For the adsorption of a gas on a solid, the plot of log(x/m) versus log p is linear with slope equal to

Options

A. k
B. logk
C. n
D. `1//n`

Answer

To solve the question regarding the adsorption of a gas on a solid and the relationship between log(x/m) and log(p), we will use Freundlich's adsorption isotherm.

### Step-by-Step Solution:

1. **Understanding Freundlich's Isotherm**:
The Freundlich adsorption isotherm is given by the equation:
\[ \frac{x}{m} = k \cdot p^{\frac{1}{n}} \]
where \(x\) is the amount of gas adsorbed, \(m\) is the mass of the solid, \(p\) is the pressure of the gas, \(k\) is a constant, and \(n\) is a constant that indicates the adsorption intensity.

2. **Taking the Logarithm of Both Sides**:
To analyze the relationship, we take the logarithm of both sides:
\[ \log\left(\frac{x}{m}\right) = \log(k \cdot p^{\frac{1}{n}}) \]

3. **Applying Logarithmic Properties**:
Using the property of logarithms that states \(\log(ab) = \log a + \log b\) and \(\log(a^b) = b \cdot \log a\), we can rewrite the equation as:
\[ \log\left(\frac{x}{m}\right) = \log k + \frac{1}{n} \log p \]

4. **Identifying the Linear Relationship**:
The equation can be rearranged to fit the linear form \(y = mx + c\):
\[ \log\left(\frac{x}{m}\right) = \frac{1}{n} \log p + \log k \]
Here, \(y = \log\left(\frac{x}{m}\right)\), \(x = \log p\), the slope \(m = \frac{1}{n}\), and the y-intercept \(c = \log k\).

5. **Conclusion**:
From the analysis, we find that the slope of the plot of \(\log\left(\frac{x}{m}\right)\) versus \(\log p\) is equal to \(\frac{1}{n}\).

### Final Answer:
The slope of the plot of \(\log\left(\frac{x}{m}\right)\) versus \(\log p\) is equal to \(\frac{1}{n}\).