Question

At \(T\)(K), adsorption of gas \((A)\) on solid adsorbent \((B)\) follows Freundlich adsorption isotherm. On \(10\) g of \(B\), adsorption of gas \((A)\) gave the isotherm shown below. What is \(x\) (the quantity of \(A\) adsorbed in one gram of \(B\)) when the pressure of \(A\) is \(1.259\) atm?

\[ (\log 1.259=0.1,\qquad \text{Antilog}(0.1)=1.259) \]

• A. \(1.259\)
• B. \(12.59\)
• C. \(0.1259\)
• D. \(0.1\)

Hint:

Freundlich adsorption isotherm is \[ \boxed{ \frac{x}{m}=kP^{1/n} } \] or \[ \boxed{ \log\left(\frac{x}{m}\right)=\log k+\frac1n\log P. } \] The slope of the graph gives \[ \boxed{\frac1n}. \]

Answer 1

The Correct Option is B

Step 1: Write Freundlich adsorption isotherm. Freundlich adsorption isotherm is \[ \frac{x}{m}=kP^{1/n}. \] Taking logarithm, \[ \log\left(\frac{x}{m}\right) = \log k+\frac1n\log P. \] Hence, a graph of \[ \log\left(\frac{x}{m}\right) \quad \text{vs} \quad \log P \] is a straight line.

Step 2: Obtain the equation of the line. From the graph, \[ (\log P,\log(x/m)) =(0.1,0.1) \] and \[ (1,1). \] Thus, the straight line is \[ \log\left(\frac{x}{m}\right)=\log P. \] Hence, \[ \frac{x}{m}=P. \]

Step 3: Calculate the adsorption. For \[ P=1.259\ \text{atm}, \] \[ \frac{x}{m}=1.259. \] Since adsorption was measured on \[ 10\ \text{g} \] of adsorbent, \[ x=10\times1.259=12.59. \] Hence, \[ \boxed{12.59}. \] Thus, \[ \boxed{(B)} \] is the correct answer.