Animated Solution for Chemistry - States of Matter: 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 ${\text{mol dm}}^{-3}$) 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 $300\text{ K}$ and $300\text{ atm}$ is _______. Use: Universal gas constant (R) = $0.082{\text{ dm}}^3{\text{ atm mol}}^{-1}{\text{ K}}^{-1}$

The behavior of gases is one of the most fascinating topics in physical chemistry. While the ideal gas law, $PV=nRT$, provides a beautifully simple model, it assumes that gas molecules have zero volume and exert no intermolecular forces. In reality, especially at high pressures and low temperatures, these assumptions break down completely. This is where the genius of Johannes Diderik van der Waals comes into play. He introduced two crucial corrections to the ideal gas law, giving birth to the famous van der Waals equation.

In this problem, we are not just plugging numbers into a formula; we are exploring the mathematical structure of the van der Waals equation. By transforming it into a cubic polynomial, we unlock the secrets of phase transitions and critical phenomena. Let's embark on this algebraic journey and see how the physical properties of a gas are encoded in the coefficients of a cubic equation.

For one mole of a real gas, the van der Waals equation is written as:

Here, $V_m$ is the molar volume. The term $\frac{a}{V_m^2}$ accounts for the attractive forces between the gas molecules, effectively increasing the pressure they exert on the container walls. The constant $a$ is a measure of the strength of these intermolecular forces. On the other hand, the term $b$ represents the excluded volume—the actual physical space occupied by the gas molecules themselves. By subtracting $b$ from the total volume, we get the true volume available for the molecules to move around.

To analyze the roots of this equation, which correspond to the possible molar volumes at a given pressure and temperature, we need to express it as a polynomial. Let's start by expanding the brackets:

This equation looks a bit messy with $V_m$ in the denominators. To clean it up and reveal its true polynomial nature, we multiply every single term by $V_m^2$:

Now we are getting somewhere! We have eliminated the fractions and are left with a cubic equation.

The next step is to organize this equation into the standard form of a cubic polynomial, which is $A{x}^3+B{x}^2+Cx+D=0$. We do this by bringing all the terms to one side and grouping them by the descending powers of $V_m$:

Factoring out the $V_m^2$ from the second and third terms, we get our master equation:

This cubic equation is incredibly powerful. For any given pressure $P$ and temperature $T$, solving this equation yields three roots for the molar volume $V_m$. Below the critical temperature, these three roots correspond to the volume of the liquid phase, the volume of the gas phase, and a physically meaningless intermediate volume. At the exact critical point, all three roots merge into a single value, the critical volume $V_c$.

The problem asks us to find the ratio of the coefficient of $V_m^2$ to the coefficient of $V_m$. Let's look closely at our master cubic equation and identify these coefficients.

The coefficient of $V_m^2$ is:

The coefficient of $V_m$ is:

Therefore, the ratio we need to calculate is:

Notice how this ratio depends on the pressure, temperature, and the specific nature of the gas (dictated by constants $a$ and $b$).

Now comes the execution phase. We are given the following values: - Pressure, $P=300\text{ atm}$ - Temperature, $T=300\text{ K}$ - van der Waals constant, $a=6.0{\text{ dm}}^6{\text{ atm mol}}^{-2}$ - van der Waals constant, $b=0.060{\text{ dm}}^3{\text{ mol}}^{-1}$ - Universal gas constant, $R=0.082{\text{ dm}}^3{\text{ atm mol}}^{-1}{\text{ K}}^{-1}$

Let's calculate the individual terms in our ratio expression to avoid any silly mistakes. First, let's find the value of $Pb$:

Next, let's calculate the $RT$ term:

Now, we add these two values together to find the magnitude of the numerator:

Finally, we plug this back into our ratio formula, remembering the negative sign and dividing by the constant $a$:

And there we have it! The ratio of the coefficients is $-7.10$. This problem is a beautiful example of how abstract algebraic manipulation can be directly applied to physical chemistry, allowing us to extract meaningful numerical relationships from fundamental equations.