For methane ( $a = 2.25 L^{2} atm mo l^{- 2}$ , $b = 0.0428 L mo l^{- 1}$ , $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ ) find the temperature in K (nearest integer) at which the attraction term $a / RT$ is exactly twice the excluded volume $b$ , so that the second virial coefficient equals $- b$ .

Numerical answer type, answer in K, tolerance 2.

Answer: 320 K

Set

a / RT = 2 b

, so

T = \frac{a}{2 R b} = \frac{T _{B}}{2}

T_{B} = \frac{2.25}{0.0821 \times 0.0428} = 640.3 K

, so

T = 320.2 K

, i.e.

320 K