CH 4 is adsorbed on 1 g charcoal at 0 ° C following the Freundlich adsorption isotherm. 10 . 0 mL of CH 4 is adsorbed at 100 mm of Hg , whereas 15 . 0 mL is adsorbed at 200 mm of Hg . The volume of CH 4 adsorbed at 300 mm of Hg is 10 x mL . The value of x is _________ × 10 - 2 . (Nearest integer) [Use log 10 2 = 0 . 3010 , log 10 3 = 0 . 4771 ]

Step-by-step solution

x m = Kp 1 n x ∝ V   for   the   same   gas 10 1 = K 100 1 n ...(1) 15 1 = K 200 1 n ....(2) V 1 = K 300 1 n ...(3) Divide (2) to (1) = 15 10 = 2 1 n log 3 2 = 1 n log 2 1 n = 0 . 4771 - 0 . 3010 0 . 3010 = 0 . 585 Divide (3) to (1) = V 10 = 3 1 n log V 10 = 1 n log 3 log V 10 = 0 . 585 × 0 . 4771 = 0 . 2791 V 10 = 10 0 . 2791 V = 10 × 10 0 . 2791 = 10 1 . 2791 = 10 x x = 1 . 2791 = 127 . 91 × 10 - 2 ≈ 128 × 10 - 2