For oxygen $a = 1.36 L^{2} atm mo l^{- 2}$ and $b = 0.0318 L mo l^{- 1}$ . Its inversion temperature $T_{i} = 2 a / R b$ in K (nearest integer, $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ ) is

Numerical answer type, answer in K, tolerance 3.

Answer: 1042 K

T_{i} = \frac{2 a}{R b} = \frac{2 \times 1.36}{0.0821 \times 0.0318} = \frac{2.72}{0.002611} = 1041.8 K

, i.e.

1042 K

. Oxygen at room temperature is far below

T_{i}

and cools on throttling.