A membrane module performs air separation to obtain higher purity oxygen

permeate product. For simplicity, assume that the air is made of $35$ vol$\%$ O$2$ and $65$ vol$\%$ N$2.$ The air feed flow rate is $8,000$ cm$3($STP$)/$hr$.$ For this membrane module, which is $0\mathrm{.}0004$ cm in thickness, the permeability of O$2$ and N$2$ are determined to be $65\backslash$times $10^-10[$cm$3($STP$)$cm$/($cm$2$s$$cmHg$)]$ and $5\backslash$times $10^-10[$cm$3$

$($STP$)$cm$/($cm$2$s$$cmHg$)],$ respectively. The retentate side has a uniform pressure of $10\mathrm{.}5$ bar, the permeate side has a uniform pressure of $1\mathrm{.}1$ bar. Our goal is to produce a permeate stream containing $70$ vol$\%$ O$2.$ Conversion factor that might be useful: $1$ mol of ideal gas has $22\mathrm{.}4$ L at STP$.$

$($a$)$ Calculate the O$2$ selectivity of this membrane over N$2$ AND the N$2$ permeance of the membrane

$($b$)$ For the retentate stream, determine its oxygen composition $($in vol$\%)$

$($c$)$ What is the volumetric flow rate?

$($d$)$ Calculate the cut fraction.

$($e$)$ Determine the area of the membrane module.

$($f$)$ Recalculate the membrane area using the Rachford$-$Rice type equation, given that the retentate side pressure is doubled $($i$.$e$.$ Pr $=21$ bar$).$

$($g$)$ Based on the result in Part $($a$),$ qualitatively describe how and why increasing retentate side pressure affects the size of the membrane module.