Q2(a)(ii). Boyle temperature of a van der Waals gas

1. Definition. Write pVₘ=RT(1+B/Vₘ+…) or Z=1+B'p+…. At the Boyle temperature T_B the second virial coefficient B=0, so the slope of Z (or of pVₘ) against p is zero as p→0. Attractions and repulsions balance, and the gas behaves ideally over a wide range of pressure.
2. van der Waals equation: p=(RT)/(Vₘ-b)-aVₘ². Multiply by Vₘ/RT: Z=(pVₘ)/(RT)=(1)/(1-b/Vₘ)-(a)/(RTVₘ)
3. Expand for low pressure (large Vₘ, b/Vₘ≪1): (1)/(1-b/Vₘ)=1+(b)/(Vₘ)+b²Vₘ²+… Z=1+(b-a/RT)(1)/(Vₘ)+b²Vₘ²+…
4. Compare with the virial equation: B(T)=b-a/RT. The term b (repulsion) makes Z>1; the term a/RT (attraction) makes Z<1.
5. Set B=0 at T_B: b-(a)/(RT_B)=0, T_B=a/Rb
6. The same result from the slope. With 1/Vₘ≈ p/RT: pVₘ=RT+(b-a/RT)p+…, so [∂(pVₘ)/∂ p]_(T, p→0)=b-a/RT, which vanishes at T=a/Rb.
7. Below and above T_B: T<T_B: a/RT>b, B<0, Z<1 (attractions dominate, the gas is easier to compress). T>T_B: B>0, Z>1 (repulsions dominate). Since T_c=8a/27Rb, T_B=27/8T_c.
8. Numbers (van der Waals a, b): N₂, a=1.352 atm dm⁶ mol⁻², b=3.87×10⁻² dm³ mol⁻¹: T_B=1.352/(0.08206×0.0387)=426 K. Argon gives 509 K against the measured 411.5 K, and CO₂ 1025 K against 714.8 K: the van der Waals estimate is high, but the trend (stronger attraction, higher T_B) is right.

Application. near T_B a gas follows pV=nRT closely over a wide pressure range, so gas thermometers and molar-mass measurements by gas density are most accurate there; the large a of CO₂ explains its high Boyle temperature and its easy liquefaction.