Q25 / 160 · TS EAMCET 2024 (09 May Shift 1) · Chemistry

Adsorption of a gas on an adsorbent follows Freundlich adsorption isotherm. The slope and intercept (on $y$-axis) of the isotherm are 0.5 and 1.0 respectively. What is the value of $\frac{x}{m}$, when the pressure of the gas $(\mathrm{A})$ is 100 atm ?

  • A
    10
  • B
    1
  • Correct answer
    C
    100
  • D
    1000

Solution

Freundlich isotherm. $\begin{aligned} & \ln \frac{x}{m}=\ln k+\frac{1}{n} \ln p \\ & \text { or, } \log \frac{x}{m}=\log k+\frac{1}{n} \log p \\ & y=c+m x \\ & \text { Slope }(m)=\frac{1}{n} \quad \therefore \quad \frac{1}{n}=0.5 \\ & \text { Intercept }(\mathrm{c})=\log k=1 \\ & \text { Pressure }(p)=100 \text { atm } \\ & \therefore \log \frac{x}{m}=1+0.5 \log 10^2 \\ & \log \frac{x}{m}=1+0.5 \times 2 \Rightarrow \log \frac{x}{m}=2 \\ & \frac{x}{m}=10^2=100\end{aligned}$

Hence, the correct answer is option: 'C'.

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