The Boyle temperature TB = a/(Rb). Given a = 0.5 Pa·m⁶/mol², b = 0.03 m³/mol, R = 8.314 J/mol·K:

TB = 0.5 / (8.314 × 0.03) = 0.5 / 0.24942 ≈ 200.2 K.

The van der Waals equation is (P + a/V²)(V – b) = RT. Given P = 10⁵ Pa, T = 300 K, R = 8.314 J/mol·K, a = 0.4 Pa·m⁶/mol², b = 0.02 m³/mol:

Solve (10⁵ + 0.4/V²)(V – 0.02) = 8.314 × 300 = 2494.2. Assume V ≈ RT/P (ideal gas approximation) = 2494.2 / 10⁵ = 0.02494 m³. Iterating:

(10⁵ + 0.4/0.02494²)(0.02494 – 0.02) ≈ (10⁵ + 643)(0.00494) ≈ 2494, which is close. Thus, V ≈ 0.0247 m³.

The Boyle temperature is the temperature at which a real gas behaves ideally at low pressures, where the second virial coefficient B = 0. For a van der Waals gas, TB = a/(Rb).

Using Pc = a/(27b²) and Tc = 8a/(27Rb), first find b: Tc = 8a/(27Rb) → b = 8a/(27RT_c). Substitute into Pc:

Pc = a / [27 (8a/(27RT_c))²]. Simplify: Pc = (27 R² T_c²) / (64 a). Thus, a = (27 R² T_c²) / (64 Pc). Given R = 8.314 J/mol·K, Tc = 150 K, Pc = 50 × 10⁵ Pa:

a = (27 × 8.314² × 150²) / (64 × 50 × 10⁵) ≈ 2.25 Pa·m⁶/mol².

: The critical volume Vc = 3b. Given b = 0.0318 m³/mol:

Vc = 3 × 0.0318 = 0.0954 m³/mol.

At the critical point, Zc = Pc Vc / (RTc). Substituting Pc = a/(27b²), Vc = 3b, Tc = 8a/(27Rb):

Zc = (a/(27b²) × 3b) / (R × 8a/(27Rb)) = (3a/(27b)) / (8a/(27b)) = 3/8 = 0.375.

 Using the critical point conditions for the van der Waals equation, the critical pressure is Pc = a/(27b²), derived alongside Vc = 3b and Tc = 8a/(27Rb).

At the critical point, (∂P/∂V)T = 0 and (∂²P/∂V²)T = 0. Solving for the van der Waals equation gives Tc = 8a/(27Rb), where a, b are van der Waals constants and R is the gas constant.

 The constant ‘b’ is the effective volume occupied by gas molecules per mole, reducing the available volume for gas motion to (V – b).

 The van der Waals equation for one mole is (P + a/V²)(V – b) = RT, where ‘a’ corrects for intermolecular attractive forces and ‘b’ accounts for the finite volume of molecules.