Questions: Virial Coefficients and Intermolecular Forces

Question 1

At a temperature below the Boyle temperature, a gas has B₂ < 0. What does this tell you about the gas at that temperature?

• A. The hard-core repulsion dominates, so the gas is harder to compress than ideal
• B. Attractive interactions dominate, so the gas is easier to compress than ideal
• C. The gas behaves exactly as an ideal gas because B₂ is small
• D. The Mayer f-function is zero throughout the integration range

Answer: (B)

B₂ < 0 means the negative (attractive) contributions to the Mayer f-function integral outweigh the positive (hard-core exclusion) contributions. Attractive forces pull molecules together, increasing effective density and reducing pressure below the ideal value — the gas is easier to compress than ideal. B₂ > 0 (not <0) indicates hard-core repulsion dominates. At the Boyle temperature, B₂ = 0 exactly, meaning the two effects cancel.

Question 2

Near the hard-core region of a pair potential where u(r) → +∞, what is the value of the Mayer f-function [e^{−u/kT} − 1]?

• A. +1, because the repulsive potential is large and positive
• B. 0, because the molecules are far apart and do not interact
• C. −1, because the Boltzmann factor e^{−u/kT} → 0 when u → +∞
• D. Undefined, because u(r) diverges

Answer: (C)

When u(r) → +∞ (hard-core repulsion), the Boltzmann factor e^{−u/kT} → 0. Therefore the Mayer f-function = 0 − 1 = −1. This negative contribution reflects excluded volume: two molecules cannot overlap, so this region of space is unavailable, reducing the effective density. The f-function vanishes (= 0) at large separations where u(r) → 0, not in the hard-core region.

Question 3

A positive second virial coefficient B₂ means the gas exerts less pressure than an ideal gas at the same conditions.

• T. True
• F. False

Answer: False

B₂ > 0 means the hard-core exclusion dominates: molecules take up space, reducing the volume available to others, which drives pressure above the ideal gas prediction. A negative B₂ indicates attractions dominate and the gas is easier to compress (lower pressure) than ideal. High-temperature gases typically have B₂ > 0 because thermal energy makes the attractive well irrelevant and only excluded volume matters.

Question 4

At the Boyle temperature, a real gas behaves nearly ideally because molecular interactions essentially vanish.

• T. True
• F. False

Answer: False

At the Boyle temperature, B₂ = 0, but this does not mean interactions are absent. It means the positive (excluded volume) and negative (attractive) contributions to the Mayer f-function integral cancel exactly. The molecules are still interacting — both attraction and repulsion are present — but their effects on the equation of state happen to cancel at this temperature, producing near-ideal behavior despite real intermolecular forces.

Question 5

Explain physically why the sign of B₂ changes from positive to negative as temperature decreases, referencing the Mayer f-function.

Model answer: At high temperature, kT is much larger than the depth of the attractive well, so the Boltzmann factor in the attractive region is only slightly above 1 (small positive f), while the excluded-volume region still contributes −1 over its volume. The hard core dominates and B₂ > 0. As temperature decreases, kT becomes comparable to the well depth; the Boltzmann factor in the attractive region grows significantly above 1 (large positive f contribution), eventually overwhelming the excluded-volume term. The net integral goes negative, giving B₂ < 0.

The Mayer f-function captures two competing effects: excluded volume (f = −1 at the hard core) and attraction (f > 0 in the attractive well). Temperature controls their relative weight because the Boltzmann factor e^{−u/kT} is temperature-sensitive in the attractive region but not in the hard-core region (where it is always ≈0). The Boyle temperature marks the crossover where these contributions exactly balance.