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}

 

Answers (1)

\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}

Most Viewed Questions

Preparation Products

Knockout JEE Main (Six Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 9999/- ₹ 8499/-
Buy Now
Knockout JEE Main (Nine Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 13999/- ₹ 12499/-
Buy Now
Test Series JEE Main 2024

Chapter/Subject/Full Mock Tests for JEE Main, Personalized Performance Report, Weakness Sheet, Complete Answer Key,.

₹ 7999/- ₹ 4999/-
Buy Now
Knockout JEE Main (One Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 7999/- ₹ 4999/-
Buy Now
Knockout JEE Main (Twelve Months Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 19999/- ₹ 14499/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
 
Exams
Articles
Questions