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 Vm as the variable. The ratio (in mol dm^–3) of the coefficient of V_m² to the coefficient of V_m for a gas having van der Waals constants a = 6.0 dm⁶ atm mol⁻² and b = 0.060 dm³ mol⁻¹ at 300 K and 300 atm is ______.

Use: Universal gas constant (R) = 0.082 dm³ atm mol⁻¹ K⁻¹

The correct answer is -7.10.

To find the ratio of the coefficient of $V_m^2$ to the coefficient of $V_m$, we first express the van der Waals equation of state for 1 mole of a gas in its cubic form:

$(P+\frac{a}{V_m^2})\left(V_{m}b\right)=RT$

Multiplying both sides by $V_m^2$ yields:

$(P{V}_m^2+a)\left(V_{m}b\right)=RT{V}_m^2$

Expanding the left-hand side of the equation:

$P{V}_m^3-Pb{V}_m^2+a{V}_{m}ab=RT{V}_m^2$

Rearranging the terms to form a standard cubic equation in terms of $V_m$:

$P{V}_m^3-(Pb+RT)V_m^2+a{V}_{m}ab=0$

From this equation, we can identify the coefficients:

The ratio of the coefficient of $V_m^2$ to the coefficient of $V_m$ is given by:

$\text{Ratio}=\frac{-(Pb+RT)}{a}$

Now, we substitute the given values into the equation:

First, calculate $Pb$:

$Pb=300\times 0.060=18{\text{dm}}^3{\text{atm mol}}^{-1}$

Next, calculate $RT$:

$RT=0.082\times 300=24.6{\text{dm}}^3{\text{atm mol}}^{-1}$

Calculate the numerator:

$-(Pb+RT)=-(18+24.6)=-42.6{\text{dm}}^3{\text{atm mol}}^{-1}$

Finally, divide by the coefficient of $V_m$ ($a$):

$\text{Ratio}=\frac{-42.6}{6.0}=-7.10{\text{mol dm}}^{-3}$