Questions: Second Virial Coefficient

5 questions to test your understanding

Score: 0 / 5
Question 1 Multiple Choice

A gas is studied at its Boyle temperature. A chemist concludes that 'the gas behaves ideally at this temperature because intermolecular forces cancel out.' What is wrong with this conclusion?

AThe Boyle temperature is where B₂ is at its maximum, not zero
BIntermolecular forces do not actually cancel — B₂ = 0 is a coincidental balance of repulsive and attractive contributions to the integral, not an absence of forces
CThe gas behaves ideally at all temperatures, not just the Boyle temperature
DAt the Boyle temperature, only repulsive forces remain, so the gas is not truly ideal
Question 2 Multiple Choice

At low temperatures, B₂(T) is negative. What physical process causes this?

AAt low temperatures, molecular velocities are too low for molecules to collide, so there is no repulsive contribution
BThe hard-core repulsion disappears at low temperatures because kT is small
CThe attractive potential well becomes significant relative to kT, causing molecules to linger near each other and reducing pressure below the ideal value
DLow temperatures reduce molecular volume, making the ideal gas approximation more accurate and B₂ smaller
Question 3 True / False

A gas with B₂ = 0 at its Boyle temperature has no intermolecular interactions at that temperature.

TTrue
FFalse
Question 4 True / False

At high temperatures, the second virial coefficient B₂(T) is typically positive because repulsive interactions dominate.

TTrue
FFalse
Question 5 Short Answer

Why is measuring B₂(T) at many temperatures more useful than a single measurement, and what can be learned from its temperature dependence?

Think about your answer, then reveal below.