Question

Molar volume ($ V_m $) of a van der Waals gas can be calculated by expressing the van der Waals equation as a cubic equation with $ V_m $ as the variable. The ratio (in mol dm$^{-3}$) of the coefficient of $ V_m^2 $ to the coefficient of $ V_m $ for a gas having van der Waals constants $ a = 6.0 \, \text{dm}^6 \, \text{atm mol}^{-2} $ and $ b = 0.060 \, \text{dm}^3 \, \text{mol}^{-1} $ at 300 K and 300 atm is ____. Use: Universal gas constant $ R = 0.082 \, \text{dm}^3 \, \text{atm mol}^{-1} \, \text{K}^{-1} $

Hint:

The van der Waals equation becomes a cubic equation in molar volume. Carefully group and identify coefficients to extract useful ratios.

Answer 1

Correct Answer: 7.1

Step 1: Write van der Waals equation in terms of molar volume
The van der Waals equation is: \[ \left( P + \frac{a}{V_m^2} \right)(V_m - b) = RT \] Step 2: Expand and simplify the equation
\[\begin{align*} \left( P + \frac{a}{V_m^2} \right)(V_m - b) &= RT \\ P V_m - P b + \frac{a}{V_m} - \frac{ab}{V_m^2} &= RT \\ \end{align*}\] Multiply throughout by \( V_m^2 \) to eliminate denominators: \[ P V_m^3 - P b V_m^2 + a V_m - a b = R T V_m^2 \] Bring all terms to one side: \[ P V_m^3 - [P b + RT] V_m^2 + a V_m - a b = 0 \] Step 3: Identify coefficients
This is a cubic equation in the form: \[ A V_m^3 + B V_m^2 + C V_m + D = 0 \] We want the ratio: \[ \text{Ratio} = \frac{\text{Coefficient of } V_m^2}{\text{Coefficient of } V_m} = \frac{-[P b + RT]}{a} \] Step 4: Substitute values

• \( P = 300 \, \text{atm} \)
• \( b = 0.060 \, \text{dm}^3 \, \text{mol}^{-1} \)
• \( R = 0.082 \, \text{dm}^3 \, \text{atm mol}^{-1} \, \text{K}^{-1} \)
• \( T = 300 \, \text{K} \)
• \( a = 6.0 \, \text{dm}^6 \, \text{atm mol}^{-2} \)

\[ P b = 300 \times 0.060 = 18, \quad RT = 0.082 \times 300 = 24.6 \Rightarrow P b + RT = 18 + 24.6 = 42.6 \] \[ \text{Ratio} = \frac{-42.6}{6} = -7.1 \] Final Answer: \( \boxed{7.1} \)