The second virial coefficient of a certain gas obeys $B = b - a / RT$ with $b = 0.0400 L mo l^{- 1}$ and $a = 1.60 L^{2} atm mo l^{- 2}$ ( $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ ). The temperature at which $B = + 0.0200 L mo l^{- 1}$ is closest to

Answer: C

b - a / RT = 0.0200

gives

a / RT = 0.0200

, so

T = \frac{a}{0.0200 R} = \frac{1.60}{0.0200 \times 0.0821} = 974 K

, i.e. about

974 K

, twice the Boyle temperature (

487 K

), where

B

reaches half of

b