Question

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 ,

• A. temperature is sufficiently high and pressure is low
• B. temperature is sufficiently low and pressure is high
• C. both temperature and pressure are very high
• D. both temperature and pressure are very low

Answer 1

The Correct Option is A

Van der Waal's equation for a real gas is given by:

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

This equation accounts for the volume occupied by gas molecules and the intermolecular forces between them, which are neglected in the ideal gas law. The ideal gas equation is:

p = \frac{nRT}{V}

To understand when Van der Waal's equation reduces to the ideal gas law, let's consider the components:

• Repulsive Forces: The term {nb} adjusts the volume to account for the finite size of gas molecules. As pressure (p) is low, the volume is larger, implying the effect of {nb} is negligible relative to V.
• Attractive Forces: The term {a\left(\frac{n}{V}\right)^{2}} corrects for intermolecular attractions. When temperature is high, these attractions become negligible because the increased kinetic energy of molecules dominates these forces.

Thus, at high temperature, the term {a\left(\frac{n}{V}\right)^{2}} becomes negligible due to enhanced kinetic energy of molecules. Similarly, at low pressures, the {nb} term becomes negligible when compared to V.

Therefore, at sufficiently high temperatures and low pressures, the Van der Waal's equation simplifies to the ideal gas equation as both modifications become insignificant. This is why the correct answer is:

temperature is sufficiently high and pressure is low.