Problema 3E18

Sabendo que no ar atmosférico temos $x_\ce{O2}=0,2$ e $x_\ce{N2}=0,8$ podemos dizer que se $\ce{P_{ar}}=\pu{0,8 atm}$ temos: $$\ce{P_{A} = x_{A}\cdot P_{total} }$$ $$\ce{P_{O_{2}}}=\pu{0,16 atm}$$ $$\ce{P_\ce{N_{2}}}=\pu{0,64 atm}$$ Cálculo da solubilidade de cada gás: $$\ce{s = k_{H} P}$$ $$\ce{s_{\ce{O2}}}=(\pu{0,0013 mol L-1 atm-1})(\pu{0,16 atm})$$ $$\ce{s_{\ce{O2}}}=\pu{2,08e-4 mol L-1}$$ $$\ce{s_{\ce{N2}}}=(\pu{7e-4 mol L-1 atm-1})(\pu{0,64 atm})$$ $$\ce{s_{\ce{N2}}}=\pu{4,48e-4 mol L-1}$$ Base de cálculo: 1 Litro de solução Temos $$n_{\ce{O2}}=\pu{2,08e-4 mol}$$ $$n_{\ce{N2}}=\pu{4,48e-4 mol}$$ Cálculo da massa de cada gás: $$m=n \cdot M$$ $$m_\ce{O2}=(\pu{2,08e-4 mol})(\pu{32 g mol-1})=\pu{6,656 mg}$$ $$m_{\ce{N2}}=(\pu{4,48e-4 mol})(\pu{28 g mol-1})=\pu{12,544 mg}$$ Cálculo da massa total de gás: $$m_\ce{\text{total}}=6,656+12,544=\pu{19,2 mg}$$ Cálculo da solubilidade do ar: $$\ce{s}=\frac{m_\ce{ar}}{V_\ce{H2O}}$$ $$\ce{s}=\frac{\pu{19,2 mg}}{\pu{1 L}}$$ $$\ce{s = \pu{19,2 mg L-1}}$$