Van der Waal's equation for a gas is stated as, $$P = \frac{nRT}{V - nb} - a\left(\frac{n}{V}\right)^2$$. This equation reduces to the perfect gas equation, $$P = \frac{nRT}{V}$$ When,

The Van der Waals equation is given as:

$$P = \frac{nRT}{V - nb} - a\left(\frac{n}{V}\right)^2$$

We need this equation to reduce to the perfect gas equation:

$$P = \frac{nRT}{V}$$

For this reduction to occur, the corrections introduced by the Van der Waals constants $$a$$ and $$b$$ must become negligible. Let's examine each term.

First, consider the volume correction term $$\frac{nRT}{V - nb}$$. In the perfect gas equation, the volume $$V$$ is assumed to be much larger than the volume occupied by the gas molecules themselves. The term $$nb$$ represents the excluded volume due to the molecules. So, when $$V$$ is very large compared to $$nb$$, we can approximate:

$$V - nb \approx V$$

Substituting this approximation:

$$\frac{nRT}{V - nb} \approx \frac{nRT}{V}$$

This approximation holds when the pressure is low because low pressure corresponds to a large volume (from Boyle's law, $$P \propto \frac{1}{V}$$ at constant temperature).

Second, consider the intermolecular forces term $$-a\left(\frac{n}{V}\right)^2$$. This term accounts for the attractive forces between molecules. For this term to become negligible, its value must be very small. This happens when $$\frac{n}{V}$$ (the concentration of gas molecules) is small, meaning the gas is dilute. A small concentration occurs when the volume $$V$$ is large (low pressure) or when the temperature is high (because high kinetic energy reduces the effect of intermolecular forces).

Specifically:

$$a\left(\frac{n}{V}\right)^2 \rightarrow 0 \quad \text{as} \quad \frac{n}{V} \rightarrow 0$$

Thus, under low pressure (large $$V$$) and high temperature, this term approaches zero.

Combining both conditions:

• Low pressure ensures $$V$$ is large, making $$V \gg nb$$ and $$\frac{n}{V}$$ small.
• High temperature ensures that even if $$\frac{n}{V}$$ is not extremely small, the increased kinetic energy overcomes intermolecular forces, making the $$a$$ term negligible.

Therefore, substituting both approximations into the Van der Waals equation:

$$P \approx \frac{nRT}{V} - 0 = \frac{nRT}{V}$$

This matches the perfect gas equation.

Now, evaluating the options:

A. Temperature is sufficiently high and pressure is low: Matches our conditions.

B. Both temperature and pressure are very low: Low pressure helps, but low temperature increases intermolecular forces, so the $$a$$ term does not vanish.

C. Both temperature and pressure are very high: High temperature helps, but high pressure reduces volume ($$V$$ small), so $$V - nb$$ is not approximately $$V$$ and $$\frac{n}{V}$$ is large, making the $$a$$ term significant.

D. Temperature is sufficiently low and pressure is high: Low temperature increases intermolecular forces, and high pressure reduces volume, worsening both corrections.

Hence, the correct answer is Option A.