For nitrogen $a = 1.39 L^{2} atm mo l^{- 2}$ and $b = 0.0391 L mo l^{- 1}$ . Its Boyle temperature in K (nearest integer), using $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , is

Numerical answer type, answer in K, tolerance 2.

Answer: 433 K

T_{B} = \frac{a}{R b} = \frac{1.39}{0.0821 \times 0.0391} = \frac{1.39}{0.003210} = 433.0 K

, i.e.

433 K

. Nitrogen at room temperature is below

T_{B}

, so

Z

dips below 1 at moderate pressures.