1. Diketahui : T = 298 K PHe1 = 0,060 atm x 1,013.105 = 0,06078.105 Pa PHe2 = 0,020 atm x 1,013.105 = 0,02026.105 Pa DAB = 0,687.10-4 m2/det Z = z2 – z1 = 0,1 m D = 5 mm P = 1,0 atm x 1,013.105 = 1,013.105 Pa

Ditanya : a. NHe dalam kgmol/det.m2 dan gmol/det.cm2= ..?

b. N(N2)= ..?

c. (PHe1 - PHe2) di titik 0,05 m =..?

Jawaban:

a. - NHe = DAB (PHe1 - PHe2) R.T(z2 – z1)

= 0,687.10−4m2/det x (0,06078.105−0,02026.105)Pa8314m3 . Pa/K .kgmol x298 K x 0,1m

= 0,2783724 Pa .m2/det247757,2m4 . Pa/kgmol

= 0,000001123 kgmol/det.m2

= 1,123 . 10-6 kgmol/det.m2

- NHe = 1,123 . 10-6 kgmol/det.m2 x 1000 : 10000 = 1,123 . 10-7 grmol/det.cm2

b. P(N2)1 = P - PHe1 = 1,013 . 105 Pa - 0,06078 . 105 Pa = 0,95222 . 105 Pa

P(N2)2 = P - PHe2 = 1,013 . 105 Pa - 0,02026.105 Pa = 0,99274 . 105 Pa

N(N2) = DABx {P (N 2) 1−P (N 2 )2}

R .T (z2 – z 1)

= 0,687.10−4m2/det x (0,95222.105Pa−0,99274. 105Pa)8314m3 .Pa /K .kgmol x298K x0,1m

= −0,2783724 Pa.m2/det247757,2m4 .Pa/kgmol

= - 0,000001123 kgmol/det.m2

= -1,123.10-6 kgmol/det.m2

c. Z = (z2 - z1) = 0,05 m

NHe = DAB x(PHe 1¿−PHe 2)

R .T (z2 – z 1)¿

(PHe1 - PHe2) = N He x Rx T x (z2−z1)

DAB

=1,123.10−6kgmol /det .m2 x 8314m3 . Pa/K .kgmol x298K x0,05m0,687.10−4m2/det

= 2024,97 Pa