$\text{CH}_4$ 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.
To solve this problem, we use the Freundlich adsorption isotherm, expressed as \( x = kP^n \), where \( x \) is the mass of the adsorbate, \( P \) is the pressure, and \( k \) and \( n \) are constants. Here, the volume of gas adsorbed is proportional to the mass, so we use this form effectively. Given the relation in log form: \[\log_{10}(x) = \log_{10}(k) + n\log_{10}(P)\] We have: 1. At 100 mm Hg, \( x_1 = 10 \) mL, so \(\log_{10}(10) = 1 = \log_{10}(k) + n\log_{10}(100)\). 2. At 200 mm Hg, \( x_2 = 15 \) mL, so \(\log_{10}(15) = \log_{10}(k) + n\log_{10}(200)\).
Subtract the first from the second: \[ \log_{10}(15) - 1 = n(\log_{10}(200) - \log_{10}(100))\] \[ \log_{10}(1.5) = n\log_{10}(2)\] Calculating \(\log_{10}(1.5)\): \(\log_{10}(1.5) = \log_{10}(3/2) = \log_{10}(3) - \log_{10}(2) = 0.4771 - 0.3010 = 0.1761\), so, \[ n = \frac{0.1761}{0.3010} \approx 0.585\]
Now compute \( \log_{10}(k) \) using the first equation: \[ 1 = \log_{10}(k) + 0.585 \times 2 \rightarrow \log_{10}(k) \approx 1 - 1.170 = -0.170 \rightarrow k \approx 10^{-0.170}\]
Predicting \( x \) at 300 mm Hg: \[ \log_{10}(x_3) = -0.170 + 0.585\log_{10}(300)\] \[ \log_{10}(300) = \log_{10}(3) + \log_{10}(100) = 0.4771 + 2 = 2.4771\] \[\log_{10}(x_3) = -0.170 + 0.585 \times 2.4771 = 1.2756\] \[ x_3 = 10^{1.2756} = 18.82\] Since \( 18.82 = 10^x \), solving gives \( x \approx 1.2756 \). Thus, \( x \approx 128 \times 10^{-2} \), confirming it lies within the range [128,128].
