For helium a = 3.46 x 10^-3 Pa m^6 mol^-2 and b = 2.38 x 10^-5 m^3 mol^-1. Its Boyle temperature in K (one decimal, R = 8.314 J K^-1 mol^-1) is

JEE Main levelNumerical answerDifficulty 2 of 3rg193

For helium $a = 3.46 \times 1 0^{- 3} Pa m^{6} mo l^{- 2}$ and $b = 2.38 \times 1 0^{- 5} m^{3} mo l^{- 1}$ . Its Boyle temperature in K (one decimal, $R = 8.314 J K^{- 1} mo l^{- 1}$ ) is

Numerical answer type, answer in K, tolerance 0.3.

T_{B} = \frac{a}{R b} = \frac{3.46 \times 1 0 ^{- 3}}{8.314 \times 2.38 \times 1 0 ^{- 5}} = \frac{3.46 \times 1 0 ^{- 3}}{1.9787 e - 04} = 17.49 K

, i.e.

17.5 K

. Helium is above

T_{B}

at every ordinary temperature.