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Help me please, - Surface Chemistry - JEE Main

Adsorption of a gas follows Freundlich adsorption isotherm. x is the mass of the gas adsorbed on mass m of the adsorbent. The plot of \log\frac{x}{m} versus \log P is shown in the given graph. \frac{x}{m} is proportional to :


 

  • Option 1)

    P^{2/3}

  • Option 2)

    P^{3}

  • Option 3)

    P^{2}

     

  • Option 4)

    P^{3/2}

 
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\frac{x}{m}\; \alpha \; \; P^{1/x}\; \; \; \; \; \; \; \; \; Let\; 1/x=a

\frac{x}{m}=kp^{1/x}\rightarrow \therefore \frac{x}{m}=kp^{a}

\log\frac{x}{m}=\log k +\frac{1}{x}\log p

slope=\frac{1}{x}=a=\frac{2}{3}

\frac{x}{m}\; \alpha \; p^{2/3}


Option 1)

P^{2/3}

Option 2)

P^{3}

Option 3)

P^{2}

 

Option 4)

P^{3/2}

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