For methane $a = 2.25 L^{2} atm mo l^{- 2}$ and $b = 0.0428 L mo l^{- 1}$ . Its critical temperature in K (nearest integer), with $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , is

Numerical answer type, answer in K, tolerance 2.

Answer: 190 K

T_{c} = \frac{8 a}{27 R b} = \frac{8 \times 2.25}{27 \times 0.0821 \times 0.0428} = \frac{18.0}{0.09487} = 189.7 K

, i.e.

190 K

. The experimental value is about

191 K