Question

The compressibility factor for a van der Waals gas at high pressure is

• A. \(1 + \frac{RT}{Pb}\)
• B. \(1 + \frac{Pb}{RT}\)
• C. \(1 - \frac{Pb}{RT}\)
• D. \(\frac{1}{16}\)

Answer 1

The Correct Option is B

The compressibility factor (Z) is defined as:

\( Z = \frac{PV}{RT} \)

For a van der Waals gas, the equation of state is:

\( \left(P + \frac{a}{V^2}\right)(V - b) = RT \)

At high pressures, the volume V decreases, making the term b (excluded volume) significant. The term involving a (intermolecular attraction) becomes negligible. Simplifying for high pressure:

\( Z \approx 1 + \frac{Pb}{RT} \)

Thus, the compressibility factor is approximately \( 1 + \frac{Pb}{RT} \) under high-pressure conditions.

Answer 2

Correct Answer:

Option 2: 1 + Pb/RT

Explanation:

Van der Waals Equation:

(P + a(n/V)²) (V - nb) = nRT

Compressibility Factor (Z):

Z = PV / nRT

High-Pressure Approximation:

At high pressure, V is small, and 'nb' is significant. 'a(n/V)²' becomes negligible compared to P.

Therefore, P(V - nb) ≈ nRT

PV - Pnb ≈ nRT

PV ≈ nRT + Pnb

PV / nRT ≈ 1 + Pnb / nRT

Z ≈ 1 + Pb / RT