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JEE Main levelSingle correct optionDifficulty 2 of 3rg018

The van der Waals constant $a$ for a gas is $1.39 L^{2} atm mo l^{- 2}$ . Its value in SI units ( $Pa m^{6} mo l^{- 2}$ ) is approximately ( $1 atm = 1.013 \times 1 0^{5} Pa$ , $1 L = 1 0^{- 3} m^{3}$ )

A $1.41 \times 1 0^{- 1}$
B $1.41 \times 1 0^{- 4}$
C $1.41 \times 1 0^{2}$
D $1.41 \times 1 0^{- 7}$

Answer: A

1 L^{2} atm = (1 0^{- 3} m^{3})^{2} \times 1.013 \times 1 0^{5} Pa = 1.013 \times 1 0^{- 1} Pa m^{6}

. So

a = 1.39 \times 0.1013 = 0.141 Pa m^{6} mo l^{- 2} \approx 1.41 \times 1 0^{- 1}

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