Question

For a real gas at 25°C temperature and high pressure (99 bar) the value of the compressibility factor is 2, so the value of van der Waal’s constant ‘b’ should be ______ × 10^–2 L mol^–1 (Nearest integer).
(Given R=0.083 L bar K-1 mol-1)

Answer 1

Correct Answer: 25

To solve for the van der Waal’s constant ‘b’ for a real gas, we begin with the compressibility factor (Z) equation for real gases:
Z = \( \frac{PV}{nRT} \)
Given:

• Temperature, T = 25°C = 298 K
• Pressure, P = 99 bar
• Compressibility factor, Z = 2
• Gas constant, R = 0.083 L bar K^–1 mol^–1

The Van der Waals equation relating Z and ‘b’ is:
Z = 1 + \( \frac{Pb}{RT} \)
Substituting the given values:
2 = 1 + \( \frac{99 \times b}{0.083 \times 298} \)
This simplifies to:
1 = \( \frac{99b}{24.734} \)
b = \( \frac{24.734}{99} \) L mol^–1
Calculate b:
b ≈ 0.2498 L mol^–1
We need to express b as × 10^–2 L mol^–1, so:
b ≈ 24.98 × 10^–2 L mol^–1
Rounded to the nearest integer, b = 25 × 10^–2 L mol^–1. This value falls within the expected range (25,25).