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JEE Main levelSingle correct optionDifficulty 2 of 3rg027

For a van der Waals gas the ratio of Boyle temperature to critical temperature, $T_{B} / T_{c}$ , equals

A $8/27$
B $27/8$
C $3/8$
D $2$

Answer: B

T_{B} = a / R b

and

T_{c} = 8 a /27 R b

, so

T_{B} / T_{c} = \frac{a / R b}{8 a /27 R b} = \frac{27}{8} = 3.375

. The Boyle temperature is always well above the critical temperature.

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