Problema 3E.07

Como $y_{\ce{CH3CHCl2}} = \pu{0,7}$, pela lei de Dalton, $$P_{\ce{CH3CHCl2}} = y_{\ce{CH3CHCl2}} \times P_\text{total} = \pu{0,7} \times \pu{160 Torr} = \pu{112 Torr}$$ Pela lei de Raoult, $P_{\ce{CH3CHCl2}} = x_{\ce{CH3CHCl2}} P_{\ce{CH3CHCl2}}^\star$, logo $$x_{\ce{CH3CHCl2}} = \dfrac{ \pu{112 Torr} }{ \pu{280 Torr} } = \pu{0,4} \implies x_{\ce{CF3CCl2F}} = \pu{0,6}$$

De $P_\text{total} = x_{\ce{CF3CCl2F}} P_{\ce{CF3CCl2F}}^\star + P_{\ce{CH3CHCl2}}$, $$\pu{160 Torr} = \pu{0,6} \times P_{\ce{CF3CCl2F}}^\star + \pu{112 Torr} \implies P_{\ce{CF3CCl2F}}^\star = \boxed{ \pu{80 Torr} }$$