Questions: Virial Expansion

Question 1

A gas has a negative second virial coefficient B₂(T). What does this imply about the compressibility factor Z = PV/NkT compared to the ideal gas?

• A. Z > 1, because the molecules repel each other and resist compression
• B. Z < 1, because attractive interactions cause molecules to cluster, reducing effective pressure
• C. Z = 1, because positive and negative contributions to B₂ always cancel
• D. Z < 1, because repulsive forces prevent the gas from expanding to fill its container

Answer: (B)

A negative B₂ means the attractive interactions dominate. The virial expansion gives Z = 1 + B₂ρ + ..., so B₂ < 0 implies Z < 1: the gas exerts less pressure than the ideal prediction. Attractive forces cause molecules to spend more time near each other rather than hitting the walls independently, reducing the apparent number of independently-acting particles. Option D confuses the mechanism — repulsive forces give B₂ > 0 and Z > 1.

Question 2

A gas is studied at its Boyle temperature, where B₂(T) = 0. What is the physical meaning of this condition?

• A. The gas has no intermolecular interactions whatsoever — it is a true ideal gas at this temperature
• B. The gas is at its critical temperature, above which it cannot be liquefied
• C. The attractive and repulsive contributions to pairwise interactions exactly cancel, producing approximately ideal behavior
• D. The density-dependent corrections diverge, making the virial expansion invalid

Answer: (C)

B₂(T) is the integral of the Mayer f-function over all pair separations, combining contributions from attractive regions (f < 0) and repulsive regions (f = −1 inside hard core, giving positive contribution). At the Boyle temperature these cancel, so the leading correction to ideal behavior vanishes. Critically, the gas still has interactions — it is not a 'true' ideal gas; at higher densities, B₃ and higher coefficients still contribute.

Question 3

The virial expansion reduces to the ideal gas law in the limit of zero density.

• T. True
• F. False

Answer: True

The virial expansion is Z = PV/NkT = 1 + B₂(T)ρ + B₃(T)ρ² + ..., where ρ is the number density. As ρ → 0, all correction terms vanish and Z → 1, recovering PV = NkT. This makes physical sense: at very low density, molecules almost never come close enough to interact, and the gas behaves ideally. The expansion is explicitly organized as a power series in density for precisely this reason.

Question 4

The ideal gas equation of state can be recovered from the virial expansion in the limit of very low temperature, where thermal energy is small compared to interaction energies.

• T. True
• F. False

Answer: False

The ideal gas limit is the low-density limit (ρ → 0), not the low-temperature limit. At low temperature, intermolecular interactions become *more* significant relative to thermal energy — B₂(T) can become large and negative as attraction dominates, and the gas may condense into a liquid. Low density is what suppresses interactions by keeping molecules far apart on average, regardless of temperature.

Question 5

Why does a negative second virial coefficient B₂(T) indicate that the net intermolecular interaction is attractive, and what happens physically to cause Z < 1?

Model answer: B₂(T) = −½ ∫ f(r) 4πr² dr, where f(r) = e^{−βu(r)} − 1. For an attractive potential, u(r) < 0 at intermediate separations, so e^{−βu(r)} > 1 and f(r) > 0 — but the integral carries a minus sign, making B₂ < 0. Physically, Z < 1 means the gas exerts less pressure than an equivalent ideal gas. This happens because attractive forces cause molecules to linger near each other rather than bouncing off walls independently; the effective number of 'free' particles hitting the walls is reduced.

The Mayer f-function is the key device: it is exactly zero for non-interacting molecules, capturing only the departure from ideal behavior. The sign of B₂ directly encodes whether the net effect is attraction (molecules cluster, Z < 1) or repulsion (molecules exclude volume, Z > 1). The Boyle temperature is where these effects cancel, and the van der Waals constants a and b can be derived directly from the components of this integral.