A gas undergoes physical adsorption on a surface and follows the givn Freundlich adsoprtion isotherm equation `x/m= kp^(0.5)`. Adsorption of the gas increases with:

Options

A. 1.Decrease in p and decrease in T
B. 2.Increase in p and increase in T
C. 3.Increase in p and decrease in T
D. 4.Decrease in p and increase in T

Explanation

To solve the problem, we will analyze the given Freundlich adsorption isotherm equation and the factors affecting physical adsorption.

### Step-by-Step Solution:

1. **Understanding the Freundlich Adsorption Isotherm**:
The Freundlich adsorption isotherm is given by the equation:
\[ \frac{x}{m} = k p^{0.5} \]
where:
- \( \frac{x}{m} \) is the amount of gas adsorbed per unit mass of the adsorbent,
- \( k \) is a constant,
- \( p \) is the pressure of the gas.

2. **Analyzing the Relationship**:
From the equation, we can see that:
\[ \frac{x}{m} \propto p^{0.5} \]
This indicates that the amount of gas adsorbed (\( \frac{x}{m} \)) is directly proportional to the square root of the pressure (\( p^{0.5} \)).

3. **Effect of Increasing Pressure**:
- As pressure (\( p \)) increases, \( p^{0.5} \) also increases.
- Therefore, \( \frac{x}{m} \) will increase, meaning that the adsorption of the gas increases with an increase in pressure.

4. **Effect of Temperature**:
- Physical adsorption is an exothermic process, which means it releases heat (\( \Delta H < 0 \)).
- According to Le Chatelier's principle, if we increase the temperature of an exothermic process, the equilibrium will shift to favor the reactants (in this case, the gas in the gas phase), thus reducing the amount of gas adsorbed.
- Therefore, increasing the temperature will decrease the adsorption of the gas.

5. **Conclusion**:
- To maximize adsorption, we should increase the pressure and decrease the temperature.
- Thus, the correct answer is that adsorption of the gas increases with an increase in pressure and a decrease in temperature.

### Final Answer:
The adsorption of the gas increases with an increase in pressure and a decrease in temperature.