Animated Solution for Chemistry - States of Matter: $a$ and $b$ are van der Waals' constants for gases. Chlorine is more easily liquefied than ethane because

For a long time, the ideal gas law, $PV=nRT$, was the holy grail of thermodynamics. It painted a beautiful, mathematically pristine picture of how gases behave. But there was a catch—it assumed that gas molecules were infinitely small point masses that never interacted with each other. In reality, if gases behaved ideally, they would never condense into liquids, no matter how much you cooled or compressed them.

But we know that gases do liquefy. Water vapor becomes rain, and the propane in your kitchen cylinder is a sloshing liquid. This physical reality meant the ideal gas law was fundamentally flawed at high pressures and low temperatures.

To fix this, van der Waals introduced two brilliant correction factors to the ideal gas equation, giving birth to the real gas equation:

These two constants, $a$ and $b$, are the stars of our current problem. Let's break down what they actually mean in the physical world.

The constant $a$ represents the intermolecular force of attraction. Real gas molecules are not antisocial; they experience London dispersion forces, dipole-dipole interactions, and sometimes hydrogen bonding. When molecules attract each other, they pull inwards, reducing the pressure they exert on the walls of the container.

If you want to liquefy a gas, you want the molecules to stick together. Therefore, a higher value of $a$ means stronger attractive forces, making the gas much easier to liquefy.

The constant $b$ represents the excluded volume, or the effective size of the gas molecules. Molecules are not point masses; they take up physical space. You cannot compress a gas down to zero volume because the electron clouds of the molecules will eventually repel each other.

For a gas to transition into a dense liquid state, the molecules need to pack closely together. If the molecules are incredibly bulky (a large $b$ value), they get in each other's way, making it harder to compress them into a liquid. Therefore, a smaller value of $b$ aids in liquefaction.

The problem states a simple empirical fact: Chlorine (${\text{Cl}}_2$) is more easily liquefied than Ethane (${\text{C}}_2{\text{H}}_6$).

Armed with our knowledge of $a$ and $b$, we can immediately decode the physical properties of these two gases:

1. Because Chlorine liquefies more easily, its molecules must have a stronger mutual attraction. This directly implies that the attraction constant for Chlorine is greater than that for Ethane: $a({\text{Cl}}_2)>a({\text{C}}_2{\text{H}}_6)$

2. Furthermore, for Chlorine to pack into a liquid state more readily, its effective molecular volume hindrance must be lower than that of Ethane: $b({\text{Cl}}_2)<b({\text{C}}_2{\text{H}}_6)$

When we look at the options provided, we are searching for the mathematical translation of our logical deduction. Option (d) states exactly this: $a$ for ${\text{Cl}}_2>a$ for ${\text{C}}_2{\text{H}}_6$ but $b$ for ${\text{Cl}}_2<b$ for ${\text{C}}_2{\text{H}}_6$.

Understanding the physical narrative behind the math turns a seemingly abstract equation into a vivid story of molecular tug-of-war!