One mole of oxygen (a = 0.138 Pa m^6 mol^-2, b = 3.19 x 10^-5 m^3 mol^-1) occupies 2.00 x 10^-3 m^3 at 300 K. Using the van der Waals equation and R = 8.314 J K^-1 mol^-1, the pressure in kPa (one decimal) is

JEE Main levelNumerical answerDifficulty 2 of 3rg125

One mole of oxygen ( $a = 0.138 Pa m^{6} mo l^{- 2}$ , $b = 3.19 \times 1 0^{- 5} m^{3} mo l^{- 1}$ ) occupies $2.00 \times 1 0^{- 3} m^{3}$ at $300 K$ . Using the van der Waals equation and $R = 8.314 J K^{- 1} mo l^{- 1}$ , the pressure in kPa (one decimal) is

Numerical answer type, answer in kPa, tolerance 0.5.

P = \frac{RT}{V _{m} - b} - \frac{a}{V _{m}^{2}} = \frac{8.314 \times 300}{2.00 \times 1 0 ^{- 3} - 3.19 \times 1 0 ^{- 5}} - \frac{0.138}{( 2.00 \times 1 0 ^{- 3} ) ^{2}} = \frac{2494.2}{1.9681 \times 1 0 ^{- 3}} - 34500 = 1267314 - 34500 = 1232814 Pa

, i.e.

1232.8 kPa