Respuesta :
The evaporation rate of the n-Hexane is [tex]7.85 \times 10^{-6} \mathrm{mol} / \mathrm{s}[/tex]
Explanation:
This is a situation regarding diffusing A through non-diffusing B.
A = n-Hexane B=Air
Where the molar flux is provided by,
[tex]N_{A}=D_{A B} P_{T}\left(P_{A 1}-P_{A 2}\right) / R T z P_{b m}[/tex]
[tex]\mathrm{D}_{\mathrm{AB}}=8.8 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}[/tex]
[tex]P_{t}=1 a t m=101325 P a\\[/tex]
[tex]\text { so, } P_{A 1}=[/tex] the vapor pressure at hexane [tex]25 \mathrm{C}[/tex] [tex]=20158.2 \mathrm{Pa}[/tex]
For wind, assume negligible hexane is present, hence [tex]P_{A 2}=0[/tex]
Now,
[tex]\mathrm{P}_{\mathrm{B} 1}=\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{\mathrm{A} 1}=101325-20158.2 \mathrm{P}_{\mathrm{a}}[/tex]
[tex]\mathrm{P}_{\mathrm{B} 2}=\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{\mathrm{A} 2}=\mathrm{P}_{\mathrm{T}}=101325 \mathrm{Pa}[/tex]
[tex]P_{B M}=\frac{\left(P_{B 2}-P_{B 1}\right)}{\log _{e}\left(P_{B 2} / P_{B 1}\right)}\\[/tex]
[tex]=\frac{101325-81166.8}{\ln \left(\frac{101325}{81166.8}\right) \mathrm{Pa}}[/tex]
[tex]=90873.57 \mathrm{Pa}[/tex]
[tex]R=8.314 \mathrm{J} / \mathrm{mol}-\mathrm{K}[/tex]
[tex]z=\text { distance }=20 \mathrm{cm}=0.2 \mathrm{m}\\[/tex]
where T = 298 K
substituting all in the equation, we get
[tex]\begin{aligned}&\mathrm{N}_{\mathrm{A}}=\\&\left(8.8 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}\right) \times 101325 \mathrm{Pa} \times(20158.2 \mathrm{Pa}) /(8.314 \mathrm{J} / \mathrm{mol}-\mathrm{K} \times 0.2 \mathrm{m} \times 298 \mathrm{K}\\&\times 90873.57 \mathrm{Pa})\end{aligned}[/tex]
[tex]=0.004 \mathrm{mol} / \mathrm{m}^{2} \mathrm{s}\\[/tex]
Now,Flux [tex]\times[/tex] area = Molar rate of evaporation
Evaporation rate = [tex]0.004 \mathrm{mol} / \mathrm{m}^{2}-5 \mathrm{x}\left(\pi \mathrm{d}^{2} / 4 \mathrm{m}^{2}\right)=0.004 \times(3.14 \times 0.05 \times 0.05 / 4)[/tex]
Evaporation rate =[tex]7.85 \times 10^{-6} \mathrm{mol} / \mathrm{s}[/tex]
