array}\right.$
$$
\Rightarrow \frac{2 \sqrt{2} \cos \theta+3}{\sin ^{2} \theta} y^{2}-\frac{2 \sqrt{2} \cos \theta+2}{\sin \theta} y-1=0 \text {. }
$$

Thus, $y_{Q_{1}}=-\frac{\sin \theta}{3+2 \sqrt{2} \cos \theta} \Rightarrow y_{P}-y_{Q_{1}}=\frac{\sin \theta(4+2 \sqrt{2} \cos \theta)}{3+2 \sqrt{2} \cos \theta}$;
Similarly, $y_{Q_{2}}=-\frac{\sin \theta}{3-2 \sqrt{2} \cos \theta} \Rightarrow y_{P}-y_{Q_{2}}=\frac{\sin \theta(4-2 \sqrt{2} \cos \theta)}{3-2 \sqrt{2} \cos \theta}$.
Therefore, $S_{\triangle P F_{1} Q_{2}}-S_{\triangle P F_{2} Q_{1}}=\frac{\sin \theta(4-2 \sqrt{2} \cos \theta)}{3-2 \sqrt{2} \cos \theta}-\frac{\sin \theta(4+2 \sqrt{2} \cos \theta)}{3+2 \sqrt{2} \cos \theta}$
$=\frac{2 \sin \theta[(2-\sqrt{2} \cos \theta)(3+2 \sqrt{2} \cos \theta)-(2+\sqrt{2} \cos \theta)(3-2 \sqrt{2} \cos \theta)]}{9-8 \cos ^{2} \theta}$
$=\frac{4 \sqrt{2} \sin \theta \cos \theta}{9-8 \cos ^{2} \theta}=\frac{4 \sqrt{2} \sin \theta \cos \theta}{9 \sin ^{2} \theta+\cos ^{2} \theta}=\frac{4 \sqrt{2} \tan \theta}{9 \tan ^{2} \theta+1}=\frac{4 \sqrt{2}}{9 \tan \theta+\frac{1}{\tan \theta}} \leqslant \frac{2 \sqrt{2}}{3}$,
equality holds when $9 \tan \theta=\frac{1}{\tan \theta} \Rightarrow \tan \theta=\frac{1}{3}$.
Therefore, the maximum value of the difference in the areas of $\triangle P F_{1} Q_{2}$ and $\triangle P F_{2} Q_{1}$ is $\frac{2 \sqrt{2}}{3}$.