Van der Waals equation takes into account **attractive forces **in *a* (polarity coefficient) and size in *b* (size coefficient):

$\overline{){\mathbf{P}}{\mathbf{=}}\frac{\mathbf{nRT}}{(\mathbf{V}\mathbf{-}\mathbf{nb})}{\mathbf{-}}{\mathbf{a}}\frac{{\mathbf{n}}^{\mathbf{2}}}{{\mathbf{V}}^{\mathbf{2}}}}$

The values of a and b are relatively low for H_{2}:

- a = 0.244 L
^{2}atm/mol^{2}→**lower**value → higher pressure - b = 0.0266 L/mol (small)

While the ideal gas equation does not:

$\mathbf{P}\mathbf{=}\frac{\mathbf{nRT}}{\mathbf{V}}$

Assuming that the van der Waals equation predictions are accurate, account for why the pressure of H_{2} is higher than that predicted for an ideal gas.

A) The pressure is higher than the ideal gas because there are strong intermolecular forces and the atoms are large.

B) The pressure is higher than the ideal gas because there are weak intermolecular forces and the atoms are small.

C) The pressure is higher than the ideal gas because there are strong intermolecular forces and the atoms are small.

D) The pressure is higher than the ideal gas because there are weak intermolecular forces and the atoms are large.

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