N a dry cleaning establishment warm dry air is blown through a revolving drum in which clothes are t

Basis: 1 lb air

If the air is saturated

P*C8H18 = 2.36 in.Hg

P*air = 29.66 - 2.36 = 27.3 in.Hg

P*C8H18 / P*air = 2.36 in.Hg/ 27.3 in.Hg = 0.0864 mol C8H18/ mol air

a) lb air/ lb C8H18 = (27.30 mol air/ 2.36 mol C8H18) x (29 lb air/ 1 lbmol air) x ( 1 lbmol C8H18/ 114.2 lb C8H18) = 2.94 lb air/ lbC8H18

b) 2.36 mol C8H18/ (27.3 + 2.36) total mol = nC8H18/ ntotal = 0.08 or 8%

c) 2.94 lb air/ 1 lb C8H18 = (lbmol air/ 29 lb air) x (359 ft3 at SC/ 1 lbmol air) x (580 oR/ 492 oR) x (29.92 in.Hg/ 29.66 in.Hg) = 43.3 ft3 air/ lb C8H18.

Detailed solutions for part c:

\(n^o=\ \frac{T_s}{T}\ X\ \frac{P}{P_S}\ x\ \frac{V}{V_S}\)

\(V\ =\ n^{o\ }x\ \frac{T}{T_S}x\ \frac{P_S}{P}\ x\ \frac{V_S}{1}\ \)

T(oR) = To(F) + 460 = 120oF + 460 = 580 oR

TSo(R) = 32oF + 460 = 492oR

PS = 1 atm = 29.921 in.Hg

\(V_s\ =\ \frac{22.4\ L}{1\ \frac{g}{mol}}\ x\ \frac{453.593\ gmol}{1\ lbmol}\ x\ \frac{1\ ft^3}{28.317\ L}=\ 358.8121\ =\ 359\ \frac{ft^3}{1\ lbmol\ air}\)

\(V\ =\ \frac{2.94\ lb\ air}{1\ lb\ C_8H_{18}}\ x\ \frac{1\ lbmol\ air}{29\ lb\ air}\ x\ \frac{359\ ft^3}{1\ lbmol\ air}\ x\ \frac{580^oR}{492^0R}x\ \frac{29.921\ in.Hg}{29.66\ in.Hg}\ =\ 43.3\ \frac{ft^3air}{\ lb\ C_8H_{18}}\)