Question

EXERCISE 6: Two chemical species, A and B are separated by isothermal gas chromatography, using a 2.00 m column having 5000 theoretical plates at a flow rate of 15.0 ml / min. The peak of unabsorbed air appears after 30 s; the peak of A appears after 5 min and that of B after 12 min (a) Calculate the dead volume VM of the column, and the retention volumes Va and VB. (b) Calculate the reduced volumes V’a and V'B (c) Calculate the retention coefficients k’ and k', (d) What are the widths at the base of peaks A and B? (e) What is the value of H for this column? (f) Determine the value of the selectivity a of this separation (g) Calculate the resolution R of the separation

Answer 1

Here Column length , L= 2m Number of plates , N= 5000 The time void volume takes to pass through the column, to = 305 Retention time of A and B are, to = 5 min = 3005 to = 12 min = 7205 Flow rate, Fr = 15 ml 1 min = 15x10 3 L 60s = 0.25x 103 Lys a) Dead volume of the column, Vm= to fr = 30x 0.25 x 103 -3 = 7.5 * 10 L Retention volume of A, Vr = to Fr = 300 x 0.25 x 103 -2, = 7.5 x 10 Similarly retention volume of B, VoR = top fo = 720 x 0.25x10 = 18 × 10 2L

b) A and B. The reduced volumes of VA= (tox-to) Fo -3 - 3 = 67.5×10 L = 270 x 0.25x 10 v B = (top-to) For = 690+ 0.25 x 10 3 = 172.5x103 , c) Retention coefficients of A and B, KA = VrA-I = 7.5x 10 2 - 1 7.58 10 3 A VM - 9 l = - 1 18x10 2 7.5x10-3 - 3 VM KB = Voo- VM = 23 d) Peak width of A WA= 16x t and B . = 16x9x104 5x 103 N m = 288 = 16x toR 16 x 518400 5ooo WB = = 1658.88 m

e) Height equivalent of a theoretical plate, It= L = 2 = 4 x 10 4 N 5000 f) selectivity, a=t&g-to = ki, tra-to K'A = 23 = 2.56 g) Resolution, R = a (top-tra) WAT WB = 2 ( 720 - 300) 288 + 1658.88 - 04314