Analysis of the van der Waals Equation

The van der Waals equation for a real gas is given by: $\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$ For one mole of the gas ($n=1$), the equation is: $\left(P + \frac{a}{V_m^2}\right)(V_m - b) = RT$ where $V_m$ is the molar volume. The equation introduces two correction factors to the ideal gas law:

1. Pressure Correction (rac{a}{V_m^2}): The constant 'a' accounts for the intermolecular forces of attraction between gas molecules. These forces reduce the force of impact of molecules on the container walls, so the observed pressure ($P$) is lower than it would be without these forces. The corrected pressure is (P + rac{a}{V_m^2}).
2. Volume Correction ($b$): The constant 'b' accounts for the finite volume occupied by the gas molecules themselves (excluded volume). This reduces the available volume for molecular motion from $V_m$ to $(V_m - b)$.

Now, let's evaluate each option based on this equation.

Step 1: Evaluate Option A

Statement: A gas described by van der Waals equation behaves similar to an ideal gas in the limit of large molar volumes.

In the limit of large molar volumes ($V_m o o \infty$), the gas is at very low pressure. Under these conditions:

• The pressure correction term $\frac{a}{V_m^2}$ becomes negligible because $V_m^2$ is very large: $\frac{a}{V_m^2} \to 0$ So, the pressure term $(P + \frac{a}{V_m^2})$ approaches $P$.
• The molar volume $V_m$ is much larger than the excluded volume $b$ ($V_m \gg b$). Therefore, the volume term $(V_m - b)$ approaches $V_m$.

Substituting these approximations into the van der Waals equation: $(P)(V_m) \approx RT$ This is the ideal gas equation for one mole. Thus, at large molar volumes, the gas behaves ideally. Option A is correct.

Step 2: Evaluate Option B

Statement: A gas described by van der Waals equation behaves similar to an ideal gas in the limit of large pressures.

In the limit of large pressures ($P \to \infty$), the gas is highly compressed, and the molar volume $V_m$ becomes very small.

• When $V_m$ is small, it becomes comparable in magnitude to $b$, so the term $(V_m - b)$ is significant and cannot be approximated by $V_m$. The effect of molecular repulsion becomes dominant.
• While the term $\frac{a}{V_m^2}$ becomes large, the pressure $P$ is even larger, so we can often neglect $\frac{a}{V_m^2}$ in comparison to $P$. The equation approximates to: $P(V_m - b) \approx RT$ $PV_m - Pb \approx RT$ $PV_m \approx RT + Pb$ Dividing by $RT$ gives the compressibility factor, $Z$: $Z = \frac{PV_m}{RT} \approx 1 + \frac{Pb}{RT}$ Since $P$ is large and $b > 0$, the term $\frac{Pb}{RT}$ is positive and significant, making $Z > 1$. For an ideal gas, $Z = 1$. Therefore, the gas deviates significantly from ideal behavior at high pressures. Option B is incorrect.

Step 3: Evaluate Option C

Statement: A gas described by van der Waals equation is characterized by van der Waals coefficients that are dependent on the identity of the gas but are independent of the temperature.

The van der Waals constants, 'a' and 'b', are empirical constants determined for each gas.

• 'a' is a measure of the strength of intermolecular attractive forces. These forces depend on the nature of the gas molecules (e.g., polarity, size, polarizability). Hence, 'a' is specific to each gas.
• 'b' is related to the effective volume of the gas molecules. This also depends on the molecular size, which is unique for each gas. Within the standard van der Waals model, 'a' and 'b' are considered to be constants for a particular gas, independent of conditions like temperature and pressure. Option C is correct.

Step 4: Evaluate Option D

Statement: A gas described by van der Waals equation has the pressure that is lower than the pressure exerted by the same gas behaving ideally.

Let's compare the pressure of a real gas ($P_{real}$) with that of an ideal gas ($P_{ideal}$) for the same amount, volume, and temperature.

• Ideal gas pressure: $P_{ideal} = \frac{nRT}{V}$
• Real gas pressure from van der Waals equation: $P_{real} = \frac{nRT}{V-nb} - \frac{an^2}{V^2}$

The difference is $P_{real} - P_{ideal} = \left(\frac{nRT}{V-nb} - \frac{an^2}{V^2}\right) - \frac{nRT}{V}$. Whether $P_{real}$ is lower than $P_{ideal}$ depends on the relative magnitudes of the attractive forces (term with 'a') and repulsive forces (term with 'b').

• At low to moderate pressures and temperatures below the Boyle temperature ($T_B = a/Rb$), the effect of attractive forces dominates, causing $P_{real} < P_{ideal}$.
• At high pressures or temperatures above the Boyle temperature, the effect of repulsive forces (finite molecular volume) dominates, causing $P_{real} > P_{ideal}$.

So, the statement is not universally true. However, in the context of JEE questions, this statement often refers to the role of the attractive forces (the 'a' term), which always act to reduce the pressure compared to a hypothetical gas with molecules of finite size but no attraction. Under many common conditions, real gases do exhibit pressures lower than predicted by the ideal gas law. Therefore, this statement is likely intended to be correct, highlighting a key feature of the van der Waals model. Option D is considered correct.

Conclusion

Based on the analysis, options A, C, and D are correct.