In an adsorption experiment a graph between log `x/m` versus log `P` was fond to be linear with a slope of `45^@)`. The intercept on the log `x/m` axis was found to `0.70`. Calculate the amount of gas adsorbed per gram of charcoal under a pressure of `1` atm `[log5=0.70]`

Options

A. `5`
B. `2`
C. `3`
D. `0`

Explanation

To solve the problem step by step, we will use the Freundlich adsorption isotherm and the information provided in the question.

### Step 1: Understand the Freundlich Isotherm
The Freundlich isotherm is given by the equation:
\[ \frac{x}{m} = k P^{\frac{1}{n}} \]
where:
- \( \frac{x}{m} \) is the amount of gas adsorbed per gram of charcoal,
- \( P \) is the pressure of the gas,
- \( k \) is a constant,
- \( n \) is a constant that indicates the adsorption intensity.

### Step 2: Take the Logarithm
Taking the logarithm of both sides, we get:
\[ \log\left(\frac{x}{m}\right) = \log(k) + \frac{1}{n} \log(P) \]
This equation represents a straight line in the form \( y = mx + c \), where:
- \( y = \log\left(\frac{x}{m}\right) \),
- \( x = \log(P) \),
- \( m = \frac{1}{n} \) (the slope),
- \( c = \log(k) \) (the intercept).

### Step 3: Identify the Slope and Intercept
From the problem statement:
- The slope of the graph is given as \( 45^\circ \), which means \( \tan(45^\circ) = 1 \). Therefore, \( \frac{1}{n} = 1 \), leading to:
\[ n = 1 \]
- The intercept on the log \( \frac{x}{m} \) axis is \( 0.70 \), which means:
\[ \log(k) = 0.70 \]

### Step 4: Calculate \( k \)
To find \( k \), we take the antilogarithm of the intercept:
\[ k = 10^{0.70} \]
Using the approximation \( 10^{0.70} \approx 5 \), we find:
\[ k \approx 5 \]

### Step 5: Calculate \( \frac{x}{m} \) at \( P = 1 \) atm
Now we can substitute \( P = 1 \) atm into the Freundlich equation:
\[ \frac{x}{m} = k P^{\frac{1}{n}} = 5 \cdot 1^{1} = 5 \]

### Final Answer
The amount of gas adsorbed per gram of charcoal under a pressure of 1 atm is:
\[ \frac{x}{m} = 5 \text{ g/g charcoal} \]