The second virial coefficient of a gas is measured as $B = - 160 mL mo l^{- 1}$ at $250 K$ and $B = 0$ at $425 K$ . Assuming van der Waals behaviour, $B = b - a / RT$ , and $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ . Which statements are correct?

Answer: A, B, C

B = 0

defines the Boyle temperature, so

T_{B} = 425 K

and

a = R b T_{B}

(A true). At

250 K

b - a / RT = b (1 - T_{B} / T) = b (1 - 425/250) = - 0.70 b = - 0.160

, giving

b = 0.2286 L mo l^{- 1}

(B true).

a = 0.0821 \times 0.2286 \times 425 = 7.98 \approx 8.0 L^{2} atm mo l^{- 2}

(C true). Above

T_{B}

the coefficient

B

is positive, so

Z

exceeds 1 at low pressure (D false).