$\text{CH}_4$ is adsorbed on 1 g charcoal at $0^\circ\text{C}$ following the Freundlich adsorption isotherm. 10.0 mL of $\text{CH}_4$ is adsorbed at 100 mm of Hg, whereas 15.0 mL is adsorbed at 200 mm of Hg. The volume of $\text{CH}_4$ adsorbed at 300 mm of Hg is $10^x$ mL. The value of $x$ is ________ $\times 10^{-2}$. (Nearest integer) [Use $\log_{10} 2 = 0.3010, \log_{10} 3 = 0.4771$]
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In numerical adsorption problems, taking the ratio of two different conditions eliminates the constant $k$, allowing you to find the exponent $1/n$ easily.
Step 1: Understanding the Concept:
The Freundlich adsorption isotherm relates the amount of gas adsorbed to the pressure of the gas at a constant temperature: $\frac{x}{m} = k P^{1/n}$. Since the mass of charcoal is constant, we can use volume ($V$) as proportional to the amount adsorbed ($x$). Step 2: Detailed Explanation:
1. Let the relation be $V = k P^{1/n}$.
- At 100 mm: $10 = k (100)^{1/n} \dots (\text{i})$
- At 200 mm: $15 = k (200)^{1/n} \dots (\text{ii})$
2. Divide (ii) by (i):
\[ 1.5 = (2)^{1/n} \implies \log 1.5 = \frac{1}{n} \log 2 \]
\[ \log(3/2) = \log 3 - \log 2 = 0.4771 - 0.3010 = 0.1761 \]
\[ 0.1761 = \frac{0.3010}{n} \implies \frac{1}{n} = \frac{0.1761}{0.3010} \approx 0.585 \]
3. Find volume at 300 mm:
\[ V_{300} = k (300)^{1/n} \]
Divide by (i): $\frac{V_{300}}{10} = (300/100)^{1/n} = (3)^{1/n}$
\[ \log(V_{300}/10) = \frac{1}{n} \log 3 = 0.585 \times 0.4771 \approx 0.2791 \]
\[ \log V_{300} - \log 10 = 0.2791 \implies \log V_{300} = 1 + 0.2791 = 1.2791 \]
4. Since $V = 10^x$, then $x = \log V$.
\[ x = 1.2791 \]
5. Express as $x \times 10^{-2}$:
\[ 1.2791 = 127.91 \times 10^{-2} \implies x \approx 128 \] Step 3: Final Answer:
The value of $x$ is 128.
