First slide

Gaseous state

Question

The pressure exerted by 1 mol of CO₂ at 273 K is 34.98 atm. Assuming that volume occupied by CO₂ molecules is negligible, the value of van der Waals' constant for attraction of CO₂ gas is ----------------dm⁶atm mol^-2

Moderate

Solution

$[P + \frac{a}{V^{2}}] [V - b] = RT$
$\begin{array}{l} ∴ [P + \frac{a}{V^{2}}] V = RT \\ or V^{2} P - RTV + a = 0 \\ V = \frac{+ RT \pm \sqrt{R^{2} T^{2} - 4 Pa}}{2 P} \end{array}$
Since, V is constant at given P and T, V can have only one value or discriminant = 0.
$\begin{array}{l} ∴ R^{2} T^{2} = 4 Pa or a = \frac{R^{2} T^{2}}{4 P} = \frac{(0 .821)^{2} \times (273)^{2}}{4 \times 34 .98} \\ = 3 .59 {dm}^{6} atm {mol}^{- 2} \end{array}$

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