For a van der Waals gas, the term (a)/(V_m^2) is sometimes called the internal pressure. For CO2 (a = 3.59 L^2 atm mol^-2) at a molar volume of 0.500 L mol^-1 the internal pressure is

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For a van der Waals gas, the term $\frac{a}{V _{m}^{2}}$ is sometimes called the internal pressure. For $CO X_{2}$ ( $a = 3.59 L^{2} atm mo l^{- 2}$ ) at a molar volume of $0.500 L mo l^{- 1}$ the internal pressure is

\frac{a}{V _{m}^{2}} = \frac{3.59}{( 0.500 ) ^{2}} = \frac{3.59}{0.250} = 14.36 atm

. A common error is to divide by

V_{m}

instead of

V_{m}^{2}

, giving

7.18 atm