Difficulty: Easy
Correct Answer: Boyle’s temperature
Explanation:
Introduction / Context:
Real gases deviate from ideal behavior due to intermolecular forces and finite molecular size. However, at a particular temperature characteristic of each gas, second-order deviations in the pressure–volume relation vanish over a pressure range, and the gas approximately satisfies Boyle’s law (PV ≈ constant). This special temperature is known as the Boyle’s temperature.
Given Data / Assumptions:
Concept / Approach:
From the virial expansion, Z = 1 + B(T)P/(RT) + … . If B(T) = 0, then Z ≈ 1 to first order in P, meaning the gas follows Boyle’s law over a finite pressure interval. This temperature, where attractive and repulsive contributions balance such that B(T) vanishes, is the Boyle’s temperature. It is distinct from inversion temperature (related to the sign change in Joule–Thomson coefficient) and from phase-equilibrium points like the triple point or eutectic point.
Step-by-Step Solution:
Verification / Alternative check:
Experimental P–V–T data show near-ideal compressibility at the Boyle’s temperature for many gases (e.g., N2), consistent with B(T_B) ≈ 0.
Why Other Options Are Wrong:
Common Pitfalls:
Confusing Boyle’s temperature with inversion or critical temperatures due to similar-sounding “special” temperatures.
Final Answer:
Boyle’s temperature
Discussion & Comments