The graph between log Xm v/s log P is straight-line inclined at an angle 45o with intercept 0.30. What will be the rate of adsorption at pressure of 0.4 atm?
0.4
0.6
0.8
0.9
A
0.6
B
0.8
C
0.9
D
0.4
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Solution
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Slope in equation is 1n = tan45o =1
Intercept is logk = 0.30, that implies k=2
Hence by equation xm=kp1/n
k=2, n=1 and p=0.4
∴ Rate of adsorption, x/m is 0.8.
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