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 slope of the plot of log(x/m) versus log P for the adsorption of a gas on a solid, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The relationship between the amount of substance adsorbed (x) per unit mass of the adsorbent (m) and the pressure (P) is given by the Prandtl's adsorption isotherm. This relationship can be expressed as:
\[ \frac{x}{m} \propto P^{\frac{1}{n}} \]
where \( n \) is a constant.

2. **Removing the Proportionality**:
To express this relationship in a more usable form, we introduce a constant \( k \):
\[ \frac{x}{m} = k P^{\frac{1}{n}} \]

3. **Taking Logarithms**:
We take the logarithm of both sides to linearize the equation:
\[ \log\left(\frac{x}{m}\right) = \log(k) + \frac{1}{n} \log(P) \]

4. **Identifying the Linear Form**:
The equation can be rearranged to fit the linear equation form \( y = mx + c \):
- Let \( y = \log\left(\frac{x}{m}\right) \)
- Let \( x = \log(P) \)
- The slope \( m \) of this linear equation is \( \frac{1}{n} \) and the intercept \( c \) is \( \log(k) \).

5. **Conclusion**:
Therefore, when plotting log(x/m) versus log(P), the slope of the line is:
\[ \text{slope} = \frac{1}{n} \]

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

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