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In an adsorption experiment, a graph between log (x/m) versus log P was found to be linear with a slope of 45°. The intercept on the y axis was found to be 0.301 . Calculate the amount of the gas adsorbed per gram of charcoal under a pressure of 3.0 atm.

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Detailed Solution

$\log \frac{x}{m} = \log K + \frac{1}{n} \log P$

$∴ Plot of \log \frac{x}{m} versus \log P is linear with slope = \frac{1}{n} and$ intercept - log k

$Thus \frac{1}{n} = \tan θ = \tan 45^{\circ} = 1 or n = 1$

Log K= 0.301 or K= antilog 0.301 = 2
At P=3 atm

$\frac{x}{m} = K P^{1 / n} = 2 \times (3)^{1} = 6$

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