For CO2, a = 0.364 Pa m^6 mol^-2 and b = 4.27 x 10^-5 m^3 mol^-1. Its Boyle temperature in K (nearest integer, R = 8.314 J K^-1 mol^-1) is

JEE Main levelNumerical answerDifficulty 2 of 3rg136

For $CO X_{2}$ , $a = 0.364 Pa m^{6} mo l^{- 2}$ and $b = 4.27 \times 1 0^{- 5} m^{3} mo l^{- 1}$ . Its Boyle temperature in K (nearest integer, $R = 8.314 J K^{- 1} mo l^{- 1}$ ) is

Numerical answer type, answer in K, tolerance 3.

T_{B} = \frac{a}{R b} = \frac{0.364}{8.314 \times 4.27 \times 1 0 ^{- 5}} = \frac{0.364}{3.5501 e - 04} = 1025.3 K

, i.e.

1025 K

. The experimental value is about

715 K