Question

An acetone-nitrogen mixture at 800 mm Hg and 30 ° C has a relative saturation of 80%. Calculate: a) Molar humidity. b) Absolute humidity. c) The percentage relative humidity. d) The specific volume. e) The mass of acetone contained in 1m '^ 3 of mixture.

Answer 1

Given that Total Pressure Pt = 800 mm ing Temparature (T)= 30°C Now Vapour pressure of acetone is given by en plat 2975.95 = 14.7171 T-34.5228 where psat in vapour pressure (kpa) T = Temparature is ink. T=30+27= 303k. in sat Pro 2975.95 = 14.7171 - 303-34.5228 Astes sowing Ipsat 37.89 kPa A vapour pressure of acetone in niteagers mixture atzo * Total pressure R = 800mming latungs 760 mph Hgh lata 101.325 kPa х f = 106.65 kPa.

Given that Total Pressure Pt = 800 mm ing Temparature (T)= 30°C Now Vapour pressure of acetone is given by en plat 2975.95 = 14.7171 T-34.5228 where psat in vapour pressure (kpa) T = Temparature is ink. T=30+27= 303k. in sat Pro 2975.95 = 14.7171 - 303-34.5228 Astes sowing Ipsat 37.89 kPa A vapour pressure of acetone in niteagers mixture atzo * Total pressure R = 800mming latungs 760 mph Hgh lata 101.325 kPa х f = 106.65 kPa.

a) Molar Hundry at saturation 37.89 Pasat Ys Pr - pasat 106.65-37.89 YS=0.551 kmol acetone vapour kinol Niteogeni Now percentage saturation Humidity Humidity at saturation X100 80 = у Ys X100 y = 80 XYs= 80 Xo.55) Tod 100 Molas Hunctity :. Y=o.uu Kind acetone vapour Kinol Nitergen b) Absolute Humidity Y"=Y MA MB where MA- Molecular weight of acetone MB molecular weight of An Nileogen ya 58 = o 44Х 28 y"=0.91 kgocetone kg N₂

Now Pastral pressure of acetone & Y= PA this P-PA After arranging gives - PA YPT Y+I 0.44 PA SISAL. X 106.65 0.444 +1. 32.58 PA=828kpa PA X100 c) Relative Humschitis (RA)= Pe Sat R.H= 3258 37.89 X100 = 85-98 . Ra T specific volune o v= RI P sat 8.314 X 303 m3 Va kg day No 106.65 - 37.89 V= 36.63 m Kg deg lail (N2) *9=1 ㅗ Kg 36.63 mo = 0.027 Kg m3 e) =im3 mass=Sxu= 0.027 kg mass of acetone