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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    Plot of\[log\text{ }x/m\]against log p is a straight line inclined at an angle of\[45{}^\circ \]. When the pressure is 0.5 arm and Freundlich parameter. \[k\]is 10, the amount of solute adsorbed per gram of adsorbent will be\[(log\text{ }5=0.6990)\]

    A) 1 g                                         

    B) 2 g

    C) 3 g                         

    D)        5 g

    E)  2.5 g

    Correct Answer: D

    Solution :

    Freundlich adsorption isotherm equation is \[\frac{x}{m}=k{{p}^{1/n}}\] On taking log both sides \[\log \frac{x}{m}=\log k+\frac{1}{n}\log \,p\] \[\log \frac{x}{m}=\log \,10+\frac{1}{n}\log \,0.5\] (\[\because \]\[slope=\frac{1}{n}=tan\,\theta =tan\text{ }45{}^\circ =1\]) \[\log \frac{x}{m}=1+\frac{1}{1}\log (5\times {{10}^{-1}})\] \[\log \frac{x}{m}=1-0.6990+1\] \[=0.6990\] \[\frac{x}{m}=5.00\] \[=5g.\]


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