Nitrogen has a = 0.1358 Pa m^6 mol^-2 and b = 3.85 x 10^-5 m^3 mol^-1. Using R = 8.314 J K^-1 mol^-1, the critical temperature in K (one decimal) is

JEE Advanced levelNumerical answerDifficulty 3 of 3rg206

Nitrogen has $a = 0.1358 Pa m^{6} mo l^{- 2}$ and $b = 3.85 \times 1 0^{- 5} m^{3} mo l^{- 1}$ . Using $R = 8.314 J K^{- 1} mo l^{- 1}$ , the critical temperature in K (one decimal) is

Numerical answer type, answer in K, tolerance 0.3.

T_{c} = \frac{8 a}{27 R b} = \frac{8 \times 0.1358}{27 \times 8.314 \times 3.85 \times 1 0 ^{- 5}} = \frac{1.0864}{8.64240 e - 03} = 125.71 K

, i.e.

125.7 K

. The experimental value is

126.2 K