Question

Adsorption of a gas (A) 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 (A) is 100 atm?

• A. 10
• B. 100
• C. 1000
• D. 10000

Hint:

For the Freundlich adsorption isotherm, the value of \( \frac{x}{m} \) can be calculated by using the slope and intercept from the graph of \( \log \left( \frac{x}{m} \right) \) vs. \( \log P \).

Answer 1

The Correct Option is C

To solve this problem, we will use the Freundlich adsorption isotherm, which is given by: \[ \frac{x}{m} = k \cdot P^{1/n} \] where: - \( \frac{x}{m} \) is the amount of gas adsorbed per unit mass of the adsorbent, - \( P \) is the pressure of the gas, - \( k \) and \( n \) are constants. Step 1: Freundlich Isotherm in Logarithmic Form The Freundlich isotherm can be expressed in logarithmic form as: \[ \log \left( \frac{x}{m} \right) = \log k + \frac{1}{n} \log P \] This is a linear equation of the form \( y = mx + c \), where: - \( y = \log \left( \frac{x}{m} \right) \), - \( m = \frac{1}{n} \) (slope), - \( c = \log k \) (intercept). Step 2: Given Data - Slope (\( \frac{1}{n} \)) = 0.5, - Intercept (\( \log k \)) = 1.0, - Pressure (\( P \)) = 100 atm. Step 3: Calculate \( n \) and \( k \) From the slope: \[ \frac{1}{n} = 0.5 \implies n = 2 \] From the intercept: \[ \log k = 1.0 \implies k = 10^1 = 10 \] Step 4: Calculate \( \frac{x}{m} \) Using the Freundlich isotherm: \[ \frac{x}{m} = k \cdot P^{1/n} \] Substitute the values: \[ \frac{x}{m} = 10 \cdot (100)^{1/2} \] Simplify: \[ \frac{x}{m} = 10 \cdot 10 = 100 \] Final Answer: The value of \( \frac{x}{m} \) is: \[ \boxed{100} \] This corresponds to option (2).