Question

A 1 litre vessel contains 2 moles of a vanderwaal's gas.

Given data: a = 2.5 atm-L² mole⁻² b = 0.4 L-mole⁻¹ T = 240 K RT = 20 L-atm mole⁻¹

Identify the correct options about the gas sample :

• A. Pressure of gas = 190 atm
• B. Compressibility factor = 4.75
• C. Attraction forces are dominant in the gaseous sample
• D. TB (Boyle temperature) = 75 K

Answer 1

Answer: (A), (B), (D)

(A), (B), and (D)

Answer 2

1. Calculate nRT: nRT = 2 mol * 20 L-atm/mol = 40 L-atm.
2. Calculate Pressure (P): Using Van der Waals equation, $(P + \frac{an^2}{V^2})(V - nb) = nRT$. $P = \frac{nRT}{V-nb} - \frac{an^2}{V^2} = \frac{40}{1 - (2 \times 0.4)} - \frac{2.5 \times 2^2}{1^2} = \frac{40}{0.2} - 10 = 200 - 10 = 190$ atm.
3. Calculate Compressibility Factor (Z): $Z = \frac{PV}{nRT} = \frac{190 \times 1}{40} = 4.75$.
4. Analyze forces: Since Z > 1, repulsive forces dominate.
5. Calculate Boyle Temperature (TB): $T_B = \frac{a}{Rb}$. R = RT/T = 20/240 = 1/12 L-atm/mol-K. $T_B = \frac{2.5}{(1/12) \times 0.4} = \frac{2.5 \times 12}{0.4} = 75$ K.