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 $$a = 6.0$$ dm$$^6$$ atm mol$$^{-2}$$ and $$b = 0.060$$ dm$$^3$$ mol$$^{-1}$$ at 300 K and 300 atm is ________.

Use: Universal gas constant (R) = 0.082 dm$$^3$$ atm mol$$^{-1}$$ K$$^{-1}$$

The van der Waals equation for one mole is$$(P+\frac{a}{V_m^{\,2}})(V_m-b)=RT\;.$$

Expand and rearrange to obtain a cubic polynomial in $$V_m$$:

$$P V_m - P b + \frac{a}{V_m}-\frac{a b}{V_m^{\,2}}=RT$$

Multiply every term by $$V_m^{\,2}$$ to clear the denominators:

$$P V_m^{\,3}-P b V_m^{\,2}+a V_m-a b = R T V_m^{\,2}$$

Collect like terms on one side:

$$P V_m^{\,3}-\left(P b + R T\right) V_m^{\,2}+a V_m-a b = 0\;.$$

Thus, in the cubic $$A V_m^{\,3}+B V_m^{\,2}+C V_m + D=0$$ we have
  • $$A = P$$
  • $$B = -\left(P b + R T\right)$$
  • $$C = a$$
  • $$D = -a b$$

The required ratio is the coefficient of $$V_m^{\,2}$$ divided by the coefficient of $$V_m$$:

$$\frac{B}{C}= \frac{-\left(P b + R T\right)}{a}$$

Insert the given data (units already consistent in dm and atm):
  • $$P = 300\ \text{atm}$$
  • $$b = 0.060\ \text{dm}^3\ \text{mol}^{-1}$$ ⇒ $$P b = 300 \times 0.060 = 18.0$$
  • $$R = 0.082\ \text{dm}^3\ \text{atm}\ \text{mol}^{-1}\ \text{K}^{-1}$$, $$T = 300\ \text{K}$$ ⇒ $$R T = 0.082 \times 300 = 24.6$$
  • $$a = 6.0\ \text{dm}^6\ \text{atm}\ \text{mol}^{-2}$$

Calculate the numerator:
$$P b + R T = 18.0 + 24.6 = 42.6$$

Finally, the ratio is
$$\frac{B}{C}= -\frac{42.6}{6.0}= -7.1\ \text{mol dm}^{-3}$$

Therefore, the required ratio lies in the range −7.2 to −7.